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Anna35 [415]
2 years ago
13

What transformation is shown below? (Look carefully at the vertices.)

Mathematics
1 answer:
ser-zykov [4K]2 years ago
6 0
The answer is reflection
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A projectile is launched from ground level with a initial velocity of Vo feet per second. Neglecting air resistance, its height
antiseptic1488 [7]
The height at t seconds after launch is
s(t) = - 16t² + V₀t
where V₀ = initial launch velocity.

Part a:
When s = 192 ft, and V₀ = 112 ft/s, then
-16t² + 112t = 192
16t² - 112t + 192 = 0
t² - 7t + 12 = 0
(t - 3)(t - 4) = 0
t = 3 s, or t = 4 s

The projectile reaches a height of  192 ft at 3 s on the way up, and at 4 s on the way down.

Part b:
When the projectile reaches the ground, s = 0.
Therefore
-16t² + 112t = 0
-16t(t - 7) = 0
t = 0 or t = 7 s

When t=0, the projectile is launched.
When t = 7 s, the projectile returns to the ground.

Answer: 7 s

4 0
3 years ago
14. Of the 400 students in the 8th grade, four-
In-s [12.5K]

Answer:

8 or 9

Step-by-step explanation:

So, you first need to take 4/5 of 400 (320).  Then, you can count by 75s for a bit

75students; 2chaperones

150; 4

225; 6

300; 8

now, you have 20 students left over...since one chaperone would watch about 37 kids, you can either divide the remaining kids up among the other chaperones, or you can add one more chaperone to watch the 20 kids.  i don't know what the teacher wants, so either 8 or 9 chaperones

7 0
3 years ago
Read 2 more answers
Please help me answer the assignment in the picture.
Arturiano [62]

Mean: 52.2 because its the average of all the numbers

Range: 2   because (60-58)

Median: 53 because its in the middle of all the numbers

Mode : None

   

6 0
3 years ago
Solve for m and n<br> I been stuck on this question for an hour
Crazy boy [7]

Answer:

m = 10 , n = 5\sqrt{2}

Step-by-step explanation:

using the cosine and tangent ratio in the right triangle and the exact values

cos45° = \frac{1}{\sqrt{2} } and tan45° = 1 , then

cos45° = \frac{adjacent}{hypotenuse} = \frac{5\sqrt{2} }{m} = \frac{1}{\sqrt{2} } ( cross- multiply )

m = 5\sqrt{2} × \sqrt{2} = 5 × 2 = 10

-----------------------------------------

tan45° = \frac{opposite}{adjacent} = \frac{n}{5\sqrt{2} } = 1 , then

n = 5\sqrt{2}

3 0
2 years ago
Name the parts of the expression then, write it in expanded form. a. 14b b.30jk
Makovka662 [10]

Answer:

a. 14 is a coefficient, and b is a variable.

This can be expanded to 14*b

b. 30 is a coefficient, and j and k are variables.

This can be expanded to 30*j*k

3 0
3 years ago
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