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3 years ago

What is the greatest common factor of 60w, 36w2, and 24w4?

Mathematics

2 answers:

blsea [12.9K]3 years ago

5 0

Answer: 12

Step-by-step explanation:

faust18 [17]3 years ago

4 0

The answer is 12 because 60 goes into 12 36 goes into 12 and 24goes into12 and the rest are even so goes into 12

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Amy fills an ice cream cone to the top edge. The diameter of the ice cream cone is 2 inches and its height is 6 inches. Don fill

sattari [20]

Answer:

1.28 in³

Step-by-step explanation:

Volume of Amy's cone

Cone = 1/3πr²h
Cone = 1/3 x 3.14 x 1² x 6
Cone = 2 x 3.14 = 6.28 in³

Volume of Don's bowl

It's given to be 5 in³

Difference

Amy's cone - Don's bowl
6.28 - 5
1.28 in³

6 0

1 year ago

Read 2 more answers

How many times larger is 3.19 x 10^8 than 8.23 x 10^5? SHOW YOUR WORK!!!

svet-max [94.6K]

In order to find the number times something is larger than something, you have to use division where you divide the larger number by the smaller number.

So:

$\frac{3.19*10^8}{8.23*10^5} =387.6063183$

I'm not sure how much you want to round to, but the answer is 387.6063183 times.

You can make sure by multiplying the smaller number by 387.6063183 and you should get the larger number:

$8.25*10^5*387.6063183=3.19*10^8$

8 0

3 years ago

Solve the equation below for y 5 — y =12 6

Oxana [17]

Answer:14.4,72/6,14 2/5

4 0

3 years ago

NO LINKS OR ANSWERING QUESTIONS YOU DON'T KNOW!!! THIS IS NOT A TEST OR AN ASSESSMENT!! Please help me with these math questions

Art [367]

Answer:

See below

Step-by-step explanation:

3. What are two ways that a vector can be represented?

Considering a vector $\vec{v}$ in some vector space $\mathbb R^n$ we have

$\vec{v} = \langle a,b\rangle$

This is the component form. I don't like that way. It is probably used in high school, but

$\vec{v} = \begin{pmatrix} a\\ b\\ \end{pmatrix}$

is preferable because the inner product on $\mathbb R^n$ is defined to be

$\langle a,b\rangle := \sum_{i = 1}^n a_i b_i$

You can also write it using linear form such as $\vec{v} = 2i+2j$

For this question, I think you meant

vectors

$\vec{u_1} = (-8, 12)$

$\vec{u_2} = (13, 15)$

Once

$\cos(\theta)=\dfrac{\vec{u_1} \cdot\vec{u_2}}{||\vec{u_1}||||\vec{u_2}||}$

Considering that the dot product is

$\vec{u_1}\cdot \vec{u_2} = (-8)\cdot 13 + 12\cdot 15 = -104+180= 76$

and the norm of $\vec{u_1}$ is $||\vec{u_1}|| = \sqrt{(-8)^2 + 12^2} = \sqrt{64 + 144}= \sqrt{208}$

and the norm of $\vec{u_2}$ is $||\vec{u_2}|| = \sqrt{13^2 + 15^2} = \sqrt{169 + 225}= \sqrt{394}$

Thus,

$\cos(\theta)=\dfrac{76}{\sqrt{208} \sqrt{394}} = \dfrac{19}{\sqrt{13}\sqrt{394}}=\dfrac{19}{\sqrt{5122}}$

$\therefore \theta = \arccos \left(\dfrac{19}{\sqrt{5122}} \right)$

3 0

2 years ago

Plz help this is all the points i have i am still on this question roght now plz what is the answer

madreJ [45]

Answer:

I think the answer is 109.752 ♡

Step-by-step explanation:

8 0

3 years ago