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Rashid [163]
3 years ago
14

Parker ran at 6 miles per hour for 27 minutes and 39 seconds. How far did Parker run?

Mathematics
1 answer:
DedPeter [7]3 years ago
7 0

Answer:

Parker ran 2.76 miles

Step-by-step explanation:

distance = speed x time

First, we have to convert 27 minutes and 39 seconds to minutes only we do this with 39/60 because then the seconds divide by total seconds and we can use it as a decimal

39/60 = .65

So it took 27.65 total minutes now we must convert it to hours, so that the 6 MILES per hour has matching terms, we do this by dividing 27.65 by 60 (theres 60 total minutes in an hour)

So it took 0.460833333 hours at a 6 mph pace, we can round this to 0.46

Now we can plug in the time and speed

Distance = 6mph x 0.46 hours

Distance = 2.76 miles

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Two times the sum of a number and 9 is half the number
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Answer:

Equation:

2(a+9) = a/2

Solve:

The umber is:

-12

Step-by-step explanation:

Equation:

2(a+9) = a/2

Solve:

2*a + 2*9 = a/2

2a + 18 = a/2

2a - a/2 = -18

4a/2 - a/2 = -18

3a/2 = -18

a = -18*2/3

a = -12

Check:

2(-12+9) = -12/2

2(-3) = -12/2 = -6

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3 years ago
Solve for x:<br> 3(x+1)=5 (x-2)+7
Aleonysh [2.5K]

Step-by-step explanation:

3( x +1) = 5 (x - 2) + 7

3x + 3 = 5x - 10 + 7

5x - 3x = 3 + 10 - 7

2x = 13 - 7

2x = 6

Therefore x = 3

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3 years ago
If the graph of the second equation in the system passes through the points (-12, 20) and
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3 years ago
Jamie needs to build a fence around his garden, as illustrated by polygon ABCDEF on the coordinate grid below. If each unit repr
NNADVOKAT [17]

Answer:

C. 26 yards

Step-by-step explanation:

Jamie's fence total length = perimeter of the polygon

Perimeter of the polygon = AB + BC + CD + DE + EF + FA

AB, FA and DE can be worked accordingly as shown below:

AB = |-5 - 0| = 5 units

FA = |5 - 2| = 3 units

DE = |1 -(-2)| = 3 units

BC, CD, and EF can be calculated using the formula d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

Distance between B(0, 5) and C(4, 2):

BC = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

Let,

B(0, 5) = (x_1, y_1)

C(4, 2) = (x_2, y_2)

BC = \sqrt{(4 - 0)^2 + (2 - 5)^2}

BC = \sqrt{(4)^2 + (-3)^2}

BC = \sqrt{16 + 9} = \sqrt{25}

BC = 5 units

Distance between C(4, 2) and D(1, -2)

CD = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

Let,

C(4, 2) = (x_1, y_1)

D(1, -2) = (x_2, y_2)

CD = \sqrt{(1 - 4)^2 + (-2 - 2)^2}

CD = \sqrt{(-3)^2 + (-4)^2}

CD = \sqrt{9 + 16} = \sqrt{25}

CD = 5 units

Distance between E(-2, -2) and F(-5, 2):

EF = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

Let,

E(-2, -2) = (x_1, y_1)

F(-5, 2) = (x_2, y_2)

EF = \sqrt{(-5 -(-2))^2 + (2 -(-2))^2}

EF = \sqrt{(-3)^2 + (4)^2}

EF = \sqrt{9 + 16} = \sqrt{25}

EF = 5 units

Total length of the wall in yards = 5 + 5 + 5 + 3 + 5 + 3 = 26 yards

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Angelina_Jolie [31]
If a=1 and b=2 , then 1+1=2 and 2>1. No it isnt possible to make the statement false because no matter what b will always be 1 greater than a
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