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Xelga [282]
4 years ago
10

What is this answer Plzzz

Mathematics
1 answer:
sammy [17]4 years ago
4 0

Answer:5/8 inch

Step-by-step explanation:

You have to find the common denominator for the two most frequent lengths which would be 8. Then you have to multiply the numerator by the same number you multiplied the denominator by and you would get 2/8 plus 3/8 and you get 5/8

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10) Alayah's puppy weighed 12 lbs. when they brought it home from the Humane Society.
TEA [102]
It gained 4 lbs:) 6*4 = 24
24+12 = 36
3 0
3 years ago
Questions eleven all the way to twenty
Margarita [4]
11=27
12=27
13=4
14=16
15=24
16=? sorry I couldn't solve them all, hope that helped :D
6 0
3 years ago
−(−49) = −49 true or false?
vova2212 [387]

I hope this helps you

false

cause two negative numbers multiple must be positive

(-1).(-49)

+49

5 0
3 years ago
The sea ice area around the North Pole fluctuates between about 6 million square kilometers in September to 14 million square ki
Aleksandr [31]

Answer:

there are approximately 5.035 months when there is less than 9 million square meters of sea ice around the North Pole in a year.

Step-by-step explanation:

Given the data in the question;

Let S(t) represent the amount sea ice around the North Pole in millions of square meters at a given time t,

t is the number of months since January.

Now, we use a cosine curve to model this scenario

Vertical shift will be;

D = ( 6 + 14 ) / 2 = 20 / 2

D = 10

Next is the Amplitude;

|A| = ( 6 - 14 ) / 2

|A| = 4

Now, the horizontal stretch factor will be;

B = 2π / 12

B = π/6

Hence;

S(t) = 4cos( π/6 × t ) + 10 ----------- let this be equation 1

Now we find when there will be less than 9 million square meters of sea ice;

S(t) = 9

so we have

9 = 4cos( π/6 × (t-2) ) + 10

9 - 10 = 4cos( π/6 × (t-2) )

-1 = 4cos( π/6 × (t-2) )

-1/4 = cos( π/6 × (t-2) )

so we have;

cos⁻¹( -1/4 ) = π/6 × (t₁-2)  -------- let this be equation 2

2π - cos⁻¹( -1/4 ) = π/6 × (t₂-2)  -------- let this be equation 3

so we solve equation 2 and 3

we have'

t₁ - t₂ = 6/π × ( 2π - cos⁻¹( -1/4 ) - cos⁻¹( -1/4 ) )

t₁ - t₂ = 6/π × ( 2π - 2cos⁻¹( -1/4 )  

t₁ - t₂ = 6/π × ( π - cos⁻¹( -1/4 )  

t₁ - t₂ = 6/π × ( π - 104.4775 )

t₁ - t₂ = 6/π × ( π - 104.4775 )  

t₁ - t₂ = 5.035

therefore, there are approximately 5.035 months when there is less than 9 million square meters of sea ice around the North Pole in a year.

6 0
3 years ago
Write an equation and use it to solve this question.
HACTEHA [7]
C. five sixth meters
5 0
3 years ago
Read 2 more answers
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