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Natalija [7]
2 years ago
6

A snowplow removes three fourths of the snow each time that it goes down a road. If a road starts with 24

Mathematics
1 answer:
olga nikolaevna [1]2 years ago
6 0
The answer I think is 47 times I’m not certain though
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Write an equation in point-slope form of a line that goes through the points (1,4) and (3,-10). Do not use any spaces in your an
olasank [31]

Answer:

y-4=-7(x-1) OR y-(-10)=-7(x-3)

Step-by-step explanation: The point-slope form is y_-y_{1}=m_(x_-x_{1})

In slope-intercept form, the equation would be y = -7x + 11

You could find the slope by using the slope equation, \frac{y_{2} -y_{1} }{x_{2} -x_{1} }.

Using that, you would get -7. Thus by inserting one of the points and the slope in the point-slope equation, you will get your answer.

6 0
2 years ago
What is the slope of a line that goes<br> through the points (2,5) and (8, 6)?
kirza4 [7]

Answer:

1/6

Step-by-step explanation:

6-5/8-2=1/6

5 0
2 years ago
Two parallel lines, cut by transversal, are shown. One of the acute angles formed measures 80. Find the measures of all the othe
musickatia [10]
The one across it is 100 and on both sides 

4 0
3 years ago
If 3 is added to twice a number, the result is 17. Find the number
Crank

Answer:

7

Step-by-step explanation:

The text of this expression is:

"If 3 is added to twice a number, the result is 17. Find the number"

Let's call x the number we want to find.

Twice the number is equal to

2x

So, if we add 3 to 2x, we get

3+2x

And we know that this is equal to 17:

3+2x=17

Now we can solve the equation to find x: first, we subtract 3 from both sides

3+2x-3=17-3\\2x=14

Now we divide both sides by 2:

\frac{2x}{2}=\frac{14}{2}\\x=7

7 0
2 years ago
1. How much heat is absorbed by a 50g iron skillet when its temperature rises from 10oC to 124oC? Joules
hoa [83]
<span>1. How much heat is absorbed by a 50g iron skillet when its temperature rises from 10oC to 124oC? Joules

Formula: Heat = mass * specific heat * ΔT

Data:
mass = 50 g = 0.050 kg
specific heat of iron = 450 J/ kg °C
ΔT = 124°C - 10°C ¿ 114 °C

=> heat = 0.050kg * 450 J / kg°C * 114°C ≈ 2.6 J

2. If a refrigerator is a heat pump that follows the first law of thermodynamics, how much heat was removed from food inside of the refrigerator if it released 492J of energy to the room? Joules

The firs law of thermodynamics is conservation of energy => energy removed from inside of the refrigerator = energy released to the room

=> Answer = 492 J

3. How much heat is needed to raise the temperature of 45g of water by 63oC? Joules

Formula: heat = mass * specific heat * ΔT

specific heat of water = 4186 J / Kg °C

heat = 0.045 kg * 4186 J/kg °C * 63°C = 11,867.31 J
</span>
7 0
3 years ago
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