All categories

4 years ago

CAN SOMEONE HELP ME!!!!!

Mathematics

1 answer:

Ivenika [448]4 years ago

7 0

Area yellow =11 x 6
= 66
red area = 7 x 2
=14
area of yellow
66 - 14 = 52

You might be interested in

Simplify −2/3 − 3/5. <br> A) − 1/15 <br> B) − 19/15 <br> C) − 2/5 <br> D) − 5/8

Novosadov [1.4K]

The answer is B.

After writing the numerators above the common denominator, you get -10 + 9/15

Which is -19/15

Hope I helped ♡

5 0

3 years ago

Read 2 more answers

A soccer ball is kicked toward the goal. The height of the ball is modeled by the function h(t) = −16t2 + 48t where t equals the

Vitek1552 [10]

Answer:

Last one

t = 1.5; it takes 1.5 seconds to reach the maximum height of 36 meters.

Step-by-step explanation:

h(t) = −16t² + 48t

h(t) = -16(t² - 3t)

h(t) = -16(t² - 2(t)(1.5) + 1.5² - 1.5²)

h(t) = -16(t - 1.5)² + 36

Axis of symmetry:

t = 1.5

Max height = 36

7 0

3 years ago

The table below displays the enrollment of freshmen taking Algebra 1. Geometry, or Algebra 2.

Vladimir79 [104]

Answer:

for the boys it is 92 then 58 and the total is 150 an then 68 for the girls then for the total it is 100

6 0

3 years ago

Read 2 more answers

which of the following is equivalent to 3 sqrt 32x^3y^6 / 3 sqrt 2x^9y^2 where x is greater than or equal to 0 and y is greater

Nutka1998 [239]

Answer:

$\frac{\sqrt[3]{16y^4}}{x^2}$

Step-by-step explanation:

The options are missing; However, I'll simplify the given expression.

Given

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$

Required

Write Equivalent Expression

To solve this expression, we'll make use of laws of indices throughout.

From laws of indices $\sqrt[n]{a} = a^{\frac{1}{n}}$

So,

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$ gives

$\frac{(32x^3y^6)^{\frac{1}{3}}}{(2x^9y^2)^\frac{1}{3}}$

Also from laws of indices

$(ab)^n = a^nb^n$

So, the above expression can be further simplified to

$\frac{(32^\frac{1}{3}x^{3*\frac{1}{3}}y^{6*\frac{1}{3}})}{(2^\frac{1}{3}x^{9*\frac{1}{3}}y^{2*\frac{1}{3}})}$

Multiply the exponents gives

$\frac{(32^\frac{1}{3}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

Substitute $2^5$ for 32

$\frac{(2^{5*\frac{1}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

$\frac{(2^{\frac{5}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

From laws of indices

$\frac{a^m}{a^n} = a^{m-n}$

This law can be applied to the expression above;

$\frac{(2^{\frac{5}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$ becomes

$2^{\frac{5}{3}-\frac{1}{3}}x^{1-3}*y^{2-\frac{2}{3}}$

Solve exponents

$2^{\frac{5-1}{3}}*x^{-2}*y^{\frac{6-2}{3}}$

$2^{\frac{4}{3}}*x^{-2}*y^{\frac{4}{3}}$

From laws of indices,

$a^{-n} = \frac{1}{a^n}$ ; So,

$2^{\frac{4}{3}}*x^{-2}*y^{\frac{4}{3}}$ gives

$\frac{2^{\frac{4}{3}}*y^{\frac{4}{3}}}{x^2}$

The expression at the numerator can be combined to give

$\frac{(2y)^{\frac{4}{3}}}{x^2}$

Lastly, From laws of indices,

$a^{\frac{m}{n} = \sqrt[n]{a^m}$ ; So,

$\frac{(2y)^{\frac{4}{3}}}{x^2}$ becomes

$\frac{\sqrt[3]{(2y)}^{4}}{x^2}$

$\frac{\sqrt[3]{16y^4}}{x^2}$

Hence,

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$ is equivalent to $\frac{\sqrt[3]{16y^4}}{x^2}$

8 0

3 years ago

4,713 people the other received 2/3 how many votes did B receied

iogann1982 [59]

Answer: 3142

Step-by-step explanation:

4713 x 2/3

3 0

3 years ago