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Iteru [2.4K]
3 years ago
8

Nathan is trying out to be the kicker on his high school's football team. He notices that for every hour he practices he can kic

k the ball 15 feet further. On his first try he kicked the ball 30 feet, on the second try he kicked the ball 15 yards. What could be a possible distance for his third kick?
A.15 feet
B.30 feet
C.20 yards
D.30 yards
Mathematics
1 answer:
Pani-rosa [81]3 years ago
5 0
The answer is <span>C.20 yards

1 yd = 3 ft
1 ft = 1/3 yd

</span><span>He notices that for every hour he practices he can kick the ball 15 feet further.
15 ft further = 15/3 yd further = 5 yd further

The first try: 30 ft = 30/3 yd = 10 yd
The second try: 30 ft + 15 ft = 45 ft = 45/3 yd = 15 yd
The third try: 30 ft + 15 ft + 15 ft = 60 ft = 60/3 yd = 20 yd</span>
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Four points are drawn on the coordinate plane and connected with straight lines to form a rectangle. Three of the vertices of th
trasher [3.6K]

Answer:

(-4, 1)

Step-by-step explanation:

If you draw it, that will make it easier.

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3 years ago
Solve for s in the proportion.<br> s=<br> Submit<br> 40/30=8/s
vitfil [10]

Answer:

s =6

Step-by-step explanation:

40            8

--------- = ---------

30            s

Using cross products

40*s = 30*8

40s = 240

Divide each side by 40

40s/40 = 240/40

s = 6

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3 years ago
A graphing calculator is recommended. A crystal growth furnace is used in research to determine how best to manufacture crystals
Sophie [7]

Answer:

a) w_1 = \frac{-2.156 -\sqrt{2.156^2 -4(0.1)(-183)}}{2*0.1}=-54.896

w_2 = \frac{-2.156 +\sqrt{2.156^2 -4(0.1)(-183)}}{2*0.1}=33.336

And since the power can't be negative then the solution would be w = 33.34 watts.

b) 202= 0.1w^2 +2.156 w +20

0.1w^2 +2.156 w-182=

w_2 = \frac{-2.156 +\sqrt{2.156^2 -4(0.1)(-182)}}{2*0.1}=33.222

204= 0.1w^2 +2.156 w +20

0.1w^2 +2.156 w-184=

w_2 = \frac{-2.156 +\sqrt{2.156^2 -4(0.1)(-184)}}{2*0.1}=33.449

So then the range of voltage would be between 33.22 W and 33.45 W.

c)For this case \epsilon = \pm 1 since that's the tolerance 1C

d) \delta_1 =|33.222-33.336|=0.116

\delta_2 =|33.449-33.336|=0.113

So then we select the smalles value on this case \delta =0.113

e) For this case if we assume a tolerance of \epsilon=\pm 1C for the temperature and a tolerance for the power input \delta =0.113 we see that:

lim_{x \to a} f(x) =L

Where a = 33.34 W, f(x) represent the temperature, x represent the input power and L = 203C

Step-by-step explanation:

For this case we have the following function

T(w)= 0.1 w^2 +2.156 w +20

Where T represent the temperature in Celsius and w the power input in watts.

Part a

For this case we need to find the value of w that makes the temperature 203C, so we can set the following equation:

203= 0.1w^2 +2.156 w +20

And we can rewrite the expression like this:

0.1w^2 +2.156 w-183=

And we can solve this using the quadratic formula given by:

w =\frac{-b \pm \sqrt{b^2 -4ac}}{2a}

Where a =0.1, b =2.156 and c=-183. If we replace we got:

w_1 = \frac{-2.156 -\sqrt{2.156^2 -4(0.1)(-183)}}{2*0.1}=-54.896

w_2 = \frac{-2.156 +\sqrt{2.156^2 -4(0.1)(-183)}}{2*0.1}=33.336

And since the power can't be negative then the solution would be w = 33.34 watts.

Part b

For this case we can find the values of w for the temperatures 203-1= 202C and 203+1 = 204 C. And we got this:

202= 0.1w^2 +2.156 w +20

0.1w^2 +2.156 w-182=

w_2 = \frac{-2.156 +\sqrt{2.156^2 -4(0.1)(-182)}}{2*0.1}=33.222

204= 0.1w^2 +2.156 w +20

0.1w^2 +2.156 w-184=

w_2 = \frac{-2.156 +\sqrt{2.156^2 -4(0.1)(-184)}}{2*0.1}=33.449

So then the range of voltage would be between 33.22 W and 33.45 W.

Part c

For this case \epsilon = \pm 1 since that's the tolerance 1C

Part d

For this case we can do this:

\delta_1 =|33.222-33.336|=0.116

\delta_2 =|33.449-33.336|=0.113

So then we select the smallest value on this case \delta =0.113

Part e

For this case if we assume a tolerance of \epsilon=\pm 1C for the temperature and a tolerance for the power input \delta =0.113 we see that:

lim_{x \to a} f(x) =L

Where a = 33.34 W, f(x) represent the temperature, x represent the input power and L = 203C

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4 years ago
4x + 3y = 6<br> -4x + 2y = 14<br> Solve the system of equations.
pshichka [43]

Answer:

x = -1.5    y = 4

Step-by-step explanation:

4x + 3y = 6

-4x + 2y = 14                  Add both equations together

       5y = 20                  Divide both sides by 5

        y = 4

Plug this into one of the original equations

4x + 3(4) = 6                Multiply

4x + 12 = 6

     -12  - 12                  Subtract 12 from both sides

4x = -6                         Divide both sides by 4

x = -1.5

If this answer is correct, please make me Brainliest!

5 0
3 years ago
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Wewaii [24]

Step-by-step explanation:

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