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pogonyaev
3 years ago
6

Find the value of c that makes t^2– 22t+ c a perfect square trinomial.

Mathematics
1 answer:
riadik2000 [5.3K]3 years ago
3 0

Answer:

ZkgJHhHJ HHha sorry i am not sure

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| 10x | &gt; -2 ?<br> Solving absolute value equations and inequalities
Alex777 [14]

The answer is x>−1/5 or x<1/5

3 0
3 years ago
I need an Answer Immediately pls!!!!!!!!!
Salsk061 [2.6K]

Answer:

2,033-993.50= 1039.50/74.25= 14

Step-by-step explanation:

not sure the exact equation but p would be 14 and you'd do something similar to that shown above

8 0
3 years ago
A bicycle wheel has a circumference of 62 inches. Find the diameter of the wheel. If necessary, round to the tenths place. A. 39
san4es73 [151]

Answer:

The answer to your question is        diameter  = 19.7 in

Step-by-step explanation:

Data

Circumference = 62 in

diameter = ?

Process

1.- Write the formula of Circumference, remember that circumference = perimeter.

    Perimeter = 2πr

2.- Equal the equation to the value given

                        2πr = 62

-Solve it for r

                            r = 62/2(3.14)

-Result

                           r = 62/6.28

                           r = 9.87 in

3.- Find the diameter

             diameter = 2r

             diameter = 2(9.87)

             diameter  = 19.7 in

4 0
3 years ago
Read 2 more answers
Please, I need it ASAP!!!! I will give brainliest if correct!!!
LenKa [72]

Answer:

recursive: f(0) = 7; f(n) = f(n-1) -8

explicit: f(n) = 7 -8n

Step-by-step explanation:

The sequence is an arithmetic sequence with first term 7 and common difference -8. Since you're numbering the terms starting with n=0, the generic case will be ...

recursive: f(0) = first term; f(n) = f(n-1) + common difference

explicit: f(n) = first term + n·(common difference)

To get the answer above, fill in the first term and common difference values.

4 0
3 years ago
Given sec(theta)= 5... find cot(90-theta)
Mumz [18]
\sec(\theta)=\frac{1}{\cos(\theta)}&#10;\\&#10;\\\cos(\theta)=\frac{1}{\sec(\theta)}=\frac{1}{5}&#10;\\&#10;\\ \sin^2\theta+\cos^2\theta=1&#10;\\&#10;\\\sin\theta= \sqrt{1-\cos^2\theta}= \sqrt{1-( \frac{1}{5})^2 }  = \sqrt{1- \frac{1}{25} } = \sqrt{ \frac{25}{25} -\frac{1}{25} }=\frac { \sqrt{24}}{5} &#10;\\&#10;\\ \cot(90^o-\theta)=\tan{\theta}= \frac{\sin\theta}{\cos\theta} = \frac{\frac { \sqrt{24}}{5} }{ \frac{1}{5} } = \sqrt{24} = \sqrt{4\times6}=2 \sqrt{6}  &#10;\\&#10;
3 0
3 years ago
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