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maks197457 [2]
3 years ago
5

What is the answer!?

Mathematics
2 answers:
prisoha [69]3 years ago
4 0
I think something wrong 
the answer must be 18
umka21 [38]3 years ago
4 0
The answer to this question is A.
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Lyla i am not cheating they are practice questions stop deleting my question Guys i need help
Tanzania [10]

Answer:

C

Step-by-step explanation:

If you simplify the equation you get \sqrt{\frac{x+3}{1-x} }.  To solve a square root function, you need to set up\sqrt{\frac{x+3}{1-x} }\geq 0. You can take the square root of both sides to get \frac{x+3}{1-x} \geq 0. Then, we identify the intervals that will satisfy the equation, which is -3 ≤ x < 1.

Sorry if the explanation is kind of confusing, I wasn't sure how to describe it.

7 0
3 years ago
What tests are used to determine the radius of convergence of a power series? A. Divergence TestB. Root Test C. Comparison Test
REY [17]

Answer:

The radius of convergence is half of the length of the interval of convergence. If the radius of convergence is R then the interval of convergence will include the open interval: (a − R, a + R). To find the radius of convergence, R, you use the Ratio Test.

7 0
3 years ago
A set of equations is given below: Equation E: m = 4n + 1 Equation F: m = 6n + 8 Which statement describes a step that can be us
Lelu [443]
8-6=2 n=0 your very welcome!
6 0
3 years ago
What is 3356983+10000004565
Leto [7]
The answer is: 1003361548
4 0
3 years ago
A(x) = 2 (3x - 2)(x - 5)
PSYCHO15rus [73]

Answer:

Quadratic polynomial can be factored using the transformation ax  

2

+bx+c=a(x−x  

1

​  

)(x−x  

2

​  

), where x  

1

​  

 and x  

2

​  

 are the solutions of the quadratic equation ax  

2

+bx+c=0.

−x  

2

−3x+5=0

All equations of the form ax  

2

+bx+c=0 can be solved using the quadratic formula:  

2a

−b±  

b  

2

−4ac

​  

 

​  

. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

x=  

2(−1)

−(−3)±  

(−3)  

2

−4(−1)×5

​  

 

​  

 

Square −3.

x=  

2(−1)

−(−3)±  

9−4(−1)×5

​  

 

​  

 

Multiply −4 times −1.

x=  

2(−1)

−(−3)±  

9+4×5

​  

 

​  

 

Multiply 4 times 5.

x=  

2(−1)

−(−3)±  

9+20

​  

 

​  

 

Add 9 to 20.

x=  

2(−1)

−(−3)±  

29

​  

 

​  

 

The opposite of −3 is 3.

x=  

2(−1)

3±  

29

​  

 

​  

 

Multiply 2 times −1.

x=  

−2

3±  

29

​  

 

​  

 

Now solve the equation x=  

−2

3±  

29

​  

 

​  

 when ± is plus. Add 3 to  

29

​  

.

x=  

−2

29

​  

+3

​  

 

Divide 3+  

29

​  

 by −2.

x=  

2

−  

29

​  

−3

​  

 

Now solve the equation x=  

−2

3±  

29

​  

 

​  

 when ± is minus. Subtract  

29

​  

 from 3.

x=  

−2

3−  

29

​  

 

​  

 

Divide 3−  

29

​  

 by −2.

x=  

2

29

​  

−3

​  

 

Factor the original expression using ax  

2

+bx+c=a(x−x  

1

​  

)(x−x  

2

​  

). Substitute  

2

−3−  

29

​  

 

​  

 for x  

1

​  

 and  

2

−3+  

29

​  

 

​  

 for x  

2

​  

.

−x  

2

−3x+5=−(x−  

2

−  

29

​  

−3

​  

)(x−  

2

29

​  

−3

​  

)

EVALUATE

5−3x−x  

2Quadratic polynomial can be factored using the transformation ax  

2

+bx+c=a(x−x  

1

​  

)(x−x  

2

​  

), where x  

1

​  

 and x  

2

​  

 are the solutions of the quadratic equation ax  

2

+bx+c=0.

−x  

2

−3x+5=0

All equations of the form ax  

2

+bx+c=0 can be solved using the quadratic formula:  

2a

−b±  

b  

2

−4ac

​  

 

​  

. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

x=  

2(−1)

−(−3)±  

(−3)  

2

−4(−1)×5

​  

 

​  

 

Square −3.

x=  

2(−1)

−(−3)±  

9−4(−1)×5

​  

 

​  

 

Multiply −4 times −1.

x=  

2(−1)

−(−3)±  

9+4×5

​  

 

​  

 

Multiply 4 times 5.

x=  

2(−1)

−(−3)±  

9+20

​  

 

​  

 

Add 9 to 20.

x=  

2(−1)

−(−3)±  

29

​  

 

​  

 

The opposite of −3 is 3.

x=  

2(−1)

3±  

29

​  

 

​  

 

Multiply 2 times −1.

x=  

−2

3±  

29

​  

 

​  

 

Now solve the equation x=  

−2

3±  

29

​  

 

​  

 when ± is plus. Add 3 to  

29

​  

.

x=  

−2

29

​  

+3

​  

 

Divide 3+  

29

​  

 by −2.

x=  

2

−  

29

​  

−3

​  

 

Now solve the equation x=  

−2

3±  

29

​  

 

​  

 when ± is minus. Subtract  

29

​  

 from 3.

x=  

−2

3−  

29

​  

 

​  

 

Divide 3−  

29

​  

 by −2.

x=  

2

29

​  

−3

​  

 

Factor the original expression using ax  

2

+bx+c=a(x−x  

1

​  

)(x−x  

2

​  

). Substitute  

2

−3−  

29

​  

 

​  

 for x  

1

​  

 and  

2

−3+  

29

​  

 

​  

 for x  

2

​  

.

−x  

2

−3x+5=−(x−  

2

−  

29

​  

−3

​  

)(x−  

2

29

​  

−3

​  

)

EVALUATE

5−3x−x  

2

Step-by-step explanation:

7 0
3 years ago
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