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1 year ago

I quite don’t get this question and I have 4 minutes remaining

Mathematics

1 answer:

gregori [183]1 year ago

7 0

The length of one of the sides of the larger pentagon is 24 cm. The length of one of the sides of the smaller pentagon is 8 cm.

$24$ ÷ $8=3$

So, if the larger pentagon is 3x the size of the smaller pentagon, we can find the area of the smaller pentagon by dividing the area of the larger pentagon by 3.

$135$ ÷ $3=45$

The area of the smaller pentagon is 45in²

Hope this helps! Pwease mark me brainliest! Have a great day!

−xXheyoXx

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F(x) = x3 – 4x² - 6x – 32 at x = 6

scoundrel [369]

Answer:

X= 140

Step-by-step explanation:

f(x) = x3 – 4x² - 6x – 32 at x = 6

(6)(6)(6) -4²=72

-6(6)-32= -68

72+68=140

3 0

2 years ago

How do You graph : y-10= -5x?

miv72 [106K]

Equation:

y - 10 = -5x

Solve:

(y - 10) + 10 = (-5x) + 10

y = -5x + 10

Remember the formula (Slope Intercept Form):

y = mx + b

m = Slope (Rise/Run)

b = Y-intercept

So your graph would look something like the attached image:

(Need further elaboration? Let me know!)

5 0

2 years ago

Find the arc length of the given curve between the specified points. x = y^4/16 + 1/2y^2 from (9/16), 1) to (9/8, 2).

lutik1710 [3]

Answer:

The arc length is $\dfrac{21}{16}$

Step-by-step explanation:

Given that,

The given curve between the specified points is

$x=\dfrac{y^4}{16}+\dfrac{1}{2y^2}$

The points from $(\dfrac{9}{16},1)$ to $(\dfrac{9}{8},2)$

We need to calculate the value of $\dfrac{dx}{dy}$

Using given equation

$x=\dfrac{y^4}{16}+\dfrac{1}{2y^2}$

On differentiating w.r.to y

$\dfrac{dx}{dy}=\dfrac{d}{dy}(\dfrac{y^2}{16}+\dfrac{1}{2y^2})$

$\dfrac{dx}{dy}=\dfrac{1}{16}\dfrac{d}{dy}(y^4)+\dfrac{1}{2}\dfrac{d}{dy}(y^{-2})$

$\dfrac{dx}{dy}=\dfrac{1}{16}(4y^{3})+\dfrac{1}{2}(-2y^{-3})$

$\dfrac{dx}{dy}=\dfrac{y^3}{4}-y^{-3}$

We need to calculate the arc length

Using formula of arc length

$L=\int_{a}^{b}{\sqrt{1+(\dfrac{dx}{dy})^2}dy}$

Put the value into the formula

$L=\int_{1}^{2}{\sqrt{1+(\dfrac{y^3}{4}-y^{-3})^2}dy}$

$L=\int_{1}^{2}{\sqrt{1+(\dfrac{y^3}{4})^2+(y^{-3})^2-2\times\dfrac{y^3}{4}\times y^{-3}}dy}$

$L=\int_{1}^{2}{\sqrt{1+(\dfrac{y^3}{4})^2+(y^{-3})^2-\dfrac{1}{2}}dy}$

$L=\int_{1}^{2}{\sqrt{(\dfrac{y^3}{4})^2+(y^{-3})^2+\dfrac{1}{2}}dy}$

$L=\int_{1}^{2}{\sqrt{(\dfrac{y^3}{4}+y^{-3})^2}dy}$

$L= \int_{1}^{2}{(\dfrac{y^3}{4}+y^{-3})dy}$

$L=(\dfrac{y^{3+1}}{4\times4}+\dfrac{y^{-3+1}}{-3+1})_{1}^{2}$

$L=(\dfrac{y^4}{16}+\dfrac{y^{-2}}{-2})_{1}^{2}$

Put the limits

$L=(\dfrac{2^4}{16}+\dfrac{2^{-2}}{-2}-\dfrac{1^4}{16}-\dfrac{(1)^{-2}}{-2})$

$L=\dfrac{21}{16}$

Hence, The arc length is $\dfrac{21}{16}$

6 0

2 years ago

Which is a perfect square? 6^1 6^2 6^3 6^5 Need the answer asap

oksian1 [2.3K]

Answer:

The answer is 6²

Step-by-step explanation:

Since 6² = 36

√36 = 6 making 6² a perfect square.

Hope this helps you

4 0

2 years ago

Read 2 more answers

Solve the following problems. Reduce your answers to their simplest form. a. 13⁄41 + 27⁄82 b. 3 5⁄24 + 6 7⁄24 + 4 9⁄24 c. 5 2⁄3

KATRIN_1 [288]

13⁄41 + 27⁄82 = 26/82 + 27/82 = 53/82

3 5/24 + 6 7/24 + 4 9/24 = 13 20/24 = 13 5/6

5 2⁄3 + 29⁄69 + 6 21⁄23 = 5 46/69 + 29/69 + 6 63/69 = 11 138/69 = 13

3 9⁄10 + 4⁄9 + 7⁄45 + 4 = 3 81/90 + 40/90 + 14/90 + 4 = 7 135/90 = 8 1/2

6 – 7⁄15 = 5 15/15 - 7/15 = 5 6/15

11 3⁄8 – 7⁄8 = 10 11/8 - 7/8 = 10 4/8 = 10 1/2

 7 1⁄6 – 3 4⁄9 = 7 9/54 - 3 18/54 = 6 63/54 - 3 18/54 = 3 45/54 = 3 5/6

5 3⁄8 – 3 2⁄5 = 5 15/40 - 3 16/40 = 4 55/40 - 3 16/40 = 1 39/40

4 0

2 years ago

Read 2 more answers