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Grace [21]
3 years ago
9

K=5r-11s solve for r​

Mathematics
2 answers:
Savatey [412]3 years ago
7 0

Answer:

      11s + k

r = ------------

          5

Step-by-step explanation:

k=5r-11s needs to be manipulated so that the 5r term is isolated / alone.

Adding 11s to both sides, we get:

5r = 11s + k

and then dividing all three terms by 5 yields:

      11s + k

r = ------------

          5

Illusion [34]3 years ago
6 0

Answer:

r=\frac{k+11s}{5}

Step-by-step explanation:

k=5r-11s

Add 11s on both sides.

k+11s=5r-11s+11s

k+11s=5r

Divide both sides by 5.

\frac{k+11s}{5} =\frac{5r}{5}

\frac{k+11s}{5} =r

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3 years ago
Select all the equations that are true, and you can only pick 3...
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Answer:

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5 0
2 years ago
Match each function with the corresponding function formula when h(x)=5-3x and g(x)=-3+5
Grace [21]

Answer:

k(x) = (3g + 5h)(x) ⇒ (1)

k(x) = (5h - 3g)(x) ⇒ (3)

k(x) = (h - g)(x) ⇒ (2)

k(x) = (g + h)(x) ⇒ (4)

k(x) = (5g + 3h)(x) ⇒ (5)

k(x) = (3h - 5g)(x) ⇒ (6)

Step-by-step explanation:

* To solve this problem we will substitute h(x) and g(x) in k(x) in the

  right column to find the corresponding function formula in the

  left column

∵ h(x) = 5 - 3x

∵ g(x) = -3^x + 5

- Lets start with the right column

# k(x) = (3g + 5h)(x)

∵ g(x) = -3^x + 5

∵ 3g(x) = 3[-3^x + 5] = [3 × -3^x + 3 × 5]

- Lets simplify 3 × -3^x

 take the negative out -(3 × 3^x), and use the rule a^n × a^m = a^(n+m)

∴ -3(3 × 3^x) = -(3^x+1)

∴ 3g(x) = -3^x+1 + 15

∵ h(x) = 5 - 3x

∵ 5h(x) = 5[5 - 3x] = [5 × 5 - 5 × 3x] = 25 - 15x

- Now substitute 3g(x) and 5h(x) in k(x)

∵ k(x) = (3g + 5h)(x)

∴ k(x) = -3^x+1 + 15 + 25 - 15x ⇒ simplify

∴ k(x) = 40 - 3^x+1 - 15x

∴ k(x) = 40 - 3^x+1 - 15x ⇒ k(x) = (3g + 5h)(x)

* k(x) = (3g + 5h)(x) ⇒ (1)

# k(x) = (5h - 3g)(x)

∵ 5h(x) = 25 - 15x

∵ 3g(x) = -3^x+1 + 15

∵ k(x) = (5h - 3g)(x)

∴ k(x) = 25 - 15x - (-3^x+1 + 15) = 25 -15x + 3^x+1 - 15 ⇒ simplify

∴ k(x) = 10 + 3^x+1 - 15x

∴ k(x) = 10 + 3^x+1 - 15x ⇒ k(x) = (5h - 3g)(x)

* k(x) = (5h - 3g)(x) ⇒ (3)

# k(x) = (h - g)(x)

∵ h(x) = 5 - 3x

∵ g(x) = -3^x + 5

∵ k(x) = (h - g)(x)

∴ k(x) = 5 - 3x - (-3^x + 5) = 5 - 3x + 3^x - 5 ⇒ simplify

∴ k(x) = 3^x - 3x

∴ k(x)= 3^x - 3x ⇒ k(x) = (h - g)(x)

* k(x) = (h - g)(x) ⇒ (2)

# k(x) = (g + h)(x)

∵ h(x) = 5 - 3x

∵ g(x) = -3^x + 5

∵ k(x) = (g + h)(x)

∴ k(x) = -3^x + 5 + 5 - 3x ⇒ simplify

∴ k(x) = 10 - 3^x - 3x

∴ k(x)= 10 - 3^x - 3x ⇒ k(x) = (g + h)(x)

* k(x) = (g + h)(x) ⇒ (4)

# k(x) = (5g + 3h)(x)

∵ g(x) = -3^x + 5

∵ 5g(x) = 5[-3^x + 5] = [5 × -3^x + 5 × 5] = 5(-3^x) + 25

∴ 5g(x) = -5(3^x) + 25

∵ h(x) = 5 - 3x

∵ 3h(x) = 3[5 - 3x] = [3 × 5 - 3 × 3x] = 15 - 9x

- Now substitute 5g(x) and 3h(x) in k(x)

∵ k(x) = (5g + 3h)(x)

∴ k(x) = -5(3^x) + 25 + 15 - 9x ⇒ simplify

∴ k(x) = 40 - 5(3^x) - 9x

∴ k(x) = 40 - 5(3^x) - 9x ⇒ k(x) = (5g + 3h)(x)

* k(x) = (5g + 3h)(x) ⇒ (5)

# k(x) = (3h - 5g)(x)

∵ 3h(x) = 15 - 9x

∵ 5g(x) = -5(3^x) + 25

∵ k(x) = (3h - 5g)(x)

∴ k(x) = 15 - 9x - [-5(3^x) + 25] = 15 - 9x + 5(3^x) - 25 ⇒ simplify

∴ k(x) = 5(3^x) - 9x - 10

∴ k(x) = 5(3^x) - 9x - 10 ⇒ k(x) = (3h - 5g)(x)

* k(x) = (3h - 5g)(x) ⇒ (6)

4 0
3 years ago
4-2x=11 how do I solve
Bingel [31]

Hey there!☺

Answer:\boxed{x=\frac{-7}{2}}

Explanation:

4-2x=11

In our first step, we will simplify both sides of the equation:

4+-2x=11\\-2x+4=11

Now subtract 4 from both sides:

-2x+4-4=11-4\\-2x=7

In our last step, we will divide both sides by -2:

\frac{-2x}{2}=\frac{7}{-2}\\  x=\frac{-7}{2}

x=-7/2 is your answer.

Hope this helps!☺

3 0
3 years ago
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