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Ganezh [65]
3 years ago
5

The average high temperature was 86 in August then 74 in September. What the decrease?

Mathematics
2 answers:
Rzqust [24]3 years ago
7 0

Answer:

12

Step-by-step explanation:

86-74=12

You have to subtract the highest by the lowest to find the decrease.

AnnyKZ [126]3 years ago
6 0
86-74= 12

just subtract and you’ll find your answer
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Step-by-step explanation:

3 0
3 years ago
Solve equation 3 - 8c = 35
NeTakaya

We would subtact 3 from both sides allowing the equation to simplify to -8c=32

We would then divide both sides by -8 to equal -4

The final answer C=-4

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3 years ago
What is the range of the function on the graph?
Marta_Voda [28]

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Read 2 more answers
Geometry! Please help!!
krek1111 [17]
Problem One
Find AM
AM = 71.5 - 22 = 49.5

Step Two
State the Givens.
AM = 49.5
MN = 71.5
MB = x
MP = 97.5

Step Three
Set up the Proportion

AM : NM :: x : PM
49.5 : 71.5 :: x : 97.5

Substitute and solve
49.5 / 71.5 = x / 97.5    Cross Multiply
49.5 * 97.5 = 71.5 * x   Combine the numbers on the left.
4860.375 = 71.5 * x     Divide by 71.5
4860.375 / 71.5 = x      
x = 67.98

Problem Two
Remark
This is just a straight application of the Pythagorean Theorem
a^2 + b^2 = c^2
a = 10
b = 24
c = ??

10^2 + 24^2 = c^2
100 + 576 = c^2
676 = c^2
sqrt(c^2) = sqrt(676)
c = 26  <<<< answer
6 0
3 years ago
Jana ran the first 3 and a half miles of a 5 mile race in 1 3rd of an hour. what was her average rate, in mph, for the first par
Lisa [10]

Answer:

always try to draw a diagram first

Step-by-step explanation:

Jana ran The first 3.5 miles in 1/3 of an hour. remember the problem tells us that we want our answer in miles per hour or mph. so we have to set up our problem so that our units match in the end. The diagram above shows 3.5 mi over 1/3 hour. so we must divide 3.5 miles by 1/3 hour

note:

3 \frac{1}{2} mi =  \frac{7mi}{2}

therefore

\frac{7mi}{2}  \:  \div  \:  \frac{1hr}{3}  =

\frac{7mi}{2}  \times  \frac{3}{1hr}  =  \frac{21mi}{2hr}  = 10 \frac{1}{2}  \frac{mi}{hr}

7 0
3 years ago
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