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Thepotemich [5.8K]
3 years ago
5

HELP ME ASAP WITH 8! PLS!! I beg u..

Mathematics
1 answer:
lakkis [162]3 years ago
7 0
7. x=3 is the midpoint between the roots. The other root is x = 2*3 -(-5) = 11.

8a) f(x) = (x +3)^2 -49. The vertex is (-3, -49). The roots are -10, 4.

8b) y = (x+4)^2 -1. The vertex is (-4, -1). The roots are -5, -3.

8c) f(x) = 2(x +3)^2 -34. The vertex is (-3, -34). The roots are -3±√17.
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Choose what the expressions below best represent within the context of the word problem. The length of a rectangle is 12 inches
nexus9112 [7]
The answer is 4. Since 12+ 12 = 24, I added 4 + 4, which made 42.
So the length of the width of the rectangle is 4.
Hope this helps!
3 0
3 years ago
Read 2 more answers
If a farmer wants to plough a farm field on time, he must plough 120 hectares a day. For technical reasons he ploughed only 85 h
natta225 [31]

Answer:

the farmer planned to have the work done in 6 days, and the area of the farm field is 120 times 6 = 720 hectares.

6 0
3 years ago
Which of the following is correct?
Brums [2.3K]
C) it is a closed figure made up of straight lines.
6 0
2 years ago
Polygon ABCD is a parallelogram, and . The length of is 10 units, and the length of is 5 units. The perimeter of the parallelogr
Aliun [14]
Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.
Below are the choices that can be found elsewhere:

A. rectangle; P = 26 linear units 

B. square; P = 42 units2 

C. parallelogram; P = 42 linear units 

D. trapezoid; P = 26 linear units


The answer would be A. 

3 0
3 years ago
Read 2 more answers
The average student-loan debt is reported to be $25,235. A student believes that the student-loan debt is higher in her area. Sh
yulyashka [42]

Answer:

Step-by-step explanation:

We would set up the hypothesis test. This is a test of a single population mean since we are dealing with mean

For the null hypothesis,

µ = 25235

For the alternative hypothesis,

µ > 25235

This is a right tailed test.

Since the population standard deviation is not given, the distribution is a student's t.

Since n = 100,

Degrees of freedom, df = n - 1 = 100 - 1 = 99

t = (x - µ)/(s/√n)

Where

x = sample mean = 27524

µ = population mean = 25235

s = samples standard deviation = 6000

t = (27524 - 25235)/(6000/√100) = 3.815

We would determine the p value using the t test calculator. It becomes

p = 0.000119

Since alpha, 0.05 > than the p value, 0.000119, then we would reject the null hypothesis. There is sufficient evidence to support the claim that student-loan debt is higher than $25,235 in her area.

6 0
3 years ago
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