Can someone please help me fill this out I generally need help

Will give brainliest Of the following sets, which numbers in {1, 2, 3, 4, 5} make the inequality 5x + 2 > 12 true? (5 points)

Radda [10]

Answer:

x = 3, 4, 5

Step-by-step explanation:

Solving the inequality

5x + 2 > 12 ( subtract 2 from both sides )

5x > 10 ( divide both sides by 5 )

x > 2

Thus the only values from the given set which make the inequality true are

x = 3, 4, 5

4 0

2 years ago

Read 2 more answers

Which of the following is equivalent to a real number?

Aleksandr-060686 [28]

Answer:

C) -144^¹/₃

Step-by-step explanation:

Having a number to the power of a fraction is the equivalent of rooting by its reciprocal. So, the sixth root, fourth root, cube root, and square root. Only odd roots can have negative answers (∛x, x can be negative; √y, y can not be negative unless we talk about imaginary numbers).

Since we are looking for a real number, C, showing a cube root of a negative number, must be the answer.

4 0

2 years ago

Show the number on a number line:<br> -12 ÷ 4 =

Gwar [14]

Answer:

well if we take -12 and divide it by four we will get -3.

Step-by-step explanation:

If we have a total of negative 12, just divide it by 4 and the difference between each division is the sum.

Your answer will be -3 on the number line.

:)

3 0

1 year ago

Read 2 more answers

Least to greatest !?

Bad White [126]

The lengths are $\frac{5}{6} \frac{5}{12} and \frac{1}{3}$
finding the LCM we get
$\frac{10,5,4}{12}$
arranging them
$\frac{1}{3} \frac{5}{12} and \frac{5}{6}$

6 0

2 years ago

Help please!!! I dont understand these questions<br><br><br>currently attaching photos dont delete

Katyanochek1 [597]

Answer:

b/a
16a²b²
n¹⁰/(16m⁶)
y⁸/x¹⁰
m⁷n³n/m

Step-by-step explanation:

These problems make use of three rules of exponents:

$a^ba^c=a^{b+c}\\\\(a^b)^c=a^{bc}\\\\a^{-b}=\dfrac{1}{a^b} \quad\text{or} \quad a^b=\dfrac{1}{a^{-b}}$

In general, you can work the problem by using these rules to compute the exponents of each of the variables (or constants), then arrange the expression so all exponents are positive. (The last problem is slightly different.)

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1. There are no "a" variables in the numerator, and the denominator "a" has a positive exponent (1), so we can leave it alone. The exponent of "b" is the difference of numerator and denominator exponents, according to the above rules.

$\dfrac{b^{-2}}{ab^{-3}}=\dfrac{b^{-2-(-3)}}{a}=\dfrac{b}{a}$

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2. 1 to any power is still 1. The outer exponent can be "distributed" to each of the terms inside parentheses, then exponents can be made positive by shifting from denominator to numerator.

$\left(\dfrac{1}{4ab}\right)^{-2}=\dfrac{1}{4^{-2}a^{-2}b^{-2}}=16a^2b^2$

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3. One way to work this one is to simplify the inside of the parentheses before applying the outside exponent.

$\left(\dfrac{4mn}{m^{-2}n^6}\right)^{-2}=\left(4m^{1-(-2)}n^{1-6}}\right)^{-2}=\left(4m^3n^{-5}}\right)^{-2}\\\\=4^{-2}m^{-6}n^{10}=\dfrac{n^{10}}{16m^6}$

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4. This works the same way the previous problem does.

$\left(\dfrac{x^{-4}y}{x^{-9}y^5}\right)^{-2}=\left(x^{-4-(-9)}y^{1-5}\right)^{-2}=\left(x^{5}y^{-4}\right)^{-2}\\\\=x^{-10}y^{8}=\dfrac{y^8}{x^{10}}$

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5. In this problem, you're only asked to eliminate the one negative exponent. That is done by moving the factor to the numerator, changing the sign of the exponent.

$\dfrac{m^7n^3}{mn^{-1}}=\dfrac{m^7n^3n}{m}$

3 0

2 years ago