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ycow [4]
3 years ago
9

Help plz and thank have a nice day

Mathematics
1 answer:
dusya [7]3 years ago
6 0

This is for the first page

Answer: answer to question 17 is r=32/7, answer to question 18 is m=-12, answer to question 19 is k=-9, answer to question 20 is p= No solution, answer to question 21 is x= infinite solutions, answer to question 22 is x= -4.

Step-by-step explanation:

This picture is for the second page

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Which of the following is an example of an exponential function?
Orlov [11]

Answer: Choice D p(\text{x}) = 500(1.02)^{\text{x}}

Reason:

Exponential functions have the variable in the exponent.

Choice A is a polynomial, which means we can rule it out. Choice B is a log function, so we can rule it out also.

Choice C is a rational function because it is a ratio of two polynomials. This rules out choice C.

Choice D is the only function with the variable in the exponent. That is assuming you meant to say p(\text{x}) = 500(1.02)^{\text{x}} instead of p(\text{x}) = 500(1.02){\text{x}}

4 0
2 years ago
The two dot plots show the number of miles ran by 14 students at the beginning and at the end of the school year compare each me
Naddika [18.5K]

The two dot plots are missing, so i have attached it.

Answer:

The mean at the beginning of the school year was 9.5 miles and the mean at the end of the school year was 10.2 miles

Step-by-step explanation:

From the attached image, we are told to compare the means for each plot to the nearest tenth.

Mean = Σx/n

Now, from the image, total number of miles run by the 14 students at the beginning of the school year is;

(1 × 7) + (2 × 8) + (4 × 9) + (4 × 10) + (2 × 11) + (1 × 12) = 133

Mean of miles run at the beginning of the school year = 133/14 = 9.5 miles

Again, from the table, total miles run at the end of the school year = (2 × 8) + (2 × 9) + (4 × 10) + (3 × 11) + (3 × 12) = 143

Mean of miles run at the end of the school year = 143/14 = 10.2 miles

Thus;

The mean at the beginning of the school year was 9.5 miles and the mean at the end of the school year was 10.2 miles

5 0
3 years ago
A man flies a kite with a 100 foot string the angle of elevation of the string is 52 how high off the ground is the kite
Slav-nsk [51]
The kite is approximately 78.80 feet off of the ground.


Explanation: If we assume that the string his held taut without any swag, we can use a right triangle to solve for the height of the kite. sin
θ
=
o
p
p
h
y
p

sin
52
o
=
h
100
f
t

100
sin
52
o
=
h

100
(
0.7880
)
=
h

78.80
f
t
=
h
4 0
3 years ago
Read 2 more answers
7. Find the slope. Make sure to reduce.
MakcuM [25]

Answer:

The slope is zero

Step-by-step explanation:

.........

4 0
3 years ago
Read 2 more answers
Find The number to the angle QPR
butalik [34]

Step-by-step explanation:

you know that QPS adds up to 47

so the sum of RPS and QPR shoukd = 47

(3x - 38) + (7x - 95) = 47

you then solve for x then you can use that to find QPR

8 0
3 years ago
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