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timofeeve [1]
3 years ago
8

4.7+x=3.1 Solve for x . x=

Mathematics
2 answers:
inn [45]3 years ago
7 0
X=  -1.5 

4.7 + X = 3.1 
move the x and it becomes 
4.7 = 3.1- X 
then divide 4.7 by 3.1. then you get 1.5 but you cannot have a negative variable so you make the 1.5 negative !
Harman [31]3 years ago
5 0
<span>4.7 + x = 3.1

Subtract 4.7 from both sides to isolate x

x = 3.1 - 4.7 = -1.6</span>
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Please actually help because my anxiety sucks. "Joe is seven years younger than Bill. The sum of their ages is less than 49. At
kherson [118]
Joe is at most 20 and Bill would be at most 27 because it has to be less than 49 combined
5 0
2 years ago
Bowl A contains 6 apples and 3 mangoes this can be written as, Bowl B contain 8 apples and 2 mangoes this can be written as. Wha
AlexFokin [52]

Answer:

2 apples and 1 mango

Step-by-step explanation:

Bowl A can be written as 6a + 3m

Bowl B can be written as 8a + 2m

The difference between the bown is 2 apples and 1 mango

6 0
3 years ago
Karl bought ​ 3 1/2 ​ gallons of milk that cost $2.70 per gallon. What was the total cost of the milk? Enter your answer in the
pav-90 [236]

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9.45

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7 0
3 years ago
Read 2 more answers
Solve this quadratic equation by completing the square.<br><br> x^2+10x=7
bogdanovich [222]
Ok, so the quadratic coefient is 1, so great

take 1/2 of the linear coefient and square it
10/2=5, (5)^2=25
add that to both sides
x^2+10x+25=7+25
factor perfect square trionomial
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3 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
2 years ago
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