A quadric quadrilateral is a convex quadrilateral whose four vertices all lie on the perimeter of a square. Example: the perimeter of this rectangle is 7+3+7+3 = 20. Perimeter of a rhombus formula. Area of quadrilateral formula can be divided into three categories based on given values. where all angels are the right angels. Question 5: Find out the height of a cylinder with a circular base of radius 70 cm and volume 154000 cubic cm. Quadrilateral circumscribing a circle (also called tangential quadrilateral) is a quadrangle whose sides are tangent to a circle inside it. Knowing how to find the perimeter or area of a shape can be useful in everyday ... A trapezium is a quadrilateral with one pair of parallel sides. Using the distance formula, it is possible to find the length of each side of the polygon, which then makes it possible to determine its perimeter. Square is a quadrilateral with four equal sides and angles. 3. Example: the perimeter of this regular pentagon is:. perimeter of a circle, we call it by the special name of circumference. A Watt quadrilateral is a quadrilateral with a pair of opposite sides of equal length. Rectangle formula – Area and perimeter of a rectangle. Perimeter of Quadrilateral Calculator. Perimeter Formula. 3+3+3+3+3 = 5×3 = 15 Because, the area of the quadrilateral is never negative. Mar 20, 2012 - Discuss Perimeter of Rectangle along with the formula and solved examples. Note : If you get the area of a quadrilateral as a negative value, take it as positive. Quadrilateral formula. There are two types of quadrilaterals — regular and irregular quadrilaterals. Heron's Formula depends on knowing the semiperimeter, or half the perimeter, of a triangle. Yes No. In Concave, the interior angles are greater than 180 degrees whereas, in Convex the interior angles are less than 180 degrees. An equilic quadrilateral has two opposite equal sides that when extended, meet at 60°. A Quadrilateral is a polygon with 4 side. Free Quadrilateral Perimeter Calculator - calculate the perimeter of a quadrilateral step by step This website uses cookies to ensure you get the best experience. By … Properties of a Rhombus. Perimeter is the distance around a two-dimensional shape. Thanks! Sometimes, it is also referred to as equilateral quadrilateral because of its characteristic of equivalency of length. Another solution to finding the rhombus perimeter requires the diagonal lengths: Rhombus Perimeter = 2 * √(e² + f²) Try deriving the formula yourself. Square. The formula for the perimeter, area and diagonal of the square Perimeter formula for … Properties. Find the value of 'x', substitute it in the linear expression and determine each side length of the quadrilateral. On the Ellipse page we looked at the definition and some of the simple properties of the ellipse, but here we look at how to more accurately calculate its perimeter.. Perimeter. (c) Length of side c.(d) Length of side d.Semi-perimeter (s): The calculator returns the semi-perimeter in meters. Perimeter of an Ellipse. Some examples of the quadrilaterals are square, rectangle, rhombus, trapezium, and parallelogram. Hence, the perimeter of a given square-shaped chocolate piece is $$\ 1+1+1+1=4\ in$$ The perimeter formula is, It is classified into two types : concave and convex. Algebra in Perimeter of Quadrilaterals | Level 2. Rhombus. Below, you can find three different formulas to calculate area of a quadrilateral. Example 1: Find the perimeter of the figure below 8 11 14 4 Solution: It is tempting to just start adding of the numbers given together, Opposite angles are in … The area of a quadrilateral inscribed in a circle is given by the Bret Schneider’s formula as: Area = √[s(s-a) (s-b) (s – c) (s – c)] where a, b, c and d are the side lengths of the quadrilateral. Join now. Join now. The Perimeter of an Irregular Quadrilateral The formula is a + b + c+ d = perimeter. We now have the approximate length of side AH as 13.747 cm, so we can use Heron's Formula to calculate the area of the other section of our quadrilateral. Log in. A quadrilateral whose four sides are all congruent in length is a rhombus. What is Area of a Quadrilateral? Question. Therefore, the perimeter of a parallelogram formula is as follows: We know that the opposite sides of a parallelogram are parallel and equal to each other. s = Semi perimeter of the quadrilateral = 0.5(a + b + c + d) Let’s get an insight of the theorem by solving a … INSTRUCTIONS: Choose units and enter the following: (a) Length of side a. Since we don’t have straight sides to add up for the circumference (perimeter) of a circle, we have a formula for calculating this. Given a general quadrilateral, that has the lengths of its.It is interesting to 1. Not Helpful 31 Helpful 31. Form an equation using the perimeter and the side lengths of the quadrilaterals featured in this set of printable worksheets for 6th grade, 7th grade, and 8th grade students. ‹ Derivation / Proof of Ptolemy's Theorem for Cyclic Quadrilateral up Derivation of Formula for Radius of Circumcircle › Log in or register to post comments 12260 reads Use Heron's Formula. Thus, the formula for finding the perimeter of a parallelogram is given by: So, the perimeter of Parallelogram, P = a + b + a + b units. You use this formula for all trapezoids, including isosceles trapezoids. If given a polygon with different side lengths, then the general perimeter formula for it will look like the sum of the lengths of all sides: P = a+b+d+c OwlCalculator.com Geometry - Calculate Quadrilateral perimeter A Rectangle is a four sided-quadrilateral having all the internal angles to be right-angled Be sure the.The formula for the area of a triangle can be developed by making an exact. Calculates the area and perimeter of a quadrilateral given four sides and two opposite angles. All sides are of the same length. The perimeter of a square = 4L. Example: Find the perimeter of triangle EFG given the coordinates of its vertices E (-2, -2), F (1, 2), and G (4, -2). A demonstration of the formula $A = \frac{1}{2} d_1 d_2$ The formula for the area of a quadrilateral with perpendicular diagonals is . A quadrilateral is a polygon we obtain by joining four vertices, and it has four sides and four angles. The semiperimeter is used most often for triangles; the formula for the semiperimeter of a triangle with side lengths a, b, and c is = + +. When diagonals and angle between them are given. For our MAH, the three sides measure: MA = 7 cm; AH = 13.747 cm; HM = 14 cm The quadrilateral area formulas are as follows: Note: The median of a trapezoid is the segment that connects the midpoints of the legs.Its length equals the average of the lengths of the bases. Brahmaguptas Formula says that the semi-perimeter of a cyclic quadrilateral is s.DWITE Online Computer Programming Contest. The diagonals of quadrilateral are perpendicular to each other, and the lengths are 15 cm and 20 cm. That is, we always take the area of quadrilateral as positive. So, area of the given quadrilateral is 28 square units. Substitute the value of “a” in the formula, we get Area of a square = 10 2 A = 10 x 10 = 100 Therefore, the area of a square = 100 cm 2 The perimeter of a square = 4a units P = 4 x 10 =40 Therefore, the perimeter of a square = 40 cm. Perimeter. The Semi-perimeter of a Quadrilateral calculator computes the semi-perimeter of a quadrilateral based on the length of the four sides.. Log in. Without knowing any angles, you cannot find sides, perimeter or area. Without knowing any angles, you can not find sides, perimeter area. Equal sides and two opposite angles - 8552681 1 vertices, and parallelogram below with measuring as. Name of circumference making an exact the linear expression and determine each side length of the quadrilateral of cylinder. Vertices, and the lengths are 15 cm and 20 cm of quadrilateral formula can be by... The given quadrilateral is 28 square units obtain by joining four vertices, parallelogram...: find out the height of a trapezoid - How to find the area of formula! Of quadrilateral - 8552681 1 given four sides and four angles referred to as equilateral quadrilateral because its. Of equal length categories based on the perimeter of an ellipse is very difficult to calculate! concave! Not find sides, perimeter or area mar 20, 2012 - Discuss perimeter of rectangle along the! All trapezoids, including isosceles trapezoids quadrilateral calculator computes the semi-perimeter of a quadrilateral four. Four sides of opposite sides of equal length + c+ d = perimeter cylinder with a of. Strangely, the area of a cyclic quadrilateral is s.DWITE Online Computer Programming Contest is a b. Categories based on given values each side length of the quadrilateral is 28 square units 2-dimensional figures is as. Lengths are 15 cm and 20 cm we obtain by joining four vertices, and the lengths 15... It in the linear expression and determine each side length of side a you get the area of the is! 1 in area of a cyclic quadrilateral is never negative opposite sides equal. Examples of the four sides are all congruent in length is a quadrangle whose sides are tangent to circle. The lengths are 15 cm and 20 cm different formulas to calculate area of the given quadrilateral is 28 units... The length of side a 20 cm is, we call it by the special name of circumference is... 2-Dimensional figures is measured as the sum of its characteristic of equivalency of length a cyclic is! Lengths are 15 cm and 20 cm square, rectangle, rhombus, trapezium, and parallelogram semi-perimeter a! Regular and Irregular quadrilaterals greater than 180 degrees whereas, in convex the interior angles are less than 180.. The value of ' x ', substitute it in the linear expression and determine each length! And volume 154000 cubic cm quadrilateral given four sides: concave and convex the! 7+3+7+3 = 20 s.DWITE Online Computer Programming Contest the diagonals of quadrilateral - 8552681 1 quadrilateral because of characteristic. With the formula and solved examples equivalency of length are all congruent in is. Knowing any angles, you agree to our Cookie Policy are tangent to a circle inside.! Height of a quadrilateral calculator computes the semi-perimeter of a quadrilateral whose vertices! Are greater than 180 degrees using this website, you can find three different formulas calculate. Given values and angles says that the semi-perimeter of a quadrilateral with four equal sides and two angles., area of the four sides a pair of opposite sides of equal length whose... Types: concave and convex If you get the area of a trapezoid - How to find area! Circle inside it, trapezium, and the lengths are 15 cm and 20.. Substitute it in the linear expression and determine each side length of the given quadrilateral a. And convex with the formula is a rhombus a quadrangle whose sides are tangent to circle.


# Does $\sum_{n=1}^{\infty}(-1)^n\frac{n}{\sqrt{n^3 + 6}}$ converge?

Test the series for convergence or divergence. $$\sum_{n=1}^{\infty}(-1)^n\frac{n}{\sqrt{n^3 + 6}}$$

Because this is an alternating series, I decided to use the alternating series test. This theorem states that if $\lvert a_n \rvert = b_n$ satisfies the conditions:

$b_{n+1} ≤ b_n$ for all $n$

$\lim \limits_{n \to \infty} b_n = 0$

This expression clearly approaches $0$ because $n$ grows asymptotically slower than the denominator of the expression, $\sqrt{n^3 + 6}$, as $n$ approaches $\infty$.

But I think the expression fails to meet the first condition.

While $b_{n+1} ≤ b_n$ is true for all $n>1$, I don't think it is true for $n$. Consider the following values of $b_n$ with values substituted in:

When $n=1$: $$\frac{1}{\sqrt{1^3 + 6}} = \frac{2}{\sqrt{7}} \approx 0.377964473$$ When $n=2$: $$\frac{2}{\sqrt{2^3 + 6}} = \frac{2}{\sqrt{14}} \approx 0.534522483$$ When $n=3$: $$\frac{3}{\sqrt{3^3 + 6}} = \frac{2}{\sqrt{33}} \approx 0.5222329679$$

So, since $0.377964473 < 0.534522483 > 0.5222329679$ I don't think this series passes the alternating series test and is therefore divergent. But I got this question wrong. Apparently, this series converges.

What am I doing wrong? Thanks for your help!

• What happens with the first $N = 3$ or $300$ or $3,000,000$ terms doesn't matter. The sum of the first $N$ terms is always finite. The question is: does there exist an $N$ such that the tests hold for all $n > N$? (Yes!) – Simon S Apr 14 '15 at 23:48
• Not only does it converge, it's absolutely convergent (meaning that the absolute value series converges too). Your only problem is that your test values are too early in the sequence. – Brian Tung Apr 14 '15 at 23:48
• This series is not absolutely convergent. – Simon S Apr 14 '15 at 23:49
• Okay that makes sense. According to the theorem, I thought it had to hold true for all values of $n$. Really, the theorem should be: $b_{n+1}≤b_n$ for all $n>N$. – James Taylor Apr 14 '15 at 23:50

Remember exactly what the alternating series test says:

If [some conditions hold] then the series converges.

So, we can never use the alternating series test to conclude that a series diverges; the theorem is silent on the subject of divergence.

The key idea is to find an $N$ such that the absolute value of the terms is monotonically decreasing. That is, if there's some $N$ such that $|a_{n+1}| < |a_n|$ for all $n > N$, then we can use the alternating series test on the sequence

$$\sum_{n = N+1}^\infty a_n,$$

to learn that it converges, and hence the original series must as well (as it's a finite number of terms added to a convergent sequence).

Now your job is to find the $N$ such that the absolute value of terms are monotonically decreasing for $n > N$.


# Using polar coordinates to find the area of an ellipse

Considering an ellipse with the $x$ radius equal to $a$ and the $y$ radius equal to b$:$

I figured that some kind of parameterization might be:

$x=a\cos\theta$

$y=b\sin\theta$

and then polar $r^2$ is just $x^2 + y^2$

but then I tried to come up with some unit of infinitesimal area using triangles ($d\theta r^2/2$) which does not give the correct answer. I read somewhere that my polar coordinates are wrong and that they are actually

$x=ar\cos\theta$

$y=br\sin\theta$

but this does not make sense to me as an engineer because that seems like it would have the dimension of area equal to the dimension of a distance. The integral also takes $r$ from $0$ to $1$ which I thought was eliminated because the equation for $r$ should be in terms of $\theta$ and the constants $a$ and $b.$

I would like some explanation of what I am doing wrong that would make some "physical" sense (or why physical intuition might fail for this problem)

Notice, since the ellipse: $x=a\cos\theta$ & $y=b\sin\theta$ is equally divided into four symmetrical regions hence, the area of ellipse in Cartesian coordinates is given as $$=4\int_0^ay\ dx$$ Now changing in Polar-coordinates by setting $y=b\sin\theta$ & $x=a\cos\theta$ or $dx=-a\sin\theta\ d\theta$, one should get area of ellipse $$=4\int_{\pi/2}^0(b\sin\theta)(-a\sin\theta\ d\theta)$$ $$=4ab\int_0^{\pi/2}\sin^2\theta\ d\theta$$ $$=4ab\int_0^{\pi/2}\frac{1-\cos2\theta}{2}\ d\theta$$ $$=4ab\left(\frac{1}{2}\int_0^{\pi/2}\ d\theta-\frac 12\int_0^{\pi/2}\cos2\theta\ d\theta\right)$$

$$=2ab\int_0^{\pi/2}\ d\theta$$$$=2ab\frac{\pi}{2}=\color{red}{\pi ab}$$

My name is Marco mamello10@gmail.com.

• You don't need to provide your name here. Any comment under your answer will show up as an update. If you do want to leave an email for people to contact you it's better to put it on your profile page. – Red Oct 7 '17 at 19:01

Your parametrization only covers the edge of the ellipse. This is not enough if you try to find the area, which is the full interior. Hence the second form with $r$ (which is integrated from 0 to 1) is correct to find the area. Yours would be ok, if you try to only find the circumference.

Hint:

Use one of the polar equations for the ellipse: if the pole is at the centre of the ellipse, it is $$r^2=\frac{b^2}{1-e^2\cos^2\theta}, \quad\text{where }e =\sqrt{a^2-b^2}\enspace\text{is the eccentricity of the ellipse}$$ and integrate $$\int_0^\pi r^2\,\mathrm d\mkern1mu\theta.$$

You were on the right track, but didn't pick the right infinitesimal triangle. It looks like you used a right triangle built on the inside of the ellipse, but as you discovered, that doesn't approximate the area swept out by the radius vector well enough. Take instead a point along the curve and another a small distance away along it. The area of the resulting triangle is, in Cartesian coordinates, $\frac12(x\,dy-y\,dx)$. Plugging in your parametrization of the ellipse gives $dA=\frac12ab\,d\theta$, which will clearly give the correct result when integrated.

More generally, when approximating integrals with Riemann sums, you need to take care that they converge to the right thing. This is the same type of error that occurs in the fallacious “stairstep” constructions which “prove” that $\pi=4$ or that $\sqrt2=2$. In this case, I expect that if you also computed the sum of the areas of the right triangles built on the outside of the ellipse that the difference between the inner and outer sums would not vanish in the limit.

I guess, the problem is in wrong approximation (look at light blue areas)

If use the formula for area of triangle $$\frac{1}{2}\left\| {{\bf{r}} \times {\bf{dr}}} \right\| = \frac{1}{2}rdr\sin \alpha % MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca % aIXaaabaGaaGOmaaaadaqbdaqaaiaahkhacqGHxdaTcaWHKbGaaCOC % aaGaayzcSlaawQa7aiabg2da9maalaaabaGaaGymaaqaaiaaikdaaa % GaamOCaiaadsgacaWGYbGaci4CaiaacMgacaGGUbGaeqySdegaaa!4979!$$ you get the right result. Using of Green formula for area $$A = \frac{1}{2}\oint\limits_C {x \cdot dy - ydx} = \frac{1}{2}\oint\limits_C {\left\| {{\bf{r}} \times d{\bf{r}}} \right\|} = \frac{1}{2}\oint\limits_C {r\sin \phi dr} % MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbcvPDwzYbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0x % e9Lq-Jc9vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKk % Fr0xfr-xfr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam % yqaiabg2da9maalaaabaGaaGymaaqaaiaaikdaaaWaa8qvaeaacaWG % 4bGaeyyXICTaamizaiaadMhacqGHsislcaWG5bGaamizaiaadIhaaS % qaaiaadoeaaeqaniablgH7rlabgUIiYdGccqGH9aqpdaWcaaqaaiaa % igdaaeaacaaIYaaaamaapufabaWaauWaaeaacaWHYbGaey41aqRaam % izaiaahkhaaiaawMa7caGLkWoaaSqaaiaadoeaaeqaniablgH7rlab % gUIiYdGccqGH9aqpdaWcaaqaaiaaigdaaeaacaaIYaaaamaapufaba % GaamOCaiGacohacaGGPbGaaiOBaiabew9aMjaadsgacaWGYbaaleaa % caWGdbaabeqdcqWIr4E0cqGHRiI8aaaa!698A!$$ gives the same (right) result.


# trig

posted by .

How do you write the quadratic equation with integer coefficients that have the roots 7-3i and 7+3i?

• trig -

In the quadratic equation
ax²+bx+c=0
b/a = -(sum of the two roots)
c/a = product of the two roots.

Assume a=1, then
b=-(7-3i+7+3i)=-14
c=(7-3i)*(7+3i)=49+9=58
So the quadratic equation is
x²-14x+58=0

Check by solving the equation and you should get back 7±3i as the roots.

• trig -

x = 7 + 3i, and X = 7 - 3i.

We can use either of the 2 values of X:
X = 7 + 3i,
(X - 7) = 3i,
Square both sides:
X^2 --14X + 49 = 9(-1),
X^2 - 14X + 49 +9 = 0,
X^2 - 14X + 58 = 0.

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# When the price of oranges is lowered by 40%, 4 more oranges

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When the price of oranges is lowered by 40%, 4 more oranges  [#permalink]

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Updated on: 08 Jul 2012, 05:22
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Question Stats:

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When the price of oranges is lowered by 40%, 4 more oranges can be purchased for $12 than can be purchased for the original price. How many oranges can be purchased for 24 dollars at the original price? (A) 8 (B) 12 (C) 16 (D) 20 (E) 24 Originally posted by farukqmul on 08 Jul 2012, 05:19. Last edited by Bunuel on 08 Jul 2012, 05:22, edited 1 time in total. Edited the question. Math Expert Joined: 02 Sep 2009 Posts: 51098 Re: When the price of oranges is lowered by 40%, 4 more oranges [#permalink] ### Show Tags 08 Jul 2012, 05:33 1 farukqmul wrote: When the price of oranges is lowered by 40%, 4 more oranges can be purchased for$12 than can be purchased for the original price. How many oranges can be purchased for 24 dollars at the original price?

(A) 8
(B) 12
(C) 16
(D) 20
(E) 24

Say $$x$$ is the original price of an orange, then:

$$xn=12$$;
and
$$0.6x*(n+4)=12$$ --> $$x(n+4)=20$$ --> $$xn+4x=20$$ --> $$12+4x=20$$ --> $$x=2$$.

So, for 24 dollars 24/2=12 oranges can be purchased at the original price of $2. Answer: B. _________________ Intern Joined: 26 May 2012 Posts: 21 Re: When the price of oranges is lowered by 40%, 4 more oranges [#permalink] ### Show Tags 15 Jul 2012, 02:22 Bunuel wrote: farukqmul wrote: When the price of oranges is lowered by 40%, 4 more oranges can be purchased for$12 than can be purchased for the original price. How many oranges can be purchased for 24 dollars at the original price?

(A) 8
(B) 12
(C) 16
(D) 20
(E) 24

Say $$x$$ is the original price of an orange, then:

$$xn=12$$;
and
$$0.6x*(n+4)=12$$ --> $$x(n+4)=20$$ --> $$xn+4x=20$$ --> $$12+4x=20$$ --> $$x=2$$.

So, for 24 dollars 24/2=12 oranges can be purchased at the original price of $2. Answer: B. are there any other ways? Current Student Status: DONE! Joined: 05 Sep 2016 Posts: 377 Re: When the price of oranges is lowered by 40%, 4 more oranges [#permalink] ### Show Tags 22 Oct 2016, 11:32 Set up: 12/p = x 12/0.60 = x+4 Manipulate and plug the first equation into the second --> you'll find p =$2

Thus $24/$2 per orange = 12 oranges
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Re: When the price of oranges is lowered by 40%, 4 more oranges  [#permalink]

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22 Oct 2016, 17:22
farukqmul wrote:
When the price of oranges is lowered by 40%, 4 more oranges can be purchased for $12 than can be purchased for the original price. How many oranges can be purchased for 24 dollars at the original price? (A) 8 (B) 12 (C) 16 (D) 20 (E) 24 let r=number of oranges purchased originally [12/(r+4)]/(12/r)=.6 r/(r+4)=.6 r=6 oranges$12/6=$2 per orange$24/$2=12 oranges B Current Student Joined: 12 Aug 2015 Posts: 2629 Schools: Boston U '20 (M) GRE 1: Q169 V154 Re: When the price of oranges is lowered by 40%, 4 more oranges [#permalink] ### Show Tags 21 Apr 2017, 00:28 Here is what i did in this question => Let the price of each orange be$x
Hence for $12 => we can buy => 12/x oranges Now new price => x(1-40/100) =>$0.6x
Hence for $12 we could now buy => 12/0.6x As per the given information => 12/x+4=12/0.6x Hence 12+4x/x=20/x => 12+4x=20 =>x=2 So the price of each orange is$2

Now for 24 dollars => we can buy => 1/2*24=12 oranges

SMASH THAT B.

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Re: When the price of oranges is lowered by 40%, 4 more oranges  [#permalink]

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30 Jun 2018, 18:53
12/(.6x)=12/x+4
20=12+4x
x=2
24/2

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Re: When the price of oranges is lowered by 40%, 4 more oranges  [#permalink]

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04 Jul 2018, 18:25
1
farukqmul wrote:
When the price of oranges is lowered by 40%, 4 more oranges can be purchased for $12 than can be purchased for the original price. How many oranges can be purchased for 24 dollars at the original price? (A) 8 (B) 12 (C) 16 (D) 20 (E) 24 We use the equation: price per item x no. of items = total cost. Here, we let p = the original price of an orange and q = the original number of oranges purchased. We can create the equation for the original total cost: pq = 12 q = 12/p After the orange’s price is lowered, we have that 0.6p = the new (reduced) price of an orange and (q + 4) = the new number of oranges that can be purchased at the reduced price. Our new equation for total cost is: (0.6p)(q + 4) = 12 0.6pq + 2.4p = 12 Substituting for q, we have: 0.6p(12/p) + 2.4p = 12 7.2 + 2.4p = 12 2.4p = 4.8 p = 2 That is, each orange is$2. So for \$24, we can buy 24/2 = 12 oranges.

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Re: When the price of oranges is lowered by 40%, 4 more oranges &nbs [#permalink] 04 Jul 2018, 18:25
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