$\frac{1}{3} n - 3 < 6 \\ add \: 3 \: to \: both \: sides \\ \frac{1}{3}n - 3 + 3 < 6 + 3 \\ \frac{1}{3} n < \\ 9 \\ multiply \: both \: sides \: by \: 3 \\ \frac{1}{3} n \times 3< 9 \times 3 \\ n < 27$

8 0

3 years ago

Rewrite the expression. Use an exponent, rather than repeated multiplication. (-4) (-4). Then simplify. Show work.

MakcuM [25]

${4}^{2}$
it is four to the power of two

7 0

2 years ago

Is the imposter sus?

sergij07 [2.7K]

Answer:

yes

Step-by-step explanation:

yes yes yes yes yes yes yes yes yes yes yes

7 0

2 years ago

Read 2 more answers

Find the indicated values, where g(t) = t^2 - t and f(x) = 1 + x g(f(3)-2f(1))

VashaNatasha [74]

Answer:

B. 0

Step-by-step explanation:

Given

g(t) = t² - t

f(x) = 1 + x

Required

Find g(f(3) - 2f(1))

First, we'll solve for f(3)

Given that f(x) = 1 + x

f(3) = 1 + 3

f(3) = 4

Then, we'll solve for 2f(1)

2f(1) = 2 * f(1)

2f(1) = 2 * (1 + 1)

2f(1) = 2 * 2

2f(1) = 4

Substitute the values of f(3) and 2f(1) in g(f(3) - 2f(1))

g(f(3) - 2f(1)) = g(4 - 4)

g(f(3) - 2f(1)) = g(0)

Now, we'll solve for g(0)

Given that g(t) = t² - t

g(0) = 0² - 0

g(0) = 0 - 0

g(0) = 0

Hence, g(f(3) - 2f(1)) = 0

From the list of given options, the correct answer is B. 0

5 0

3 years ago