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VashaNatasha [74]
3 years ago
9

(2.07) The equation below shows the relationship between the temperature in degrees Celsius, C, and degrees Fahrenheit, F: C = 5

over 9(F − 32) Which of the following formulas correctly solves for F?
Mathematics
2 answers:
Vlad [161]3 years ago
8 0
C = 5/9(F - 32)....multiply both sides by 9/5
9/5C = F - 32...add 32 to both sides
9/5C + 32 = F
Novay_Z [31]3 years ago
4 0

Answer:

Option 1

Step-by-step explanation:

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Rom4ik [11]
Hello,

Let's suppose f(x)=3x+7
f(a-1)=3(a-1)+7=3a-3+7=3x+4
Answer B with a sign "++++++++++++++++++++++++++++"


5 0
4 years ago
Please help me!! Also if you can show the steps, I’d appreciate it. Thank you very much
natita [175]
-3x=2x+19
-19 -19
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3 0
3 years ago
How would write 42 as an operation???
ArbitrLikvidat [17]
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7 0
3 years ago
Read 2 more answers
X+2y=-1 and 4x-4y=20 solve by substitution
Illusion [34]

To do this problem you would first need to factor out a variable, which in this case I would want to do the first equation because it is isolated. Now the equations would look like this:

x = -2y - 1

4x - 4y = 20

Since we know that x is now equal to -2y - 1 we can plug it in to the x value in the second equation:

4 (-2y -1) - 4y = 20

-8y -4 - 4y

-12y - 4 = 20

-12y = 24

y = -2

Now that we know the y value plug the y value to one equation to find the x, I will be using the first equation

x + 2(-2) = -1

x - 4 = -1

x = 3

Solutions:

y = -2

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8 0
3 years ago
Given F(x)=1/x-4, g(x)=1/6-x. Find f+g(x).
Yakvenalex [24]

\boxed{f(x)+g(x)=\frac{1}{x}-x-\frac{23}{6}}

<h2>Explanation:</h2>

We are given two functions:

f(x)=\frac{1}{4}x-4 \\ \\ g(x)=\frac{1}{6}-x

So, we have to find f(x)+g(x) which is the sum of these two functions:

f(x)+g(x)=\frac{1}{x}-4+\frac{1}{6}-x \\ \\ \\ Let's \ call \ h(x)=f(x)+g(x) \\ \\ Simplifying: \\ \\ h(x)=\left(\frac{1}{x}-x)+(-4+\frac{1}{6}) \\ \\ \\ \\

Finally: \\ \\ \boxed{f(x)+g(x)=\frac{1}{x}-x-\frac{23}{6}}

<h2>Learn more:</h2>

Shifting graphs: brainly.com/question/10010217

#LearnWithBrainly

3 0
3 years ago
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