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xxMikexx [17]
2 years ago
8

F(x)=3x-7 find the value of c so that f(c) =20 is true

Mathematics
2 answers:
Andreas93 [3]2 years ago
7 0

<u>Answer</u>:

c = 9

<u>Explanation</u>:

given: f(x)=3x-7

<em>if f(c) =20</em>

To solve this replace x with c and then y is 20.

<u>Solve</u>:

3c - 7 = 20

3c = 27

c = 27 ÷ 3

c = 9

Harman [31]2 years ago
3 0

Answer:  C = 9

Step-by-step explanation:

hope this helps

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I need help with #39-42, please explain.
shusha [124]

For question 39 & 40, we need to use the below equation to complete the sentence

l=m\sqrt{n}

Question 39:

When ' n ' increase, the \sqrt{n} will also increase and that multiplied with constant ' m ', the l will also increase.

Solution for question 39:

As n increases and m stays constant , l <u>increases</u>

-------

Question 40:

Solving the equation for m, we get

l = m\sqrt{n} \\ \\ m=\frac{l}{\sqrt{n}}

When ' l ' increases, the numerator increase, the denominator stays constant because 'n' stays constant, for this condition, the fraction increases.

Solution for question 40:

As l increase and n stays constant, m <em><u>increases</u></em>

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For question 41 & 42, we need to use the below equation to complete the sentence

r=s^2/t^2

Question 41:

When s is triped, the equation will be...

r=(3s)^2/t^2=\frac{3^{2}s^{2}}{t^2}   =9s^2/t^2

Solution for question 41:

If s is tripled and t stays constant, r is multiplied by <em><u>9</u></em>

--------

Question 42:

When t is doubled, the equation will be...

r=s^2/(2t)^2=\frac{s^2}{2^2 \cdot t^2}=\frac{s^2}{4t^2}\\   \\ r=0.25s^2/t^2 \; \; (or) \; \; \frac{1}{4} \cdot \frac{s^2}{t^2}

Solution for 42:

If t doubled and s stays constant, r is multiplied by <em><u>1/4 or 0.25</u></em>


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Step-by-step explanation:

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3 years ago
A and b are positive integers and a-b=2. Evaluate the following:<br> 9^1/2b/3^a
OlgaM077 [116]

Answer:

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Step-by-step explanation:

\huge \frac{ {9}^{ \frac{1}{2}b } }{ {3}^{a} }  \\  \\  =  \huge \frac{ {( {3}^{2} )}^{ \frac{1}{2}b } }{ {3}^{a} } \\  \\  =  \huge \frac{ { {3}}^{2 \times  \frac{1}{2}b } }{ {3}^{a} }  \\  \\  \huge =  \frac{ {3}^{b} }{ {3}^{a} }  \\  \\  \huge =  \frac{1}{ {3}^{a - b} }  \\  \\   \huge=  \frac{1}{ {3}^{2} } \\  ( \because \: a - b = 2)\\  \\  \huge =  \frac{1}{ 9 }

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3 years ago
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