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Margarita [4]
2 years ago
12

How much more snow fell in February 1889 than November 1888?

Mathematics
2 answers:
Sindrei [870]2 years ago
7 0

Answer:

4) 5.4 inches

Step-by-step explanation:

5.4 inches of snow fell in February 1889 than November 1888.

According to the graph:

Feb '89: 8.5 inches

Nov '88: 3 inches

8.5 - 3 = 5.5 or 5.4.

defon2 years ago
3 0

Answer:

Step-by-step explanation:

i believe it is 4.

assuming feburary it snowed about 8.5 inches and in november it is at about the 3 in line you would subtract to get about 5.5 inches

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During a sale, every item in a store is 80% of its regular price. If the regular price of a T-shirt is $10, what is its sale pri
kipiarov [429]

Answer:

Step-by-step explanation:

If the regular price of a T-shirt is 10 USD and the store sells off its wares at 80% of its regular price, then the price at which the store would still the aforementioned T-shirt would be 80% of its 10 USD price.

80% of $10 =

0.8 * 10 = $8

This means that the price at which the T-shirt will go for, due to the ongoing sales will be $8.

Also, being asked to find the 10% and 20% of $10 is just as simple

10% of $10 =

0.1 * 10 = $1

20% of $10 =

0.2 * 10 = $2

Therefore, the 10% and 20% of $10 is $1 and $2 respectively.

3 0
3 years ago
Y = -4x - 5 <br>y = 3x - 2 <br><br>Is (3, 7) a solution of the system?<br><br>Will give brainliest!​
Evgesh-ka [11]

Answer:

x=-3/7 and y=_23/7

Step-by-step explanation:

y=-4x-5 and y=3x-2

using comparison method

-4x-5=3x-2

-4x-3x=-2+5

-7x=3

x=3/-7

substituting the value of x

y=3x-2

=3×(-3/7)-2

=-9/7-2

=-9-14/7

=-23/7

6 0
3 years ago
8^(2x-1)=4^(3x+4)<br> PLEASE HELP ME
icang [17]

Answer: x=1.6

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
Please someone help me
hram777 [196]
The answer is 57 inches

5 0
3 years ago
Does anybody know the answer to these questions?
san4es73 [151]

Answer:

  1. 625,000 J

  2. 100 J

  4. 5 kg

  5. √5 ≈ 2.236 m/s

Step-by-step explanation:

You should be aware that the SI derived units of Joules are equivalent to kg·m²/s².

To reduce confusion between <em>m</em> for mass and m for meters, we'll use an <em>italic m</em> for mass.

In each case, the "find" variable is what's left after we put the numbers into the formula. It is what the question is asking for. The "given" values are the ones in the problem statement and are the values we put into the formula. The formula is the same in every case.

__

1. KE = (1/2)<em>m</em>v² = (1/2)(2000 kg)(25 m/s)² = 625,000 kg·m²/s² = 625,000 J

__

2. KE = (1/2)<em>m</em>v² = (1/2)(0.5 kg)(20 m/s)² = 100 kg·m²/s² = 100 J

__

4. KE = (1/2)<em>m</em>v²

  250 J = (1/2)<em>m</em>(10 m/s)² = 50 m²/s²

  (250 kg·m²/s²)/(50 m²/s²) = <em>m</em> = 5 kg

__

5. KE = (1/2)<em>m</em>v²

  2000 kg·m²/s² = (1/2)(800 kg)v²

  (2000 kg·m²/s²)/(400 kg) = v² = 5 m²/s²

  v = √5 m/s ≈ 2.236 m/s

7 0
3 years ago
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