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3 years ago

Plz help me plz show work thx

Mathematics

1 answer:

jonny [76]3 years ago

5 0

This is prime factorization
we need to know what number might go into 63
Most people who know their times tables remember that 9 * 7 = 63
7 is a prime number so it would go into the circle, and the 9 will go into the box

9 isn't prime, but 3 * 3 = 9. 3 is a prime number so the 3's will go into the circles below the 9.

If you need a final answer you can write 63 = 3 * 3 * 7

You can get the same factors if you use 3 *21 instead of 9*7 at the beginning. I'll let you figure out that one.

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The formula v= u+ atdescribes connections between u: starting velocity a: acceleration t: time v: final velocity a)In a particul

Bess [88]

Answer:

49 = 0 +9.8t
t = 5

Step-by-step explanation:

It usually works to follow directions.

a) Substitute the numbers into the formula:

v = u + at

49 = 0 +9.8t

b) Solve the equation:

49/9.8 = t = 5 . . . . . divide by the coefficient of t

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3 years ago

What's the slope and y-intercept?

Shtirlitz [24]

Answer:

Slope: 2; Y-Intercept: (0,3)

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2 years ago

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during a softball game kay hit a fly ball the function f(x) = -16t^2 + 64t + 4 describes the height of the softball in feet. mak

STatiana [176]

<h2>Hello!</h2>

The graph is attached.

To graph a parabola we need to know the following:

- If the parabola is open upwards or downwards

- They axis intercepts (if they exist)

- The vertex position (point)

We are given the function:

$f(t)=-16t^{2}+64t+4$

Where,

$a=-16\\b=64\\c=4$

For this case, the coefficient of the quadratic term (a) is negative, it means that the parabola opens downwards.

Finding the axis interception points:

Making the function equal to 0, we can find the x-axis (t) intercepts, but since the equation is a function of the time, we will only consider the positive values, so:

$f(t)=-16t^{2}+64t+4\\0=-16t^{2}+64t+4\\-16t^{2}+64t+4=0$

Using the quadratic equation:

$\frac{-b+-\sqrt{b^{2}-4ac } }{2a}=\frac{-64+-\sqrt{64^{2}-4*-16*4} }{2*-16}\\\\\frac{-64+-\sqrt{64^{2}-4*-16*4} }{2*-16}=\frac{-64+-\sqrt{4096+256} }{-32}\\\\\frac{-64+-\sqrt{4096+256} }{-32}=\frac{-64+-(65.96) }{-32}\\\\t1=\frac{-64+(65.96) }{-32}=-0.06\\\\t2=\frac{-64-(65.96) }{-32}=4.0615$

So, at t=4.0615 the height of the softball will be 0.

Since we will work only with positive values of "x", since we are working with a function of time:

Let's start from "t" equals to 0 to "t" equals to 4.0615.

So, evaluating we have:

$f(0)=-16(0)^{2}+64(0)+4=4\\\\f(1)=-16(1)^{2}+64(1)+4=52\\\\f(2)=-16(2)^{2}+64(2)+4=68\\\\f(3)=-16(3)^{2}+64(3)+4=52\\\\f(4.061)=-16(4.0615)^{2}+64(4.0615)+4=0.0034=0$