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3 years ago

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The area of a rectangle is 52 m2 , and the length of the rectangle is 5 m less than double the width. Find the dimensions of the

rectangle.

1 answer:

kotegsom [21]3 years ago

7 0

52m^2= 2[(x-5)+(x)]. 52m^2= 2(2x-5). 52m^2=4x-10

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the total weight of 6 pieces of butter and a bag of sugar is 3.8 Ib. if the weight of the bag of sugar is 1.4 Ib, what is the we

baherus [9]

0.4 lbs each.

<h3>Further explanation</h3>

<u>Given: </u>

The total weight of 6 pieces of butter and a bag of sugar is 3.8 Ib.
The weight of the bag of sugar is 1.4 Ib

<u>Question: </u>

What is the weight of each piece of butter?

<u>The Process: </u>

The two steps for obtaining the weight of each piece of butter are as follows:

<u>Step-1:</u> the total amount of butter is the total bag of butter and sugar subtracted by the weight of the bag of sugar.

$\boxed{ \ 3.8 \ lbs - 1.4 \ lbs = 2.4 \ lbs \ }$

<u>Step-2:</u> the total amount of butter divided by the number of pieces of butter is the the weight of each piece of butter.

$\boxed{ \ 2.4 \ lbs \div 6 \ pieces = \boxed{ \ 0.4 \ lbs/piece \ } \ }$

Let's also consider the following quick steps.

<u>Quick steps: </u>

$\boxed{ \ \frac{(3.8 - 1.4) \ lbs}{6 \ pieces} = \boxed{ \ 0.4 \ lbs/piece \ } \ }$

Therefore, the weight of each piece of butter is 0.4 lbs.

<h3>Learn more </h3>

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Keyword: the total weight, 6 pieces of butter, a bag of sugar is 3.8 Ib, if the weight of the bag of sugar is 1.4 Ib, each piece, subtracted, divide

5 0

3 years ago

Read 2 more answers

Find the mass of the lamina that occupies the region D = {(x, y) : 0 ≤ x ≤ 1, 0 ≤ y ≤ 1} with the density function ρ(x, y) = xye

Alona [7]

Answer:

The mass of the lamina is 1

Step-by-step explanation:

Let $\rho(x,y)$ be a continuous density function of a lamina in the plane region D,then the mass of the lamina is given by:

$m=\int\limits \int\limits_D \rho(x,y) \, dA$ .

From the question, the given density function is $\rho (x,y)=xye^{x+y}$ .

Again, the lamina occupies a rectangular region: D={(x, y) : 0 ≤ x ≤ 1, 0 ≤ y ≤ 1}.

The mass of the lamina can be found by evaluating the double integral:

$I=\int\limits^1_0\int\limits^1_0xye^{x+y}dydx$ .

Since D is a rectangular region, we can apply Fubini's Theorem to get:

$I=\int\limits^1_0(\int\limits^1_0xye^{x+y}dy)dx$ .

Let the inner integral be: $I_0=\int\limits^1_0xye^{x+y}dy$ , then

$I=\int\limits^1_0(I_0)dx$ .

The inner integral is evaluated using integration by parts.

Let $u=xy$ , the partial derivative of u wrt y is

$\implies du=xdy$

and

$dv=\int\limits e^{x+y} dy$ , integrating wrt y, we obtain

$v=\int\limits e^{x+y}$

Recall the integration by parts formula: $\int\limits udv=uv- \int\limits vdu$

This implies that:

$\int\limits xye^{x+y}dy=xye^{x+y}-\int\limits e^{x+y}\cdot xdy$

$\int\limits xye^{x+y}dy=xye^{x+y}-xe^{x+y}$

$I_0=\int\limits^1_0 xye^{x+y}dy$

We substitute the limits of integration and evaluate to get:

$I_0=xe^x$

This implies that:

$I=\int\limits^1_0(xe^x)dx$ .

Or

$I=\int\limits^1_0xe^xdx$ .

We again apply integration by parts formula to get:

$\int\limits xe^xdx=e^x(x-1)$ .

$I=\int\limits^1_0xe^xdx=e^1(1-1)-e^0(0-1)$ .

$I=\int\limits^1_0xe^xdx=0-1(0-1)$ .

$I=\int\limits^1_0xe^xdx=0-1(-1)=1$ .

No unit is given, therefore the mass of the lamina is 1.

3 0

3 years ago

Simple interest: I= PRT solve for r

garri49 [273]

I = PRT....for R....divide both sides by PT

I / PT = R <==

4 0

3 years ago

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Solve for x.<br> 9x = 3x - 36<br> A x = -6<br> B x = – 4<br> C X=4<br> D x = 6

Lelu [443]

Answer:

x=-6

Step-by-step explanation:

9x-3x=-36

6x=-36

x=-6

brainliest?

7 0

3 years ago

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Tyrese can eat 246 pieces of candy in 4 hours and 45 minutes. What would be the approximate number of pieces of candy that Tyres

likoan [24]

Answer:

around 51.78

Step-by-step explanation:

do 246÷4.75 which is around 51.78

7 0

3 years ago