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olga_2 [115]
3 years ago
8

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
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