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bearhunter [10]
3 years ago
8

7. How would you solve this equation: 9x = 153? * Multiply Divide Add Subtract

Mathematics
2 answers:
Kipish [7]3 years ago
5 0

You would need to DIVIDE by 9

erastova [34]3 years ago
3 0

Answer:

x=17

Step-by-step explanation:

divide both sides of the equation by 9

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Which expression is equivalent to StartRoot 120 x EndRoot?.
Irina-Kira [14]

Here for 120, the expression for the start root and end root is given  

\sqrt{120} =2\sqrt{30}

<h3>What will be the expression for start root and enroot of 120?</h3>

Here we need to find \sqrt{120}

By simplifying the given expression, we get:

120=2\times2\times2\times3\times5

120=(2^{2} )\times3\times5

by taking square roots on both sides we get

\sqrt{120} =2\sqrt{30}

Thus for 120, the expression for the start root and end root is given  \sqrt{120} =2\sqrt{30}

To know more about the Square roots follow

brainly.com/question/124481

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1 year ago
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You meant pi, pi is a irrational number, but they use 3.14 for approximation, so that is your answer.

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Type the correct answer in the box. Write your answer as a fraction, using / for the fraction bar.
zmey [24]

Answer:

4/11

Step-by-step explanation:

7 0
2 years ago
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What is f(−3) for the function f(a)=−2a2−5a+4?
Readme [11.4K]

Answer:

f(-3)

f(a) = -2a² - 5a + 4

Substitute a with its value, -3

f(-3) = -2(-3²) - 5(-3) + 4

f(-3) = -2(9) + 15 + 4

f(-3) = -18 + 19

f(-3) = 1

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7 0
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An airplane climbs 100 feet during the first second after takeoff. In each succeeding second it climbs 100 feet more than it cli
tigry1 [53]

Answer:

15 seconds.

Step-by-step explanation:

∵ The distance covered by plane in first second = 100 ft,

Also, in each succeeding second it climbs 100 feet more than it climbed during the previous second,

So, distance covered in second second = 200,

In third second = 300,

In fourth second = 400,

............, so on....

Thus, the total distance covered by plane in n seconds = 100 + 200 + 300 +400......... upto n seconds

=\frac{n}{2}(2\times 100 + (n-1)100)  ( Sum of AP )

=\frac{n}{2}(200+100n-100)

=\frac{n}{2}(100+100n)

=50n+50n^2

Suppose the distance covered in n seconds is 12,000 feet,

\implies 50n+50n^2=12000

n+n^2=240

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Hence, after 15 seconds the plane will reach an altitude of 12,000 feet above its takeoff height.

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