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Y_Kistochka [10]
3 years ago
7

How far away from home was she ?

Mathematics
1 answer:
Vlad [161]3 years ago
6 0
Approx. 5 miles I would say give or take the 1/2 mile shortcut by taking a straight path back to the house. I don't know if you need an exact distance but I hope this helps.
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Can you help me? <br><br> 4+3(x+5)=5x-7
PSYCHO15rus [73]

Answer:

13

Step-by-step explanation:

4 + 3(x + 5)=5 x - 7

3x + 19=5x - 7

3x + 19 - 19=5x - 7 - 19

3x =5x - 26

3x - 5x=5x - 26 - 5x

-2= -26

-2x/-2 = -26/-2

x= 13

It might be wrong. I may have made a mistake with typing sorry.

Hope this helps a bit.

6 0
2 years ago
Can anyone help me with this. please​
zvonat [6]

Answer:

Step-by-step explanation:

using pythagoras theorem

a^2+b^2=c^2

7^2+KL^2=25^2

49+KL^2=625

KL^2=625-49

KL=\sqrt{576

KL=24

take M as reference angle

using tan rule

tan M=/opposite/adjacent

tan M=KL/ML

tan M=24/7

5 0
3 years ago
HELP ME SOMEONE PLEASE :(
Lemur [1.5K]

Answer: The answer is A.

Step-by-step explanation:

The formula for the area of circles:

A = πr^2

Plug in the radius

A = π8^2

A = π64

A = 201.06192983/ 201 (rounded)

3 0
2 years ago
Read 2 more answers
A washer and a dryer cost 634 dollars combined. The washer cost 66 dollars less than the dryer. What is the cost of the dryer?
Nutka1998 [239]

Steps to Solve:

1.  634 / 2 = 317

2. 317 + 66 = 383

Answer:

A = $383

Notes:

Read the question, this problem is not that hard to solve!

8 0
3 years ago
Read 2 more answers
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
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