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3 years ago

Help please!!!!

Mathematics

1 answer:

MissTica3 years ago

4 0

Answer:

I'm not sure about it

Step-by-step explanation:

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10 POINTS IDENTIFY THE SLOPE, Y-INTERCEPT AND GRAPH.<br> 1. Y = 2x + 3<br> 2. Y = 3/4x - 2

Alenkasestr [34]

Answer for problem 1: slope = 2, y intercept = 3

Answer for problem 2: slope = 3/4, y intercept = -2

Explanation:

Both equations are in the form y = mx+b. This is known as slope intercept form. This is because we can read off the values of the slope and y intercept very quickly. We have y = 2x+3 match up with y = mx+b. The m is the slope and b is the y intercept. So m = 2 and b = 3 for this equation. A similar situation happens with the other equation as well. It might help to rewrite the second equation into y = (3/4)x + (-2).

4 0

3 years ago

Read 2 more answers

Perform the indicated operation. 9z^3/16xy . 4x/27z^3

ziro4ka [17]

Answer:

$\frac{1}{12y}$

Step-by-step explanation:

This is a multiplication problem.

We want to multiply $\frac{9z^3}{16xy}\cdot \frac{4x}{27z^3}$

We factor to get:

$\frac{9z^3}{4\times 4xy}\cdot \frac{4x}{9\times 3z^3}$

We now cancel out the common factors to get:

$\frac{1}{4\times y}\cdot \frac{1}{1\times 3}$

We now multiply the numerators and the denominators separately to get.

This simplifies to $\frac{1}{12y}$

Therefore the simplified expression is $\frac{1}{12y}$

8 0

3 years ago

Could you explain what the answer is and why? I am not sure of how to do this

lidiya [134]

D, because after being reflected onto itself or turning 180 degrees, you would be transforming onto itself. I hope this helps!

8 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

Find the volume of the sphere. Round your answer to the nearest tenth.

ollegr [7]

Answer:

Step-by-step explanation:

Givens

pi = 3.14

d = 3.8

r = d/2

r = 3.8/2 = 1.9

Formula

V = (4/3) * pi * r^3

Solution

V = (4/3) * 3.14 * 1.9^3

V = 4/3 * 3.14 * 6.859

V = 28.7163

6 0

2 years ago