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1 year ago

I’m not sure if I got the answer right and I need help on the table I’m lost

Mathematics

1 answer:

harkovskaia [24]1 year ago

3 0

The given information permits us to calculate the circumference (C) of the wheel as follows

$C=\frac{265}{5}=53in$

Then (as you said correctly!) the same wheel will move

$(53in)\cdot16=848in$

after 16 rolls.

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K is between J and M. L is

I am Lyosha [343]

Answer:

Step-by-step explanation:

for this you need to draw a straight line and plot the points J,K,L,M,N according to the question.

here,

given K lies between J and M, L is between K and M , M is between K and N.

JN= 30, KM=8

JK=KL=LM

MJ= ?

according to the question,

J--------K-------L--------M-------N

KM= 8

KL+LM= 8 (KL+LM=KM)

KL+KL=8 ( given, LM=KL)

2KL= 8

KL=4

so,

JK= KL=4 and LM= KL= 4

now,

MJ= ?

we know

MJ= JK+KM = 4+8= 12

7 0

3 years ago

Match each polynomial with its degree

slega [8]

Answer:

Step-by-step explanation:

degree 1 - 5x+5 - d.

degree 2- 3x^2-2x+4- b.

degree 3 - 1/2x^3+8x^2-3- a.

degree 4 - x^4+(x+4)^2- c

5 0

3 years ago

Find the greatest common factor of 10 and 125.

natulia [17]

Answer:

Step-by-step explanation:

10 = 2 * 5

125 = 5 * 5 * 5

Taking out the duplicates from the 2 lists :- GCF = 5

5 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

3 years ago

Given r= 10 and e=0.57, find the Cartesian coordinates. Check your answer.

Rudik [331]

I think it is d not sure tho

3 0

3 years ago