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ohaa [14]
3 years ago
9

I need help with these two questions

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

Answer: 47

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

Work Shown:

f(x) = x^2 - 7x + 3
f(x) = (x)^2 - 7(x) + 3
f(-4) = (-4)^2 - 7(-4) + 3 ... replace each x with -4
f(-4) = 16 - 7(-4) + 3
f(-4) = 16 + 28 + 3
f(-4) = 44 + 3
f(-4) = 47

============================================================

Problem 5

Answer: See the attached image for the table

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

Work Shown:

Plug in n = 27 and we get...
C = 26 + 10*n
C = 26 + 10*27
C = 26 + 270
C = 296
The input n = 27 leads to the output C = 296. This means that 27 people will have the cost be $296

Do the same for n = 39
C = 26 + 10*n
C = 26 + 10*39
C = 26 + 390
C = 416
The input n = 39 leads to the output C = 416. This means that 39 people will have the cost be $416

and also n = 43 as well
C = 26 + 10*n
C = 26 + 10*43
C = 26 + 430
C = 456
The input n = 43 leads to the output C = 456. This means that 43 people will have the cost be $456

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Solve using the quadratic formula<br><br> x² - 12x + 36=0
motikmotik

Answer:

6

Step-by-step explanation:

(-12)^2-4*1*36=0

144-144=0

x= -(-12)/2*1=

12/2=

6

x=6

4 0
3 years ago
The stem-and-leaf plot shows the ages of people at a family reunion.
klio [65]

Answer:

1) 11

2) 4

Step-by-step explanation:

1) It is 11 because in the first row we ignore the 0 since it is referring to the first digit, so the first row has 7. In the second row, we ignore the one so it would be 4 because the 9, or would be 19, is above 18.

2) It is 4 because there is one person over 65. In the second row we ignore the 7 and there are 3 people in the 7 row.

5 0
2 years ago
równanie zmiany prędkość autokaru poruszającego się po prostym odcinku szosy i rozpoczynającego hamowanie od szybkości 20m/s ma
Rasek [7]

Answer:

s = 22.5 m

Step-by-step explanation:

the equation for the speed change of a coach moving along a straight section of the road and starting braking at a speed of 20 m / s has the form v (t) = 25-5t. Using integral calculus, determine the coach's braking distance.

v (t) = 25 - 5 t

at t = 0 , v = 20 m/s

Let the distance is s.

s =\int v(t) dt\\\\s =\int (25 - 5t)dt\\\\s= 25 t - 2.5 t^2 \\

Let at t = t, the v = 20

So,

20 = 25 - 5 t

t = 1 s

So, s = 25 x 1 - 2.5 x 1 = 22.5 m

3 0
2 years ago
[someone help] The speed of the object 7 seconds after it is dropped is ___m/sec
nadezda [96]

9514 1404 393

Answer:

  68.6 m/s

Step-by-step explanation:

The speed at a given time is the derivative of the height at that time.

  dh/dt = -9.8t

At t=7, the speed is ...

  -9.8(7) = -68.6 . . . . . meters per second (downward)

The speed is 68.6 meters per second.

_____

<em>Additional comment</em>

Height is considered to be positive in the up direction, so velocity and acceleration are also positive in the up direction. The derivative of height will be the velocity. Its negative sign in this case indicates the object is moving in the downward direction. The speed is the magnitude (absolute value) of the velocity.

3 0
3 years ago
An engineer would like to design a parking garage in the most cost-effective manner. He reads that the average height of pickup
aleksley [76]

The answer is: Yes, all conditions for inference are met.

4 0
2 years ago
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