Ask question

Ask question

All categories

2 years ago

14

rectangles perimeter and it's area have the same numerical value. the width of the rectangle is 3 units . what is the length of

the rectangle in units?

1 answer:

castortr0y [4]2 years ago

5 0

Answer:

Step-by-step explanation:

2(x+y)=parameter

xy=area

If area=parameter then 2(x+y)=xy.

If y=3, then

2(x+3)=3x

2x+6=3x

X=6

now, let’s double check.

6*3=18

2(3+6)=18

You might be interested in

Quadrilateral ABCD is dilated by a scale factor of 2 centered around (2,2).

ANEK [815]

Answer:

B’D’ will run through (2,2) and will be longer than BD

Step-by-step explanation:

B'D' will be twice as long as BD and run through (2, 2).

6 0

2 years ago

Pleas please help me with my math problem! Thank You :)

dalvyx [7]

Check the picture below.

as you recall from the previous exercise, x = 74°, now, using the "inscribed angle theorem" as you saw already, the green intercepted arc is 160°, so y = 160° - 74°.

as far as ∡z, well, we can just use the "inscribed quadrilateral conjecture".

3 0

3 years ago

I need help if y’all want y’all can help me I need help

Snezhnost [94]

The answer is D. 42

A triangle equals 180 in total so what you would do it subtract 122 from 180 because 180 is also the angle of a straight line. You would get 58 then you add 58 and 80 to get 138. Subtract 138 from 180 and you get your answer of 42

6 0

3 years ago

<img src="https://tex.z-dn.net/?f=%20%28%7B%20%7Bx%7D%5E%7B2%7D%20%20-%204%7D%29%5E%7B5%7D%20%28%20%7B4x%20-%205%7D%29%5E%7B4%7D

Makovka662 [10]

Let $u=x^2-4$ and $v=4x-5$ . By the product rule,

$\dfrac{\mathrm d(u^5v^4)}{\mathrm dx}=\dfrac{\mathrm d(u^5)}{\mathrm dx}v^4+u^5\dfrac{\mathrm d(v^4)}{\mathrm dx}$

By the power rule, we have $(u^5)'=5u^4$ and $(v^4)'=4v^3$ , but $u,v$ are functions of $x$ , so we also need to apply the chain rule:

$\dfrac{\mathrm d(u^5)}{\mathrm dx}=5u^4\dfrac{\mathrm du}{\mathrm dx}$

$\dfrac{\mathrm d(v^4)}{\mathrm dx}=4v^3\dfrac{\mathrm dv}{\mathrm dx}$

and we have

$\dfrac{\mathrm du}{\mathrm dx}=2x$

$\dfrac{\mathrm dv}{\mathrm dx}=4$

So we end up with

$\dfrac{\mathrm d(u^5v^4)}{\mathrm dx}=10xu^4v^4+16u^5v^3$

Replace $u,v$ to get everything in terms of $x$ :

$\dfrac{\mathrm d((x^2-4)^5(4x-5)^4)}{\mathrm dx}=10x(x^2-4)^4(4x-5)^4+16(x^2-4)^5(4x-5)^3$

We can simplify this by factoring:

$10x(x^2-4)^4(4x-5)^4+16(x^2-4)^5(4x-5)^3=2(x^2-4)^4(4x-5)^3\bigg(5x(4x-5)+8(x^2-4)\bigg)$

$=2(x^2-4)^4(4x-5)^3(28x^2-57)$

7 0

3 years ago

On a number line, suppose point E has a coordinate of 4, and EG = 11. What are the possible coordinates of point G?

navik [9.2K]

Answer:

-7, 15

Step-by-step explanation:

The positive difference between E and G is 11.

<h3>Solution</h3>

G - E = 11

G = 11 + E = 11 + 4

G = 15 . . . . . . . . . . . . . when G is right of E

__

E - G = 11

E - 11 = G = 4 - 11

G = -7 . . . . . . . . . . . . when G is left of E

Possible coordinates for G are -7 and 15.

4 0

2 years ago