All categories

2 years ago

Helpp pls

Mathematics

2 answers:

Ratling [72]2 years ago

6 0

Answer:

Step-by-step explanation:

Substitute x = 5 and x = 9 into f(x) and equate

f(5) = (5 - b)² = 25 - 10b + b²

f(9) = (9 - b)² = 81 - 18b + b²

Then equating the 2 gives

b² - 10b + 25 = b² - 18b + 81 ( subtract b² - 18b from both sides )

8b + 25 = 81 ( subtract 25 from both sides )

8b = 56 ( divide both sides by 8 )

b = 7

Then

f(x) = (x - 7)² and

f(1) = (1 - 7)² = (- 6)² = 36 → B

fredd [130]2 years ago

4 0

I don’t know but try whatever you guys tell you

You might be interested in

PLEASE HELP ME I'M DESPERATE.<br> Delta Math Question!!!

makkiz [27]

Answer:

cos S = $\frac{7}{25}$

Step-by-step explanation:

this is a 7- 24- 25 right angle with RS = 7 , then

cos S = $\frac{adjacent}{hypotenuse}$ = $\frac{RS}{QS}$ = $\frac{7}{25}$

7 0

1 year ago

choose the equation that shows a step in the process of completing the square on the given quadratic. y = x2 8x – 3 y = x2 8x 8

posledela

Y = x^2 + 8x - 3
y = x^2 + 8x + (8/2)^2 - 3 - (8/2)^2
y = x^2 + 8x + 16 - 3 - 16

3 0

3 years ago

Read 2 more answers

Unit 6.6, Lesson 9: Practice Problems

7nadin3 [17]

Answer:

the answer is 798 + 789 mawdgghhji the same time as the work of the Maroons

8 0

1 year ago

student randomly receive 1 of 4 versions(A, B, C, D) of a math test. What is the probability that at least 3 of the 5 student te

alexdok [17]

Answer:

1.2%

Step-by-step explanation:

We are given that the students receive different versions of the math namely A, B, C and D.

So, the probability that a student receives version A = $\frac{1}{4}$ .

Thus, the probability that the student does not receive version A = $1-\frac{1}{4}$ = $\frac{3}{4}$ .

So, the possibilities that at-least 3 out of 5 students receive version A are,

1) 3 receives version A and 2 does not receive version A

2) 4 receives version A and 1 does not receive version A

3) All 5 students receive version A

Then the probability that at-least 3 out of 5 students receive version A is given by,

$\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{3}{4}\times \frac{3}{4}$ + $\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{3}{4}$ + $\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}$