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pychu [463]
3 years ago
8

1. When a cannonball is fired, the equation of its pathway can be modeled by h = -1612 + 128t.

Mathematics
1 answer:
sashaice [31]3 years ago
4 0

Answer:

The maximum height of the ball is 256m

Step-by-step explanation:

Given the equation of a pathway modelled as pathway can be modeled by h = -16t² + 128t

At maximum height, the velocity of the ball is zero.

velocity = dh/dt

velocity = -32t + 128

Since v = 0 at maximum height

0 = -32t+128

32t = 128

t = 128/32

t = 4seconds

The maximum height can be gotten by substituting t = 4 into the modelled equation.

h = -16t² + 128t

h = -16(4)²+128(4)

h = -16(16)+512

h = -256+512

h = 256m

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kolbaska11 [484]

Answer:

q = \frac{8}{3}

Step-by-step explanation:

The point of intersection (3, 1 ) is the solution to both equations

Using

y = qx - 7 and substituting x = 3, y = 1

1 = 3q - 7 ( add 7 to both sides )

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6 0
3 years ago
Im horrible at math and need help z and both w's
puteri [66]
The sum of all the angles in a triangle is 180°.
There is a right angle -> a 90° angle on the hypotenuse of the largest triangle (the largest being the one that contains both smaller triangles)

To solve this problem, you don't really need to solve y.
Next to the 48° angle, you have two of the same angles, both x variables.
The sum of these three angles has to be 180°48 + 2x = 180 \\ 2x = 180 - 48 \\ 2x = 132 \\ x = 66

Now that we have x, we can calculate the value of z.
We add the value of x and the 90° angle, we later subtract it from 180°

Calculating the value of z:
66 + 90 = 156 \\ 180 - 156 = 24 \\ z = 24

If you really want to know the value of y, you just add up the 90° angle and the 48°, then subtract that from 180°
90 + 48 = 138 \\ 180 - 138 = 42 \\ y = 42


6 0
3 years ago
Compare using &lt; or &gt; or =<br><br><br> -15____-21
Salsk061 [2.6K]
<h3><u><em>Answer:</em></u></h3>

-15 > -21

<h3><u><em>step by step:</em></u></h3>

We know that 15 < 21 then  -15 > -21

5 0
3 years ago
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2 years ago
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Answer: 40

Step-by-step explanation:

4 0
3 years ago
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