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rodikova [14]
3 years ago
6

Jamal drew the function below. Which explains whether or not his function is linear?

Mathematics
2 answers:
sladkih [1.3K]3 years ago
8 0

Answer:

"line" a straight line graph is always linear once the line isn't straight, it'll no longer be called a line and will never be linear anymore

Agata [3.3K]3 years ago
8 0

Answer:

it is linear because it has a constant rate of change

Step-by-step explanation:

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A particle moves with velocity vector
asambeis [7]

Answer:

\vec{s(t)}=\frac{1}{2}t^2\hat{i}-(2t+1)\hat{j}+(\frac{1}{3}t^3+1)\hat{k}

Step-by-step explanation:

We are given that velocity vector of a particle

\vec{v(t)}=t\hat{i}-2\hat{j}+t^2\hat{k}

When t=0 then the particle is at the point (0,-1,1).

We have to find the position of particle  at time t.

We know that

Velocity =\frac{Displacement }{time}=\frac{ds}{dt}

Therefore,\vec{v}=\frac{\vec{ds}}{dt}

\int{ds}=\int (t\hat{i}-2\hat{j}+t^2\hat{k})dt

Integrate on both sides then we get

\vec{s(t)}=\frac{1}{2}t^2\hat{i}-2t\hat{j}+\frac{1}{3}t^3\hat{k}+C

\int x^n dx=\frac{x^{n+1}}{n+1}+C

Substitute the value of point at time t=0 then we get

C=-\hat{j}+\hat{k}

Substitute the value of C then we get

\vec{s(t)}=\frac{1}{2}t^2\hat{i}-2t\hat{j}+\frac{1}{3}t^3\hat{k}-\hat{j}+\hat{k}

Therefore, the position of particle at time t

\vec{s(t)}=\frac{1}{2}t^2\hat{i}-(2t+1)\hat{j}+(\frac{1}{3}t^3+1)\hat{k}

8 0
3 years ago
The length of a rectangle is 6 units longer that the width , w.
Ulleksa [173]
No B

lets say W is 5

C is saying 4x5+6 right?

so all 4 side are 5 and plus a random six

W is only equal to 2 sides so C and D are gone

A. 2 x 5 + 6 is saying
2 sides are 5 + 6 and then a missing side

but B is 2 x 5 + 12
B could also say 2 x 5 + 2 x 6
2 sides are 5 and then 6 side are 5

Your answer is B
3 0
3 years ago
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