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4 years ago

If (62)x = 1 what is the value of x? Explain your answer.

1 answer:

Alex4 years ago

3 0

Whenever you solve an equation with a single variable you need to isolate the variable by itself on one side of the equal sign. Just be sure that whatever operations you perform on one side you perform on the other sides as well.

62x=1 divide both sides by 62

x=1/62

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-129 = -48 - n <br> Thank you

Flauer [41]

−129=−48−n

Simplify:

−129=−48+−n

Flip the equation.

−n−48=−129

Add 48 to both sides.

−n−48+48=−129+48

−n=−81

Divide both sides by -1.

−n/−1 = −81/−1

n=81

3 0

3 years ago

Which expression is equivalent to log w (x2-6)4

natka813 [3]

Log w (x^2-6)^4

Using log a b = log a + log b, with a=w and b=(x^-6)^4:
log w (x^2-6)^4 = log w + log (x^2-6)^4

Using in the second term log a^b = b log a, with a=x^2-6 and b=4
log w (x^2-6)^4 = log w + log (x^2-6)^4 = log w + 4 log (x^2-6)

Then, the answer is:
log w (x^2-6)^4 = log w + 4 log (x^2-6)

7 0

4 years ago

Read 2 more answers

Solve for 'p'

Darina [25.2K]

$\left(\dfrac{9a^{-4}}{25}\right)^p=\dfrac3{5a^2}$
$\left(\dfrac9{25a^4}\right)^p=\dfrac3{5a^2}$

One way to find $p$ is to notice that the part on the left hand side within the parentheses can be written as a square:

$\dfrac9{25a^4}=\dfrac{3^2}{5^2(a^2)^2}=\left(\dfrac3{5a^2}\right)^2$

So you could write

$\left(\left(\dfrac3{5a^2}\right)^2\right)^p=\left(\dfrac3{5a^2}\right)^{2p}=\dfrac3{5a^2}$

Since these two expressions are equal, and the bases are the same, you know the exponents must be equal, with the exponent on the right hand side being 1. So $2p=1$ , or $p=\dfrac12$ .

6 0

3 years ago

Use a graphing calculator to approximate the vertex of the graph of the parabola defined by the following

vivado [14]

Answer: (-0.4, 6.8)

The graph is attached

Step-by-step explanation:

We can find the axis of symmetry using the formula x = -b/2a

x = - -4/2(-5)

x = 4/-10

x = 0.04

Substitute for x and calculate y

y = $\left(0.4^{2}\right)\left(-5\right)-4\left(-0.4\right)\ +6$

y = 6.8

3 0

3 years ago

Anybody? Showing the work too plz?

shusha [124]

I can’t see it, make it clearer please

8 0

3 years ago