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

1. Evaluate:<br> 4(5) +3(7-2)<br> a. 25<br> C. 18<br> d. 40<br> e. -40

Brrunno [24]

Answer:

Step-by-step explanation:

Simplify the following:

4×5 + 3 (7 - 2)

7 - 2 = 5:

4×5 + 3×5

4×5 = 20:

20 + 3×5

3×5 = 15:

20 + 15

| 2 | 0

+ | 1 | 5

| 3 | 5:

Answer: 35

6 0

3 years ago

Algebra: Explain in words how to write an equation that is part one: parallel and then also part two: perpendicular to the equat

sweet-ann [11.9K]

A line that is parallel has the same slop where a line that is perpendicular has a slop that is negative and reciprocal.

so for the parallel one you don't need to worry about the slop because it will be 2/3x. But yous the point slope equation form

y-y1=m(x-x1)

y+5=2/3(x+2)

y+5=2/3x+ 4/3

y=2/3x-11/3

-2/3x+y=-11/3

multiple by -1 so A inst negative

2/3-y=11/3

For a line that is perpendicular you just need to flip the original 2/3x slope and make it negative.

y+5=-3/2(x+2)

y+5=-3/2x-3

y=-3/2x-8

3/2x+y=-8

8 0

3 years ago

Owen ate three-eighths of a pizza for lunch. For dinner, he ate two-thirds of what was left from lunch. How much of the whole

Step2247 [10]

Answer: 5/12 of the whole pizza

So we start we the whole pizza = 1
Then he ate 3/8:
1 - 3/8
And then 2/3 of what was left:
2/3(1 - 3/8)
2/3 - 1/4
8/12 - 3/12
5/12

8 0

3 years ago

The set of life spans of an appliance is normally distributed with a mean mu = 48 months and a standard deviation sigma = 8 mont

navik [9.2K]

Using the normal distribution, it is found that the life span of an appliance that has a z-score of –3 is of 24 months.

<h3>Normal Probability Distribution</h3>

The z-score of a measure X of a normally distributed variable with mean $\mu$ and standard deviation $\sigma$ is given by: