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3 years ago

10

Pictures below. Find x.

1 answer:

denis-greek [22]3 years ago

8 0

Hope this helps u...!!

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Sonya took out a mortgage of $65,000 at 7% for 25 years. What is the cost of the mortgage?

Vika [28.1K]

The cost of the mortgage is $81250

<h3>What are interests?</h3>

Interests are percentages of a principal

Given the following parameters

Principal = $65000

Rate = 7% = 0.07
Time = 5 years

<h3>Calculate the interest</h3>

I = PRT/100
I = 65000*0.07*25
I = 16,250

<h3>Determine the cost of the mortgage</h3>

Cost of mortgage = Principal + Interest

Cost of mortgage = 65000 + 16250
Cost of mortgage = 81,250

Hence the cost of the mortgage is $81250

Learn more on mortgage here: brainly.com/question/22846480

5 0

2 years ago

Read 2 more answers

Write each of the following decimals in words.

Sergeeva-Olga [200]

A. nine hundredths

B. Thirty five hundredths

C. Seven Tenths

D. Three Thousandths

E. One Hundred Fourty five Thousandths

F. Fifty Nine Ten Thousandths

8 0

3 years ago

If u help me your dreams will come true ya know

wariber [46]

Answer:

Area of larger circle = $\pi *r^2=3.14(11^2)=379.94$

Area of smaller circle = $\pi *r^2=3.14(3^2)=28.26$

Probability that it lands in the shaded area (smaller circle):

28.26 ÷ 379.94 = 0.07438... = 0.07 (nearest hundredth)

8 0

3 years ago

How many milliliters are in 5 liters?

blsea [12.9K]

Answer:

There are 5000 mililitres in 5 litres

Therefore option A.

Hope u got the answer............

5 0

2 years ago

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