Quadratic functions are second-order equations of the form y=ax^2+bx+c. Their graphs form parabolas. The defining characteristic of a quadratic is that the acceleration of the outputs is a constant.

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2 years ago

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Which of these choices show a pair of equivalent expressions? Check all that apply.

AleksAgata [21]

The main rule we use in this exercise is

$( \sqrt{a} ) ^{m} =( a^{ \frac{1}{2}} )^{m} = a^{ \frac{1}{2}*m}=a^{ \frac{m}{2}}$

and the more general case:

$( \sqrt[n]{a} ) ^{m} =( a^{ \frac{1}{n}} )^{m} = a^{ \frac{1}{n}*m}=a^{ \frac{m}{n}}$
Using these rules, we make all the indices, or exponents, look like $a^{ \frac{m}{n}}$ , and compare them to each other.

A. $( \sqrt{8} )^{9} = ( 8^{ \frac{1}{2}} )^{9} = 8^{ \frac{1}{2}*9}= 8^{ \frac{9}{2}}$

B. $( \sqrt{4} ) ^{5} =( 4^{ \frac{1}{2}} )^{5} = 4^{ \frac{1}{2}*5}=4^{ \frac{5}{2}}$

C. $( \sqrt[3]{125} ) ^{7} =( 125^{ \frac{1}{3}} )^{7} = 125^{ \frac{1}{3}*7}= 125^{ \frac{7}{3}}$

D. $( \sqrt{12} ) ^{7} =( 12^{ \frac{1}{2}} )^{7} = 12^{ \frac{1}{2}*7}=12^{ \frac{7}{2}}$

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2 years ago

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Georgie joins a gym. She pays $25 to sign up and then $15 each month. Graph it. Label the y-intercept,the slope, and your axes.

Firlakuza [10]

The first thing you should do for this case is to find the equation of the line that best suits the problem and then plot it.
Let
x: number of months
y: amount paid.
The equation of the line is
y = 15x + 25
y-intercept = 25
the slope = 15
answer
Georgie pays (y axis) $ 15 dollars (the spole) monthly (x-axis) in the gym with a $ 25 registration (y-intercept)

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2 years ago

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Fill in the blank with a number to make the expression a perfect square.

Sati [7]

Answer:

Step-by-step explanation:

Substituting 16 will make the expression valid and correct since the square of -4 is 16 and the addition of it is -8.

$v^2-8v+16=(v-4)^2$

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2 years ago