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3 years ago

Please help me with this

Mathematics

1 answer:

Eduardwww [97]3 years ago

5 0

Answer:

$r^{6}$

Step-by-step explanation:

Using the rule of exponents

$\frac{a^{m} }{a^{n} }$ = $a^{(m-n)}$ , then

$\frac{r^{9} }{r^{3} }$ = $r^{(9-3)}$ = $r^{6}$

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svp [43]

Answer:

the 2 images are the solutions to the questions

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4 years ago

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topjm [15]

Answer:

The answer is option 3.

Step-by-step explanation:

The steps are :

$3 \sqrt{2} (5 \sqrt{6} - 7 \sqrt{3})$

$= (3 \sqrt{2} )(5 \sqrt{6} ) - (3 \sqrt{2} )(7 \sqrt{3} )$

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3 years ago

Help me please !!!!!!

Mrac [35]

Answer:

z = -44

Step-by-step explanation:

3/4 (x - 8) = 4

3(x - 8) = 16

3x - 24 = 16

3x = 40

x = 40/3

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4 = -3x - z

z = -3x - 4

z = -3(40/3) - 4

z = -40 - 4

z = -44

8 0

3 years ago

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A ball is thrown upward from a height of 15 feet with an initial velocity of 64 feet per second. Use indefinite integrals to wri

FrozenT [24]

Answer:

The equation of the position at time t will be:

$y(t)=-\frac{1}{2}9.81t^{2}+64t+15$

Step-by-step explanation:

We can start by saying that the acceleration here is g = -9.81 m/s². The minus sign is because the gravity acceleration is a vector downward (negative value), and the ball is going upward (positive value).

And we know that acceleration can be written as a second derivative:

$g=\frac{d^{2}y}{dt^{2}}=y''$ (1)

Now, we can take the integral in each side:

$\int gdt=\int y''dt$

$\int y''dt=\int -9.81dt$ (2)

Solving the integral in each side we have:

$y'(t)=-9.81t+C$ (3)

Where y'(t) is the velocity at t time (v = dy/dt = y' ) and c is a constant value.

Now, the initial conditions are:

initial height y(0) = 15 feet

initial velocity y'(0) = v(0) = 64 feet/s

Using this condition we can find C. Let's evaluate equation (3) at t = 0.

$y'(0)=-9.81(0)+C$

$C=64\: feet/s$

So we have:

$y'(t)=-9.81t+64$ (4)

Now, we need to take the integral of equation (4) to get the position function:

$\int y'(t)dt=\int (-9.81t+64)dt$

Solving this new integral we have:

$y(t)=-\frac{1}{2}9.81t^{2}+64t+D$ (5)

Using the same method above, we can find D evaluating (5) at t = 0, we have:

$y(0)=-\frac{1}{2}9.81(0)^{2}+64(0)+D$

$D=y(0)$

$D=15\: feet$

Finally, the equation of the position at time t will be:

$y(t)=-\frac{1}{2}9.81t^{2}+64t+15$

I hope it helps you!

3 0

3 years ago

Hola! alguien me podría ayudar con la solución de este problema?

AysviL [449]

Answer:

Step-by-step explanation:

6 0

3 years ago