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3 years ago

5

What is the reference angle for 225° and what is a reference angle?

1 answer:

sertanlavr [38]3 years ago

6 0

<span> the angle that the Main Angle is called reference angle
</span><span>Given angle - 180
</span><span> 225° -180
</span>=45
eference angle for 225° is 45

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The function g(x) = (x + 2y2 + 4 is a transformation of the parent function f(x) = x? Which of the following statements are true

Nonamiya [84]

Answer:

I think its D

Step-by-step explanation:

3 0

3 years ago

(-6,1) is a point on the graph of y=g(x)

NNADVOKAT [17]

Answer:

$(-7, -4)$

$(-4, 2)$

$(-4, 1)$

Step-by-step explanation:

We know that the point (-6, 1) belongs to the main function g(x)

The transformation

$y = g (x + 1) -5$

add 1 to the input variable (x) and subtract 5 to the output variable (y)

So the point in the graph of $y = g (x + 1) -5$ is

$x + 1 =-6\\x = -7$

$y= 1-5\\y = -4$

The point is: $(-7, -4)$

The transformation

$y = -2g(x -2) +4$

subtract two units from the input variable (x), multiply the output variable (y) by -2 and then add 4 units

So the point in the graph of $y = -2g(x -2) +4$ is

$x -2 =-6\\x = -4\\\\y = -2(1)+4\\y = 2$

The point is: $(-4, 2)$

The transformation

$y=g(2x+2)$

Multiply the input variable (x) by 2 and then add two units

So the point in the graph of $y=g(2x+2)$ is

$2x +2 =-6\\2x = -8\\x=-4$

$y=1$

The point is: $(-4, 1)$

3 0

3 years ago

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 midpoint of the segment with the following endpoints.<br> (-3,-3) and (1, 2)

Vlad [161]

$\left(\frac{-3+1}{2}, \frac{-3+2}{2} \right)=\boxed{\left(-1, -\frac{1}{2} \right)}$

6 0

2 years ago

Read 2 more answers

PLS HELP ME!!!!!!!!!!

KonstantinChe [14]

Answer:

1) yes, SAS

2) yes, AAS

3) yes, SAS

4) not enough information to know

Step-by-step explanation:

6 0

3 years ago