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irakobra [83]
3 years ago
5

How many feet are in 108 inches?

Mathematics
2 answers:
MissTica3 years ago
7 0
Well first we need to know how many inches are in one foot which is 12 inches for every foot. So now we can take 108 inches / 12 because there are 12 inches to a foot and we need to convert inches to feet. Which gives us are end answer of 9 feet in 108 inches.
iris [78.8K]3 years ago
5 0
There are 9 feet because there are 12 inches in a foot so 108÷12=9
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7/7q+21= x /5q^2-45 then x=?​
anzhelika [568]

Answer:

x = 5q - 15

Step-by-step explanation:

\frac{7}{7q+21}=\frac{x}{5q^{2}-45}\\\\\frac{7}{7(q+3)}=\frac{x}{5 (q^{2} -9)}\\\\frac{1}{q+3}=\frac{x}{5*(q^{2}-3^{2})}\\\\\frac{1}{q+3}=\frac{x}{5(q+3)(q-3)}\\\\\frac{1}{q+3}*5*(q+3)(q-3)=x\\\\5(q-3)=x\\\\x= 5q-15

3 0
3 years ago
Find the measure of each numbered angle in the rectangle
azamat

m∠1=31

m∠2=59

m∠3=90

m∠4=31

m∠5=90

hope this helps brainliest please!

6 0
3 years ago
Meredith deposits $1,400 into an account that earns 3% annual simple interest. How many years will it take Meredth to earn $294
Valentin [98]

Answer:

21 years

Step-by-step explanation:

8 0
3 years ago
I don't know what the answer is to divide 180 in a ratio of 4:5:9
Karolina [17]
Hello there,

First you need to add together the numbers.
4+5+9 = 18
Divide the total by this,
180 / 18 = 10
Multiply each number by 10.
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4 0
3 years ago
Read 2 more answers
a small rocket is shot from the edge of a cliff suppose that after t seconds the rocket is y meters above the cliff where y=30t-
Natalka [10]

Answer:

Greatest height: 45 meters

Time for greatest height: 3 seconds

Height after 5 seconds: 25 meters above the cliff

Time for height of 40 meters: 7.123 seconds

Height after 7 seconds: -35 meters (35 meters below the cliff)

Step-by-step explanation:

to find the maximum height, we need to calculate the derivative of y in relation to t and then find when dy/dt = 0:

dy/dt = 30 - 10t = 0

10t = 30

t = 3 seconds

In this time, the height is:

y = 30*3 - 5*3^2 = 45 meters

After 5 seconds, the height is:

y = 30*5 - 5*5^2 = 25 meters

The time for the height of 40 meters is:

40 = 30t - 5t^2

t^2 - 6t - 8 = 0

Using Bhaskara's formula, we have:

Delta = 6^2 + 4*8 = 68

sqrt(Delta) = 8.246

t1 = (6 + 8.246) / 2 = 7.123 seconds

t2 = (6 - 8.246) / 2 = -1.123 seconds (negative value for time is not valid)

So the time when the rocket reaches 40 meters is 7.123 seconds

After 7 seconds, the height is:

y = 30*7 - 5*7^2 = -35 meters

The rocket will be 35 meters below the cliff.

5 0
3 years ago
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