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Charra [1.4K]
3 years ago
12

What is the equation of an exponential function with base=2 that has a vertical stretch by a factor of 3, a shift to the right 5

and a shift up 6 ?
Mathematics
1 answer:
il63 [147K]3 years ago
7 0

Answer:

y=3(2)^{x-5}+6

Step-by-step explanation:

Let the equation of the exponential function with base 2 is,

y=a(2)^{x+c}+d

Here, a = vertical stretch

c = Horizontal shift

d = Vertical shift

If this function has a vertical stretch by a factor of 3,

Equation will be,

y=3(2)^{x+c}+d

If this function has a shift of 5 to the right,

Equation of the function will be,

y=3(2)^{x-5}+d

If this function has a shift of 6 units upwards,

y=3(2)^{x-5}+6

You might be interested in
2<br> .<br> 2<br> 12n = 42<br> NEED HELP ASAP⬆️
Anni [7]

Answer:

  6  

 —————

 n + 8

Step-by-step explanation:

Step by Step Solution:

More Icon

STEP

1

:

Equation at the end of step 1

 ((12•(n3))-(24•(n2)))       (12n-42)      

 —————————————————————•———————————————————

  (((4•(n2))-22n)+28)  ((6•(n3))+(24•3n2))  

STEP  

2

:

Equation at the end of step

2

:

 ((12•(n3))-(24•(n2)))      (12n-42)      

 —————————————————————•——————————————————

  (((4•(n2))-22n)+28)  ((2•3n3)+(24•3n2))  

STEP

3

:

            12n - 42  

Simplify   ——————————

           6n3 + 48n2

STEP

4

:

Pulling out like terms

4.1     Pull out like factors :

  12n - 42  =   6 • (2n - 7)  

STEP

5

:

Pulling out like terms

5.1     Pull out like factors :

  6n3 + 48n2  =   6n2 • (n + 8)  

Equation at the end of step

5

:

 ((12•(n3))-(24•(n2)))  (2n-7)  

 —————————————————————•————————

  (((4•(n2))-22n)+28)  n2•(n+8)

STEP  

6

:

Equation at the end of step

6

:

 ((12•(n3))-(24•(n2)))  (2n-7)  

 —————————————————————•————————

    ((22n2-22n)+28)    n2•(n+8)

STEP  

7

:

Equation at the end of step

7

:

 ((12•(n3))-(23•3n2))  (2n-7)  

 ————————————————————•————————

     (4n2-22n+28)     n2•(n+8)

STEP  

8

:

Equation at the end of step

8

:

 ((22•3n3) - (23•3n2))      (2n - 7)  

 ————————————————————— • ————————————

   (4n2 - 22n + 28)      n2 • (n + 8)

STEP

9

:

             12n3 - 24n2  

Simplify   ——————————————

           4n2 - 22n + 28

STEP

10

:

Pulling out like terms

10.1     Pull out like factors :

  12n3 - 24n2  =   12n2 • (n - 2)  

STEP

11

:

Pulling out like terms

11.1     Pull out like factors :

  4n2 - 22n + 28  =   2 • (2n2 - 11n + 14)  

Trying to factor by splitting the middle term

11.2     Factoring  2n2 - 11n + 14  

The first term is,  2n2  its coefficient is  2 .

The middle term is,  -11n  its coefficient is  -11 .

The last term, "the constant", is  +14  

Step-1 : Multiply the coefficient of the first term by the constant   2 • 14 = 28  

Step-2 : Find two factors of  28  whose sum equals the coefficient of the middle term, which is   -11 .

     -28    +    -1    =    -29  

     -14    +    -2    =    -16  

     -7    +    -4    =    -11    That's it

Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above,  -7  and  -4  

                    2n2 - 7n - 4n - 14

Step-4 : Add up the first 2 terms, pulling out like factors :

                   n • (2n-7)

             Add up the last 2 terms, pulling out common factors :

                   2 • (2n-7)

Step-5 : Add up the four terms of step 4 :

                   (n-2)  •  (2n-7)

            Which is the desired factorization

Canceling Out :

11.3    Cancel out  (n-2)  which appears on both sides of the fraction line.

Equation at the end of step

11

:

   6n2      (2n - 7)  

 —————— • ————————————

 2n - 7   n2 • (n + 8)

STEP

12

:

Canceling Out

12.1    Cancel out  (2n-7)  which appears on both sides of the fraction line.

Canceling Out :

12.2    Canceling out n2 as it appears on both sides of the fraction line

Final result :

   6  

 —————

 n + 8

8 0
4 years ago
No links or anything
Soloha48 [4]

Answer:

1. 38, 80, 89 and 4. 14, 15, 29

I don't know if there is supposed to be only one, but both of those do not form right triangles.

Step-by-step explanation:

Evaluate all of them and see if they meet the requirements of the Pythagorean Theorem, a² + b² = c².

1. 38, 80, 89

a² + b² = c²

38² + 80² = 89²

1444 + 6400 = 7921

7844 ≠ 7921.

This is an answer because it doesn't satisfy the Pythagorean Theorem.

2. 16, 63, 65

a² + b² = c²

16² + 63² = 65²

256 + 3969 = 4225

4225 = 4225

This isn't the answer because it satisfies the Pythagorean Theorem.

3. 36, 77, 85

a² + b² = c²

36² + 77² = 85²

1296 + 5929 = 7225

7225 = 7225

This isn't the answer because it satisfies the Pythagorean Theorem.

4. 14, 15, 29

a² + b² = c²

14² + 15² = 29²

196 + 225 = 841

421 ≠ 841

This is an answer because it does not satisfy the Pythagorean Theorem.

6 0
3 years ago
#2 <br> #3 <br> Need help for my discrete math class
miss Akunina [59]

For (2), start with the base case. When n = 2, we have

(n + 1)! = (2 + 1)! = 3! = 6

2ⁿ = 2² = 4

6 > 4, so the case of n = 2 is true.

Now assume the inequality holds for n = k, so that

(k + 1)! > 2ᵏ

Under this hypothesis, we want to show the inequality holds for n = k + 1. By definition of factorial, we have

((k + 1) + 1)! = (k + 2)! = (k + 2) (k + 1)!

Then by our hypothesis,

(k + 2) (k + 1)! > (k + 2) 2ᵏ = k•2ᵏ + 2ᵏ⁺¹

and k•2ᵏ ≥ 2•2² = 8, so

k•2ᵏ + 2ᵏ⁺¹ ≥ 8 + 2ᵏ⁺¹ > 2ᵏ⁺¹

which proves the claim.

Unfortunately, I can't help you with (3). Sorry!

3 0
3 years ago
PLEASE HELP:
Semmy [17]

Answer:

A' = { 2,4,6,8,10,12,14,16,17,18,19}

A' U B =  {2,4,6,8,10,12,14,16,17,18,19,7,11,13}

Step-by-step explanation:

A = {1,3,5,7,9,11,13,15}

B = {7,11,13,17,19}

3 0
3 years ago
Please help I'm so lost!!!<br><br> Factor. <br><br> 9x2−49y2
pantera1 [17]

Answer:

(3x+7y)(3x−7y)

Step-by-step explanation:

Since both terms are perfect squares, factor using the difference of squares formula, a2−b2=(a+b)(a−b) where a=3x and b=7y

Umm if you need more details just ask.

Hope this helps .-.

4 0
4 years ago
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