All categories

2 years ago

What is the value of the expression when x = 12 and p = -4?

Mathematics

1 answer:

goblinko [34]2 years ago

4 0

Answer:

-48

Step-by-step explanation:

You might be interested in

What’s the difference between numbers that are to the power of with parentheses and no parentheses. For example (-2/5)^3 and -2/

Verdich [7]

Answer:

See below

Step-by-step explanation:

The first one 'cubes' the denominator AND the numerator:

(-2/5)^3 = -2/5 * -2/5 * -2/5 = -8/125

The second one 'cubes' only the denominator:

-2/5^3 = -2 / (5 * 5 * 5) = - 2 / 125

8 0

2 years ago

PLEASE HELP ME!!!! I NEED THE HELP!!

makvit [3.9K]

Answer:

try D it seems right to me

4 0

2 years ago

Read 2 more answers

Please help me solve. <br> Thanks.

Slav-nsk [51]

Answer:

The last one is correct

4 0

3 years ago

Read 2 more answers

Select the equivalent expression.

Fittoniya [83]

Answer:

Step-by-step explanation:

8 0

3 years ago

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 $y>2$

7 0

3 years ago