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2 years ago

8

Which set of ordered pairs could be generated by an exponential function?

1 answer:

Ede4ka [16]2 years ago

6 0

Answer:

(1,1), (2, 1/4), (3, 1/9), (4, 1/16)

Step-by-step explanation:

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Equations by substitution<br> y=3x-7; 2x+4y=14

Bad White [126]

Let's plug in y=3x-7 into 2x+4y=14
2x+4 (3x-7)=14
Distribute
2x+12x-28=14
Add
14x-28=14
Add 28 to both sides
14x=42
Divide
X=3
Y=3 (3)-7
Y=9-7
Y=2
(3,2) A the solution is (3,2)

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3 years ago

The graph shows the cube root parent function.<br>Which statement is true? <br>A.<br>B.<br>C.<br>D.

viktelen [127]

Answer:

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Step-by-step explanation:

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3 years ago

Factor out:<br> 1/8x - 49/8

bonufazy [111]

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3 years ago

I will give out brainliest!

asambeis [7]

Answer:

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Step-by-step explanation:

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3 years ago

Read 2 more answers

Please help asap, brainliest,thanks, and 50 points. Thank you soooo much! <3

Anuta_ua [19.1K]

Answers:

1) $x^{8} y^{8}$

2) $y^{3} \sqrt{y}$

3) $5x^{4} \sqrt{6}$

4) $\sqrt{7}$

5) $\frac{\sqrt{z}}{z}$

Step-by-step explanation:

1) $\sqrt{x^{16} y^{36}}$

Rewriting the expression:

$(x^{16} y^{36})^{\frac{1}{2}}$

Multiplying the exponents:

$x^{\frac{16}{2}} y^{\frac{36}{2}}$

Simplifying:

$x^{8} y^{8}$

2) $\sqrt{y^{7}}$

Rewriting the expression:

$\sqrt{y^{6} y}=(y^{6} y)^{\frac{1}{2}}$

Multiplying the exponents:

$y^{\frac{6}{2}} y^{\frac{1}{2}}$

Simplifying:

$y^{3} y^{\frac{1}{2}}=y^{3} \sqrt{y}$

3) $\sqrt{150 x^{8}}$

Rewriting the expression:

$\sqrt{(6)(25) x^{8}}$

Since $\sqrt{25}=5$ :

$5x^{4}\sqrt{6}$

4) $\frac{7}{\sqrt{7}}$

Multiplying numerator and denominator by $\sqrt{7}$ :

$\frac{7}{\sqrt{7}} (\frac{\sqrt{7}}{\sqrt{7}})=\frac{7}{7\sqrt{7}}$

Simplifying:

$\sqrt{7}$

5) $\frac{5z}{\sqrt{25 z^{3}}}$

Rewriting the expression:

$\frac{5z}{5z \sqrt{z}}$

Simplifying:

$\frac{1}{\sqrt{z}}$

Since we do not want the square root in the denominator, we can multiply numerator and denominator by $\sqrt{z}$ :

$\frac{1}{\sqrt{z}}(\frac{\sqrt{z}}{\sqrt{z}})$

Finally:

$\frac{\sqrt{z}}{z}$

3 0

3 years ago