All categories

3 years ago

HELP!!! Use the properties of exponents to simplify the expression:

Mathematics

1 answer:

Dvinal [7]3 years ago

6 0

There you go the answer key

You might be interested in

Classify each polynomial by its degree and bt its number of terms (i.e. cubic or degree=3; trinomial. (a) 5x^4-8 (b) 4a^2-2a-16

lord [1]

Answer:

a. 4th degree binomial

b. 2nd degree trinomial

c. 3rd degree monomial.

Step-by-step explanation:

In order to find each of these, start by writing the degree based on the highest exponent. The work for a is done for you below.

5x^4 - 8

This means it is a 4th degree.

Now we can decide what the final word is based on how many terms there are. In this one there is 2, so we call it a binomial.

4th degree binomial.

3 0

4 years ago

0 = 2x2 - 6x + 4 What are the solutions to

Ainat [17]

The answer would be .5

6 0

3 years ago

Read 2 more answers

What is the opposite of -5.2

Shalnov [3]

Answer:

The opposite or absolute value is 5.2

5 0

4 years ago

How do I solve this math problem?

Nat2105 [25]

Answer: 4x² + 4x - 2 - $\frac{3}{2x+3}$

Step-by-step explanation:

You can use either synthetic or long division. I am using synthetic division:

2x + 3 = 0 ⇒ x = $-\frac{3}{2}$

$-\frac{3}{2}$ | 4 10 4 -6

| ↓ -6 -6 3

4 4 -2 -3 ← -3 is the remainder

↓ ↓ ↓

4x² +4x -2 ← factored polynomial

3 0

3 years ago

The average value of a function f over the interval [−2,3] is −6 , and the average value of f over the interval [3,5] is 20. Wha

Xelga [282]

Answer:

The average value of $f$ over the interval $[-2,5]$ is $\frac{10}{7}$ .

Step-by-step explanation:

Let suppose that function $f$ is continuous and integrable in the given intervals, by integral definition of average we have that:

$\frac{1}{3-(-2)} \int\limits^{3}_{-2} {f(x)} \, dx = -6$ (1)

$\frac{1}{5-3} \int\limits^{5}_{3} {f(x)} \, dx = 20$ (2)

By Fundamental Theorems of Calculus we expand both expressions:

$\frac{F(3)-F(-2)}{3-(-2)} = -6$

$F(3) - F(-2) = -30$ (1b)

$\frac{F(5)-F(3)}{5-3} = 20$

$F(5) - F(3) = 40$ (2b)

We obtain the average value of $f$ over the interval $[-2, 5]$ by algebraic handling:

$F(5) - F(3) +[F(3)-F(-2)] = 40 + (-30)$

$F(5) - F(-2) = 10$

$\frac{F(5)-F(-2)}{5-(-2)} = \frac{10}{5-(-2)}$

$\bar f = \frac{10}{7}$

The average value of $f$ over the interval $[-2,5]$ is $\frac{10}{7}$ .

4 0

3 years ago