There is a picture attached try to show work

Mathematics

2 answers:

kozerog [31]3 years ago

8 0

$\frac{5}{9}$ (F - 32)

Plug in 113 for F.
$\frac{5}{9}$ (113 - 32)
Solve inside the parentheses first.

$\frac{5}{9}$ (81) = 45

The temp is 45°C when it is 113°F.

Harlamova29_29 [7]3 years ago

5 0

Replace F in the equation with 113.

5/9(113-32)

5/9(81)

5 * 81 = 405
405 / 9 = 45 degrees Celcius

You might be interested in

Consider the function represented by the equation 6q = 3s - 9. Write the equation in function notation, where q is the independe

serg [7]

Answer:

s = 2q + 3

Step-by-step explanation:

A linear function has the form:

● y = mx + b

● y is the output of the function

● x is the variabke that we input

● b is the y-intetcept.

Focus on y and x.

Notice that y depends of the value of x. The value of y changes by changing x. So the value of x controls the output y.

y is dependent but x is not.

■■■■■■■■■■■■■■■■■■■■■■■■■■

● 6q = 3s - 9

We want q to be the independent variable wich means that q will be the input. Therefore s should be the output.

The strategy we are going to follow is separating s in one side alone.

● 6q = 3s - 9

Add 9 to both sides

● 6q + 9 = 3s -9 + 9

● 6q + 9 = 3s

Divide both sides by 3

● (6q + 9)/3 = (3s)/3

● (6q)/3 + 9/3 = s

● s = 2q + 3

So the answer is s = 2q + 3

6 0

3 years ago

I need help finishing a question. I used the explanation function on the program, but it didn't specify what to do to get the fi

leva [86]

I would compute sqrt(1 + 160pi^2) first to get approximately 39.75093337

Add this to 1 and we have 40.75093337

Then divide over 2pi to get a final approximate result of 6.48571248

So x = 6.48571248 is one approximate solution

In short, I computed $\frac{1+\sqrt{1+160\pi^2}}{2\pi}$ only focusing on the plus for now.

----------------------

If you were to compute $\frac{1-\sqrt{1+160\pi^2}}{2\pi}$ you should get roughly -6.167402596 as your other solution. Each solution can then be plugged into the original equation to check if you get 0 or not. You likely won't land exactly on 0 but you'll get close enough.