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3 years ago

12

The bomb cyclone storm dumped 3 inches of snow per hour in Providence, Rhode Island. How many inches of snow fell during a 12-ho

ur period?

2 answers:

Viktor [21]3 years ago

8 0

Answer:

36

Step-by-step explanation:

earnstyle [38]3 years ago

5 0

Answer:

36

Step-by-step explanation:

3x12=36

12+12=24

add 12 to 24 and it equals 36.

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What are three different ways to make the number 15,638 with only hundreds, tens, and ones?

Mademuasel [1]

156 hundreds 3 teens and 8 ones I believe

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Please help extra points and Mark brainlist

jenyasd209 [6]

Answer:

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Step-by-step explanation:

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3 years ago

1. What is the measure of the angle that is complementary to

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3 years ago

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Find the distance between the pair of points. give an exact answer, and, where appropiate, and apporximation to two decimal plac

atroni [7]

Distance Formula: $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$
Product rule of radicals: √ab = √a x √b

Using our two points, we can solve for the distance as such:

$\sqrt{(4-(-6))^2+(3-(-1))^2}\\ \sqrt{(10)^2+(4)^2}\\ \sqrt{100+16}\\ \sqrt{116}$

Now our exact answer may be √116, however using the product rule of radicals we can further simplify the radical as such:

√116 = √4 x √29 = 2√29.

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3 years ago

Which expression is a sum of cubes?

erik [133]

we know that

A polynomial in the form $a^{3} +b^{3}$ is called a sum of cubes

so

Let's verify each case to determine the solution

<u>case A)</u> $-64x^{6} y^{12} +125x^{16} y^{3}$

we know that

$-64=-4^{3}$

$x^{6}= (x^{2})^{3}$

$y^{12}= (y^{4})^{3}$

$125=5^{3}$

$x^{16}=x^{15} *x=x*(x^{5})^{3}$ -------> is not a perfect cube

$y^{3}= (y)^{3}$

therefore

the case A) is not a sum of cubes

<u>case B)</u> $-32x^{6} y^{12} +125x^{16} y^{3}$

we know that

$-32=-2^{5}$ -------> is not a perfect cube

$x^{6}= (x^{2})^{3}$

$y^{12}= (y^{4})^{3}$

$125=5^{3}$

$x^{16}=x^{15} *x=x*(x^{5})^{3}$ -------> is not a perfect cube

$y^{3}= (y)^{3}$

therefore

the case B) is not a sum of cubes

<u>case C)</u> $32x^{6} y^{12} +125x^{9} y^{3}$

we know that

$32=2^{5}$ -------> is not a perfect cube

$x^{6}= (x^{2})^{3}$

$y^{12}= (y^{4})^{3}$

$125=5^{3}$

$x^{9}=(x^{3})^{3}$

$y^{3}= (y)^{3}$

therefore

the case C) is not a sum of cubes

<u>case A)</u> $64x^{6} y^{12} +125x^{9} y^{3}$

we know that

$64=4^{3}$

$x^{6}= (x^{2})^{3}$

$y^{12}= (y^{4})^{3}$

$125=5^{3}$

$x^{9}=(x^{3})^{3}$

$y^{3}= (y)^{3}$

Substitute

$4^{3}(x^{2})^{3}(y^{4})^{3} +5^{3}(x^{3})^{3}(y)^{3}$

$(4x^{2}y^{4})^{3} +(5x^{3}y)^{3}$

therefore

<u>the answer is</u>

$64x^{6} y^{12} +125x^{9} y^{3}$ is a sum of cubes

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3 years ago

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