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djverab [1.8K]
3 years ago
12

Please help will give brainliest

Mathematics
1 answer:
KatRina [158]3 years ago
5 0
Hi there!

First, scientific notation is when you take a number like 15,000 or 0.0015 and turn it into a number that is less than 10 and greater than 1. What you do to make it that is you move that decimal point over to the left or right as needed. So for example, 15,000 would be 1.5 × 10^{4} and 0.0015 is 1.5 × 10^{-3}.

So if one light year is 5.88 × 10^{12} mi. that is just the scientific notation for 5880000000000 mi.

So if you take 5880000000000 mi. and multiply it by 11 then you get 64680000000000

5880000000000 × 11 = 64680000000000 or 6.468 × 10^{13}

So the answer would be \boxed{64680000000000} or \boxed{6.468*10^{13} }

Your friend, ASIAX
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Factor x2 + 29x - 30.
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3 years ago
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The function f(x) =2(3)^x is to be multiplied by the function g(x) 3(3)^(2x) to create the function h(x). Which function is prod
maria [59]

Answer C is the right answer.

Step-by-step explanation:

h(x) = f(x) * g(x)

h(x) = (2*3^x ) * ( 3*3^2x )

2 * 3 is easy, that will be 6.

The ground number 3 remains 3 in h(x), so that is easy too...

But with multiplying exponents, you can add them.

Let's concentrate only on the exponents of f(x) and g(x)... and add them...

x + 2x =3x

So, now combine the easy part with this new exponent, and you get <u>h(x)</u><u> </u><u>=</u><u> </u><u>6</u><u>*</u><u>(3)^(3x)</u>

<u>So</u><u> </u><u>answer C is the right answer.</u>

7 0
3 years ago
PLEASE I need help on these 4 problems you don’t have to do them all but if you can at least do one of them TYSM
Brrunno [24]

Answer:

For right angle triangle,

we use Pythagoras theorem that is:

c^{2} =a^{2} +b^{2}

c = \sqrt{a^{2} +b^{2} }

For question 1:

c = ?

a = 40

b = 9

putting them in formula,

c = \sqrt{40^{2} + 9^{2} }

c = 41

For question 2:

c = ?

a = 12

b = 13

putting them in formula,

c = \sqrt{12^{2} + 13^{2} }

c = approximately 17.69181

For question 3:

c = 35

a = 20

b = ?

putting them in formula,

c^{2} =a^{2} +b^{2}

35^{2} = 20^{2} + b^{2}

1225 = 400 + b^{2}

b^{2} = 1225 - 400

b^{2} = 825

\sqrt{b^{2} } = \sqrt{825}

b = 5 \sqrt{33}

For question 4:

c = 37

a = 20

b = ?

putting them in formula,

c^{2} =a^{2} +b^{2}

37^{2} = 20^{2} + b^{2}

1369 = 400 + b^{2}

b^{2} = 1369 - 400

b^{2} = 969

Taking square root on both sides

b = 31.12

Hope it helps.

4 0
3 years ago
Robert earned $10, which was 4 less thank twice what Eric made, How much mint did Eric earn?
Jobisdone [24]

Answer:

12

Step-by-step explanation:

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What is the solution set for 2x+5y&gt;-1 and 4x-3&lt;-3?
bekas [8.4K]

PROBLEM ONE

•

Solving for x in 2x + 5y > -1.

•

Step 1 ) Subtract 5y from both sides.

2x + 5y > -1

2x + 5y - 5y > -1 - 5y

2x > -1 - 5y

Step 2 ) Divide both sides by 2.

2x > -1 - 5y

\displaystyle\frac{2x}{2} > \displaystyle\frac{-1 - 5y}{2}

\displaystyle\ x > \frac{-1 - 5y}{2}

So, the solution for x in 2x + 5y > -1 is...

\displaystyle\ x > \frac{-1 - 5y}{2}

•

Solving for y in 2x + 5y > -1.

•

Step 1 ) Subtract 2x from both sides.

2x + 5y > -1

2x - 2x + 5y > -1 - 2x

5y > -1 - 1x

Step 2 ) Divide both sides by 5.

5y > -1 - 1x

\displaystyle\frac{5x}{5} > \frac{-1 -1x}{5}

\displaystyle\ x > \frac{-1 -1x}{5}

So, the solution for y in 2x + 5y > -1 is...

\displaystyle\ x > \frac{-1 -1x}{5}

•

PROBLEM TWO

•

Solving for x in 4x - 3 < -3.

•

Step 1 ) Subtract 3 from both sides.

4x - 3 < -3

4x -3 - 3 < -3 - 3

4x < 0

Step 2 ) Divide both sides by x.

4x < 0

\displaystyle\frac{4x}{4}

x < 0

So, the solution for x in 4x - 3 < -3 is...

x < 0

•

•

- <em>Marlon Nunez</em>

7 0
3 years ago
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