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3 years ago

Find the GCF of the following literal terms. m^7n^4p^3 and mn^12p^5

Mathematics

2 answers:

Nadusha1986 [10]3 years ago

5 0

Both terms have in common:
m
n⁴
p³

The GCF = mn⁴p³

Sidana [21]3 years ago

3 0

I hope this helps you

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Solve 3x + k = cfor x.

Ghella [55]

First, subtract k from both sides. That means c - k is in the left side and 3x is on the right. Then, divide by 3 on both sides. So, the answer is x = (c - k) / 3.

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3 years ago

Which point lies on the graph of f(x) = -2(x + 3) – 5 A) (-6,1) B) (-6,-8) C) (6,-13) D) (6,-4)

nalin [4]

First you'd work around the parenthesis, getting you a reduced problem of -2x-6-5.

combine like terms to receive -2x-11

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4 years ago

What is 54 3/11 as a decimal

saul85 [17]

The answer is 54.27. Hope I was halpful

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3 years ago

F(1) = 3, f(n) = [f(n-1)]?

wlad13 [49]

Answer:

No, they are not equal.

Step-by-step explanation:

f(n) = f(n-1)

Let's say, n = 3. Let's put that in the equation:

f(3) = f(3-1)

f(3) = f(2)

Are they the same? Apparently, no. So, these two functions are not equal.

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3 years ago

How much kinetic energy does a 4 kg cats have while at 9 m/s

coldgirl [10]

Answer:

162 Joules

Step-by-step explanation:

Remember, the equation for the kinetic energy of an object is:

$K=\frac{1}{2}mv^{2}$

where m is the mass of the object in kg (kilograms), v is the velocity of the object in m/s (meters per second), and K is the kinetic energy of the object in J (joules).

We are given that the cat's mass is 4 kg. We will use this for m. We are also given that the cat is moving at 9 m/s. This would be it's speed. We can use this for velocity, or v. Plugging in we get:

$K=\frac{1}{2}mv^{2}=\frac{1}{2}(4)(9)^{2}=(2)(81)=162$

So K, or the kinetic energy, would be 162J.

I hope you find my answer and explanation to be helpful. Happy studying.

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3 years ago