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4 years ago

Find the GCF of 6x^7 and 15x^3?

1 answer:

5 0

I believe the answer is 3x^3

since 3 is the greatest common factor for the whole number and both have at least x to the third

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I need help w number 5 pls !!

OleMash [197]

Answer:

Step-by-step explanation:

the parameter of B figure is

12+9+9+5

as we know the scale factor is 7/2

so the parameter of A figure is

35×7/2

122.5

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3 years ago

How to do secret message alphabet keyword

Ber [7]

#17 the answer is 54 look at the box at the bottom 54 = p 17 on the blank put p

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3 years ago

The cereal box shown below is cut from a piece of cardboard with 600 square inches. How many cereal boxes can be cut from this p

dimaraw [331]

Answer:

87 moneys

Step-by-step explanation:

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3 years ago

What are the answer to both of these. I am confused. Please Help. I will give Brainliest Answer.

Mumz [18]

Answer:

The First 1 is B. And the Second 1 is B. As well.

Step-by-step explanation:

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3 years ago

GRAVITATION The height h(t) in feet of an object t seconds after it is propelled straight up from the ground with an initial vel

zhuklara [117]

<h3>Option B</h3><h3>At 2 second and 1.75 second, the object be at a height of 56 feet</h3>

<em><u>Solution:</u></em>

Given that,

<em><u>The height h(t) in feet of an object t seconds after it is propelled straight up from the ground with an initial velocity of 60 feet per second is modeled by the equation:</u></em>

$h(t) = -16t^2 + 60t$

<em><u>At what times will the object be at a height of 56 feet</u></em>

<em><u>Substitute h = 56</u></em>

$56 = -16t^2 + 60t\\\\16t^2 - 60t + 56 = 0\\\\Divide\ the\ equation\ by\ 4\\\\4t^2 - 15t + 14=0$

Solve the above equation by quadratic formula

$\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}\\\\x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$\mathrm{For\:}\quad a=4,\:b=-15,\:c=14\\\\x =\frac{-\left(-15\right)\pm \sqrt{\left(-15\right)^2-4\cdot \:4\cdot \:14}}{2\cdot \:4}\\\\Simplify\\\\x = \frac{15 \pm \sqrt{1}}{8}\\\\x = \frac{15 \pm 1}{8}\\\\We\ have\ two\ solutions\\\\x = \frac{15+1}{8} \text{ and } x = \frac{15-1}{8}\\\\x = 2 \text{ and } 1.75$

Thus, at 2 second and 1.75 second, the object be at a height of 56 feet

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4 years ago