All categories

3 years ago

Please Solve 9(x - 11) = 9

Mathematics

1 answer:

Licemer1 [7]3 years ago

3 0

Answer:

9(x-11) = 9

9x-99 = 9

9x = 108

x = 108/9

thus,

x = 12

You might be interested in

In 9 years, Lance earned $3262.50 interest on a deposit made to an account that earned 2.7% simple interest per year: How much w

LenaWriter [7]

Answer:

$4055.28

Step-by-step explanation:

Your formula is A=P(1+rt)

P is the amount of money Lance puts in = 3262.50

R is the interest rate in a decimal = .027

T is the time in years = 9

A=3262.50(1+.027x9)

Use a calculator

A=4055.28

4 0

3 years ago

Read 2 more answers

Help.please ASAP will mark brainliest

ZanzabumX [31]

Answer:

Step-by-step explanation:

3 0

4 years ago

adult tickets to space city amusement park cost x dollars. childrens tickets cost y dollars.the henson family bought 3 adult and

ASHA 777 [7]

See the pic-the top two equations are your "system" of equations to solve

8 0

4 years ago

At 2 PM, a thermometer reading 80°F is taken outside where the air temperature is 20°F. At 2:03 PM, the temperature reading yiel

Alexeev081 [22]

Use Law of Cooling:
$T(t) = (T_0 - T_A)e^{-kt} +T_A$
T0 = initial temperature, TA = ambient or final temperature

First solve for k using given info, T(3) = 42
$42 = (80-20)e^{-3k} +20 \\ \\ e^{-3k} = \frac{22}{60}=\frac{11}{30} \\ \\ k = -\frac{1}{3} ln (\frac{11}{30})$

Substituting k back into cooling equation gives:
$T(t) = 60(\frac{11}{30})^{t/3} + 20$

At some time "t", it is brought back inside at temperature "x".
We know that temperature goes back up to 71 at 2:10 so the time it is inside is 10-t, where t is time that it had been outside.
The new cooling equation for when its back inside is:
$T(t) = (x-80)(\frac{11}{30})^{t/3} + 80 \\ \\ 71 = (x-80)(\frac{11}{30})^{\frac{10-t}{3}} + 80$
Solve for x:
$x = -9(\frac{11}{30})^{\frac{t-10}{3}} + 80$
Sub back into original cooling equation, x = T(t)
$-9(\frac{11}{30})^{\frac{t-10}{3}} + 80 = 60 (\frac{11}{30})^{t/3} +20$
Solve for t:
$60 (\frac{11}{30})^{t/3} +9(\frac{11}{30})^{\frac{-10}{3}}(\frac{11}{30})^{t/3} = 60 \\ \\ (\frac{11}{30})^{t/3} = \frac{60}{60+9(\frac{11}{30})^{\frac{-10}{3}}} \\ \\ \\ t = 3(\frac{ln(\frac{60}{60+9(\frac{11}{30})^{\frac{-10}{3}}})}{ln (\frac{11}{30})}) \\ \\ \\ t = 4.959$

This means the exact time it was brought indoors was about 2.5 seconds before 2:05 PM

8 0

4 years ago

What must be true of two rational expressions before they can be added

Kisachek [45]

Answer:

reeee

Step-by-step explanation:

4 0

3 years ago