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Archy [21]
3 years ago
12

3/4 mile each day for 8 days how far did I walk

Mathematics
2 answers:
aleksley [76]3 years ago
8 0
You walked 6 miles in 8 days
Bingel [31]3 years ago
4 0

Answer:

6 miles

Step-by-step explanation:

Since you walked 3/4 miles in 1 day then in 8 days you walked...

3/4*8=3*2=6 miles

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a jogger ran 4 miles, decreased her speed by 1 mile per hour, and then ran another 5 miles. If her total jogging time was 2 1/20
LekaFEV [45]
Let the speed of the joker initially be x mi/hr
after reduction, the new speed will be: (x-1) mi/hr
but
time=(distance)/(speed)
hence :
Total time taken will be:
4/x+5/(x+1)=41/20
solving for x we get:
x=21.585 mi/hr
thus the initial speed was 21.585 mi/hr after reduction the new speed was 20.585 mi/hr
5 0
3 years ago
What is the solution to the equation 13.7y = 6.2y + 30?
il63 [147K]

Here, we are required to determine the solution to the equation: 13.7y = 6.2y + 30?

The solution to the equation is y = 7.5

The solution is thus,

  • 13.7y = 6.2y + 30
  • 13.7y - 6.2y = 30
  • 7.5y = 30

Therefore, y = 30/7.5 = 4

Therefore, y = 4 is the solution to the equation.

Read more:

brainly.com/question/12184348

8 0
3 years ago
What is the value of x?
hammer [34]

Answer:

16.

How to find... ↓

The small triangle is 1/3 size of the big triangle.

If this is the case, find the LCF here, 4, and multiply each angle's value by the least common factor, 4.

3 x 4 (bottom) = 12

5 x 4 (right side) = 20

4 x 4 (left side <em>the missing side) </em>= 16

Therefore,

The missing side value, <em>x</em>, is 16.

6 0
2 years ago
Read 2 more answers
Solve y'' + 10y' + 25y = 0, y(0) = -2, y'(0) = 11 y(t) = Preview
svetlana [45]

Answer:  The required solution is

y=(-2+t)e^{-5t}.

Step-by-step explanation:   We are given to solve the following differential equation :

y^{\prime\prime}+10y^\prime+25y=0,~~~~~~~y(0)=-2,~~y^\prime(0)=11~~~~~~~~~~~~~~~~~~~~~~~~(i)

Let us consider that

y=e^{mt} be an auxiliary solution of equation (i).

Then, we have

y^prime=me^{mt},~~~~~y^{\prime\prime}=m^2e^{mt}.

Substituting these values in equation (i), we get

m^2e^{mt}+10me^{mt}+25e^{mt}=0\\\\\Rightarrow (m^2+10y+25)e^{mt}=0\\\\\Rightarrow m^2+10m+25=0~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{since }e^{mt}\neq0]\\\\\Rightarrow m^2+2\times m\times5+5^2=0\\\\\Rightarrow (m+5)^2=0\\\\\Rightarrow m=-5,-5.

So, the general solution of the given equation is

y(t)=(A+Bt)e^{-5t}.

Differentiating with respect to t, we get

y^\prime(t)=-5e^{-5t}(A+Bt)+Be^{-5t}.

According to the given conditions, we have

y(0)=-2\\\\\Rightarrow A=-2

and

y^\prime(0)=11\\\\\Rightarrow -5(A+B\times0)+B=11\\\\\Rightarrow -5A+B=11\\\\\Rightarrow (-5)\times(-2)+B=11\\\\\Rightarrow 10+B=11\\\\\Rightarrow B=11-10\\\\\Rightarrow B=1.

Thus, the required solution is

y(t)=(-2+1\times t)e^{-5t}\\\\\Rightarrow y(t)=(-2+t)e^{-5t}.

6 0
3 years ago
A tank pumps out liquid at a rate of 12 gallons every 20 days. What is the unit rate in gallons per week?
pashok25 [27]

Answer:

4.2

Step-by-step explanation:

simplify 12 and 20 to 6 and 10 then divide 6 by 10 and multiply by 7 to get the answer.

8 0
3 years ago
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