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rusak2 [61]
3 years ago
15

Two students in Mrs. Johnson's preschool class are stacking blocks, one on top of the other. Sydney's blocks are 4 cm high, and

Maddy's blocks are 9 cm high. How tall will their stacks be when they are the same height for the first time? Do NOT write "cm" -- only type the number.
Mathematics
1 answer:
ycow [4]3 years ago
3 0
57 because they are both gonna add up to together
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What is the range of the equation
a_sh-v [17]

The range of the equation is y>2

Explanation:

The given equation is y=2(4)^{x+3}+2

We need to determine the range of the equation.

<u>Range:</u>

The range of the function is the set of all dependent y - values for which the function is well defined.

Let us simplify the equation.

Thus, we have;

y=2 \cdot 4^{x+3}+2

This can be written as y=2^{1+2(x+3)}+2

Now, we shall determine the range.

Let us interchange the variables x and y.

Thus, we have;

x=2^{1+2(y+3)}+2

Solving for y, we get;

x-2=2^{1+2(y+3)}

Applying the log rule, if f(x) = g(x) then \ln (f(x))=\ln (g(x)), then, we get;

\ln \left(2^{1+2(y+3)}\right)=\ln (x-2)

Simplifying, we get;

(1+2(y+3)) \ln (2)=\ln (x-2)

Dividing both sides by \ln (2), we have;

2 y+7=\frac{\ln (x-2)}{\ln (2)}

Subtracting 7 from both sides of the equation, we have;

2 y=\frac{\ln (x-2)}{\ln (2)}-7

Dividing both sides by 2, we get;

y=\frac{\ln (x-2)-7 \ln (2)}{2 \ln (2)}

Let us find the positive values for logs.

Thus, we have,;

x-2>0

     x>2

The function domain is x>2

By combining the intervals, the range becomes y>2

Hence, the range of the equation is y>2

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PLEASE PLEASE HELP ASAP WILL MARK BRAINLIEST
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Answer:

The answer is 216.

Step-by-step explanation:

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AURORKA [14]
The prime factors are: 2 x 2 x 3 x 3 
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Which function is equivalent to f(x) =2x² - 8x + 3?
Afina-wow [57]

Answer:

the answer would be the last one : g(x) = 2(x-2)² - 5

Step-by-step explanation:

I I highly recommend photomath for problems like these because you can easily type in this problem, get the answer, and it gives you a step-by-step process of how it got the answer, and it's free!

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