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ozzi
3 years ago
7

Mike and his 3 friends were eating pizza. They had 2 pizzas that they were going to

Mathematics
1 answer:
Phoenix [80]3 years ago
8 0

Each person would get 1 and left overs r in the trash lel

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How many solutions can be found for the equation 3x − 2 = 3x + 5?
aksik [14]

Answer:

Zero

Step-by-step explanation:

there is no solution the sides are uneven. if you solve for it 3x and 3x cancels out each other leaving you with -2=5

7 0
3 years ago
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Pls help i dont know this one
telo118 [61]

Answer:

The answer is A :)

Step-by-step explanation:

6 0
2 years ago
HELPPP PLASSSS BRAINLIEST
Svetach [21]

Answer:

5

Step-by-step explanation:

If the parallelogram is being rotated the side length will stay the same as it is in the current state.

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2 years ago
Find the point on the line and the lines slope y+1=-3/2(x+7)
klemol [59]

Answer:

A point on the line would be (-3, -7) and the slope if -3/2

Step-by-step explanation:

We can find a point on the line by choosing any value for x and putting it in to the equation. Then we can find the y value. For the purpose of this problem, we will use x = -3.

y + 1 = -3/2(x + 7)

y + 1 = -3/2(-3 + 7)

y + 1 = -3/2(4)

y + 1 = -6

y = -7

And we can find the slope simply by looking for the number outside of the parenthesis on the right side of the equal sign (-3/2)

8 0
3 years ago
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
2 years ago
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