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Alex787 [66]
2 years ago
9

At 6:00 A.M. the temperature was 33°F. By noon the temperature had increased by 10°F and by 3:00 P.M. it had increased another 1

2°F. If at 10:00 P.M. the temperature had decreased by 15°F, how much does the temperature need to rise or fall to return to the original temperature of 33°F?
Mathematics
1 answer:
zhuklara [117]2 years ago
7 0
33 + 10 + 12 - 15 - 33 = 7°F
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The 12 boys in the class make up 15% of the class. What is the total number of students in the class?
alina1380 [7]

Answer:

80 students

Step-by-step explanation:

12/80 is 0.15 or 15%

5 0
2 years ago
-3k+6=10+k solve with steps please
Artemon [7]

Answer:

k = -1

Step-by-step explanation:

-3k + 6 = 10 + k

combine similar terms:

-3k - k = 10 - 6

simplify:

-4k = 4

simplify:

k = 4 / -4

k = -1

4 0
3 years ago
Read 2 more answers
Question:
aivan3 [116]

Answer:

part A) The scale factor of the sides (small to large) is 1/2

part B) Te ratio of the areas (small to large) is 1/4

part C) see the explanation

Step-by-step explanation:

Part A) Determine the scale factor of the sides (small to large).

we know that

The dilation is a non rigid transformation that produce similar figures

If two figures are similar, then the ratio of its corresponding sides is proportional

so

Let

z ----> the scale factor

\frac{CB}{C'B'}=\frac{CD}{C'D'}=\frac{BD}{B'D'}

The scale factor is equal to

z=\frac{CB}{C'B'}

substitute

z=\frac{4}{8}

simplify

z=\frac{1}{2}

Part B) What is the ratio of the areas (small to large)?

<em>Area of the small triangle</em>

A=\frac{1}{2}(2)(4)=4\ units^2

<em>Area of the large triangle</em>

A=\frac{1}{2}(4)(8)=16\ units^2

ratio of the areas (small to large)

ratio=\frac{4}{16}=\frac{1}{4}

Part C) Write a generalization about the ratio of the sides and the ratio of the areas of similar figures

In similar figures the ratio of its corresponding sides is proportional and this ratio is called the scale factor

In similar figures the ratio of its areas is equal to the scale factor squared

4 0
3 years ago
HELP DUE IN 10 MINS!
Scilla [17]

Answer:

\checkmark \text{B. The area of the circle is } 729\pi,\\\checkmark \text{D. The arc length of the sector is } 12\pi

Step-by-step explanation:

Area of a circle with radius r: r^2\pi

Circumference of a sector with radius r: 2r\pi

Area of sector with angle \theta:  r^2\pi\cdot \frac{\theta}{360}.

Arc length of sector with angle \theta: 2r\pi\cdot \frac{\theta}{360}

Using these equations, we get the following information:

Area of circle: 27^2\pi=729\pi

Circumference of a circle: 2\cdot 27\cdot \pi=54\pi

Area of sector: 27^2\pi\cdot \frac{80}{360}=162\pi

Arc length of sector: 2\cdot 27\cdot \pi \cdot \frac{80}{360}=12\pi

3 0
2 years ago
How do you solve 5x-2y= 18 and -5x+3y= -22 by solving a system of linear equations by elimination? Please add the steps in which
MaRussiya [10]
Write the 2 equations:

5x-2y=18
-5x+3y=-22

You can see that the coefficients of x are opposites, so you can add them(the 2 equations) to eliminate x.

Adding them, you get:

y=-3

So now that you know y, you can substitute the value of y into one of the original equations. Let's substitute it in the first one:

5x+6=18

Solving this, we get:

5x=12
x=12/5

x=12/5, y=-3
3 0
3 years ago
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