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adelina 88 [10]

3 years ago

9

HELP FAST!!! Here are two rectangles. The length and width of one rectangle are 8 and 5. The width of the other rectangle is 5,

but its length is unknown so we labeled it x. Write an expression for the sum of the areas of the two rectangles.

1 answer:

deff fn [24]3 years ago

6 0

Answer:

5•x=? and 8•5=40, personal opinion: the left rectangle is most likely a square so if you need to solve it it's 5•5=10

Step-by-step explanation:

A(area)=l (length)•w (width)

just plug in your numbers, in this case it would be 8(length) times 5(width)

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2 years ago

Pam has 90 m of fencing to enclose an area in a petting zoo with two dividers to separate three types of young animals. The thre

andrew-mc [135]

Answer:

The area function is

$A=\frac{135}{2}x-\frac{9}{2}x^2$ .

The domain and range of A is $(0,15m)$ and $(0, 253.125 m^2]$ .

Step-by-step explanation:

The given length of fencing is $90 m$ .

Let the length and width of each pen be $x$ and $y$ respectively as shown in the figure.

As there are 3 pens, so, the total area,

$A= 3 xy \;\cdots (i)$

From the figure the total length of fencing is $6x+4y$ .

Here, for a significant area for the animals, $x>0$ as well as $y>0$ as $x$ and $y$ are the sides of ben.

From the given value:

$6x+4y=90\;\cdots (ii)$

$\Rightarrow y=\frac {45}{2}-\frac{3x}{2}$

Now, from equation (i)

$A=3x\left(\frac {45}{2}-\frac{3x}{2}\right)$

$\Rightarrow A=\frac{135}{2}x-\frac{9}{2}x^2\;\cdots (iii)$

This is the required area function in the terms of variable $x$ .

For the domain of area function, from equation (ii)

$x=15-\frac{2y}{3}$

$\Rightarrow x$ [as y>0]

So, the domain of area function is $(0,15m)$ .

For the range of area function:

As $x \rightarrow 0$ or $y\rightarrow 0$ , then $A\rightarrow 0$ [from equation (i)]

$\Rightarrow A>0$

Now, differentiate the area function with respect to $x$ .

$\frac {dA}{dx}=\frac{135}{2}-9x$

Equate $\frac {dA}{dx}$ to zero to get the extremum point.

$\frac {dA}{dx}=0$

$\Rightarrow \frac{135}{2}-9x=0$

$\Rightarrow x=\frac{15}{2}$

Check this point by double differentiation

$\frac {d^2A}{dx^2}=-9$

As, $\frac {d^2A}{dx^2}$ , so, point $x=\frac{15}{2}$ is corresponding to maxima.

Put this value back to equation (iii) to get the maximum value of area function. We have

$A=\frac{135}{2}\times \frac {15}{2}-\frac{9}{2}\times \left(\frac {15}{2}\right)^2$

$\Rightarrow A=253.125 m^2$

Hence, the range of area function is $(0, 253.125 m^2]$ .

4 0

3 years ago

If a₁ = 4 and an = 5an-1 then find the value of a5.

igor_vitrenko [27]

The value of $a_{5}$ is 2500, when $a_{1}=4$ and $5a_{n-1}$ .

Given that, $a_{1}=4$ and $5a_{n-1}$ .

We need to find the value of $a_{5}$ .

<h3>What is an arithmetic sequence?</h3>

An arithmetic progression or arithmetic sequence is a sequence of numbers such that the difference between the consecutive terms is constant.

Now, to find the value of $a_{5}$ :

$a_{2} =5a_{2-1}=5a_{1}=5 \times4=20$

$a_{3} =5a_{3-1}=5a_{2}=5 \times20=100$

$a_{4} =5a_{4-1}=5a_{3}=5 \times100=500$

$a_{5} =5a_{5-1}=5a_{4}=5 \times500=2500$

Therefore, the value of $a_{5}$ is 2500.

To learn more about arithmetic sequence visit:

brainly.com/question/15412619.

#SPJ1

5 0

2 years ago