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3 years ago

11

Please help i am so confused

1 answer:

harkovskaia [24]3 years ago

3 0

Answer:

$q\geq -81$

Step-by-step explanation:

First, begin by multiplying $\frac{q}{27} \\$ by its reciprocal.

To do this, flip the fraction upside-down.

The resulting equation looks like this: $\frac{27}{1} *\frac{q}{27}$

Cancel out the two 27's and you are left with $\frac{q}{1}$ , which you can simplify to q.

Next, multiply -3 by $\frac{27}{1}$ . Remember, what you do on one side, you must do on the other.

Your equation should look like this:

$q\geq -81$

The final answer is $q\geq -81$ .

Have a nice day!

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I know you want to answer this question.

Alik [6]

Answer:

D. x = 3

Step-by-step explanation:

$\frac{1}{2} ^{x-4}$ - 3 = $4^{x-3}$ - 2

First, convert $4^{x-3}$ to base 2:

$4^{x-3}$ = $(2^{2})^{x-3}$

$\frac{1}{2} ^{x-4}$ - 3 = $(2^{2})^{x-3}$ - 2

Next, convert $\frac{1}{2} ^{x-4}$ to base 2:

$\frac{1}{2} ^{x-4}$ = $(2^{-1})^{x-4}$

$(2^{-1})^{x-4}$ - 3 = $(2^{2})^{x-3}$ - 2

Apply exponent rule: $(a^{b})^{c}$ = $a^{bc}$ :

$(2^{-1})^{x-4}$ = $2^{-1*(x-4)}$

$2^{-1*(x-4)}$ - 3 = $(2^{2})^{x-3}$ - 2

Apply exponent rule: $(a^{b})^{c}$ = $a^{bc}$ :

$(2^{2})^{x-3}$ = $2^{2(x-3)}$

$2^{-1*(x-4)}$ - 3 = $2^{2(x-3)}$ - 2

Apply exponent rule: $a^{b+c}$ = $a^{b}$ $a^{c}$ :

$2^{-1(x-4)}$ = $2^{-1x}$ * $2^{4}$ , $2^{2(x-3)}$ = $2^{2x}$ * $2^{-6}$

$2^{-1 * x}$ * $2^{4}$ - 3 = $2^{2x}$ * $2^{-6}$ - 2

Apply exponent rule: $(a^{b})^{c}$ = $a^{bc}$ :

$2^{-1x}$ = $(2^{x})^{-1}$ , $2^{2x}$ = $(2^{x})^{2}$

$(2^{x})^{-1}$ * $2^{4}$ - 3 = $(2^{x})^{2}$ * $2^{-6}$ - 2

Rewrite the equation with $2^{x} = u$ :

$(u)^{-1}$ * $2^{4}$ - 3 = $(u)^{2}$ * $2^{-6}$ - 2

Solve $u^{-1}$ * $2^{4}$ - 3 = $u^{2}$ * $2^{-6}$ - 2:

$u^{-1}$ * $2^{4}$ - 3 = $u^{2}$ * $2^{-6}$ - 2

Refine:

$\frac{16}{u}$ - 3 = $\frac{1}{64}$ $u^{2}$ - 2

Add 3 to both sides:

$\frac{16}{u}$ - 3 + 3 = $\frac{1}{64}$ $u^{2}$ - 2 + 3

Simplify:

$\frac{16}{u}$ = $\frac{1}{64}$ $u^{2}$ + 1

Multiply by the Least Common Multiplier (64u):

$\frac{16}{u}$ * 64u = $\frac{1}{64}$ $u^{2}$ + 1 * 64u

Simplify:

$\frac{16}{u}$ * 64u = $\frac{1}{64}$ $u^{2}$ + 1 * 64u

Simplify $\frac{16}{u}$ * 64u:

1024

Simplify $\frac{1}{64}$ $u^{2}$ * 64u:

$u^{3}$

Substitute:

1024 = $u^{3}$ + 64u

Solve for u:

u = 8

Substitute back u = $2^{x}$ :

8 = $2^{x}$

Solve for x:

x = 3

4 0

3 years ago

What is the length of line segment FE?

Jobisdone [24]

Answer:

(7,-1)because it is problem of vector lesson so that we have to use formula of vector

6 0

3 years ago

Read 2 more answers

If p = -2 and q = 5, is the value of p/q equivalent to 2/5?

grandymaker [24]

Answer:

No.

Step-by-step explanation:

1. Subsutute p/q with the numbers given, p = -2 and q = 5.

-2 / 5.

2. Divide 2/5 and -2/5 and compare.

2 / 5 = 0.40

-2 / 5 = -0.40

-0.40 < 0.40

If they are not =, then they are <u>NOT</u> equivalent.

Which means, they are Not equivalent.

5 0

3 years ago

Which of the following is the algebraic form for the verbal statement shown: "12 more than the product 4 and a number, n" - A. n

katrin2010 [14]

If you mean “13 more than the product 4 and a number, n”
then the answer is B

the word product means multiply

and 12 more means adding 12 to the product

6 0

3 years ago

Read 2 more answers

The initial deposit into an account?

zhuklara [117]

Answer:

B

Step-by-step explanation:

Simple interest is based on the principal amount of a loan or deposit.

8 0

3 years ago