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3 years ago

James got 5 questions wrong out of 40 questions what percent of the questions did he get correct

Mathematics

1 answer:

Arte-miy333 [17]3 years ago

8 0

8% of them right, 40/5

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2 years ago

Someone help please!!!!

LUCKY_DIMON [66]

Answer:

x = -2

TU = 4

UB = 2

Step-by-step explanation:

you can add x^2 with 4x+10 and equate it to 6:

x^2 + 4x + 10 = 6

x^2 + 4x + 4

then u can use the roots formula : x = (-b ± √ (b2 - 4ac) )/2a

so it'll be x = {-4±[√16 - 4(4)]}/2

x= -2

then u can substitute it and find TU and UB

TU= (-2)^2 = 4

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5 0

2 years ago

A ball moves in a straight line has an acceleration of a(t) = 2t + 5. Find the position function of the ball if its initial velo

Vlad1618 [11]

Answer:

$s(t) = frac{t^3}{3} + \frac{5t^2}{2} - 3t + 12$

Step-by-step explanation:

Relation between acceleration, velocity and position:

The velocity function is the integral of the acceleration function.

The position function is the integral of the velocity function.

Acceleration:

As given by the problem, the acceleration function is $a(t) = 2t + 5$

Velocity:

$v(t) = \int a(t) dt = \int (2t+5) dt = \frac{2t^2}{2} + 5t + K = t^2 + 5t + K$

In which K is the constant of integration, which is the initial velocity. So K = -3 and:

$v(t) = t^2 + 5t - 3$

Position:

$s(t) = \int v(t) dt = \int (t^2 + 5t - 3) = \frac{t^3}{3} + \frac{5t^2}{2} - 3t + K$

In which K, the constant of integration, is the initial position. Since it is 12:

$s(t) = frac{t^3}{3} + \frac{5t^2}{2} - 3t + 12$

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3 years ago