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inysia [295]
2 years ago
12

Flo sets a goal of walking 12,000 total steps per day. flo's normal routine involves walking 6,000 steps per day. if flo walks 1

00 steps per minute, how many extra minutes should she walk per day beyond her normal routine?
Mathematics
1 answer:
Oksana_A [137]2 years ago
6 0

She has to walk 60 minutes more to reach her goal.

Given: Flo sets a goal of walking 12,000 total steps per day. Flo's normal routine involves walking 6,000 steps per day.

To find: To Determine how many extra minutes should she walk per day beyond her normal routine?

Formula: Number of extra minutes = she walks per day beyond her normal routine / number of steps per minute

flo's normal routine involves walking steps per day= 6,000

If flo walks 100 steps per minute:

she walks per day beyond her normal routine = 12000-6000

                                                                             = 6000

Extra minutes = \frac{6000}{100}

                       = 60 minutes

Therefore, she has to walk 60 minutes more to reach her goal.

Learn more about time and distance here brainly.com/question/24283318

#SPJ4

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Y = -2x - 2. Slope here is -2. A parallel line will have the same slope.

y = mx + b
slope(m) = -2
(3,4)...x = 3 and y = 4
now we sub and find b, the y int
4 = -2(3) + b
4 = - 6 + b
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so ur parallel equation is : y = -2x + 10
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Which point is a solution to the system of inequalities graphed here? y ≤ 2x + 2
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Answer:

D) (5,0)

Step-by-step explanation:

-5x + 4 ≤ y

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3 years ago
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A 9,000-gallon silo of wheat is being drained at a rate of 500 gallons per hour. A 7,000-gallon silo of wheat is being filled at
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If 2(4x + 3)/(x - 3)(x + 7) = a/x - 3 + b/x + 7, find the values of a and b.
zmey [24]

Answer:

a=3 and b=5.

Step-by-step explanation:

So I believe the problem is this:

\frac{2(4x+3)}{x-3}(x+7)}=\frac{a}{x-3}+\frac{b}{x+7}

where we are asked to find values for a and b such that the equation holds for any x in the equation's domain.

So I'm actually going to get rid of any domain restrictions by multiplying both sides by (x-3)(x+7).

In other words this will clear the fractions.

\frac{2(4x+3)}{x-3}(x+7)}\cdot(x-3)(x+7)=\frac{a}{x-3}\cdot(x-3)(x+7)+\frac{b}{x+7}(x-3)(x+7)

2(4x+3)=a(x+7)+b(x-3)

As you can see there was some cancellation.

I'm going to plug in -7 for x because x+7 becomes 0 then.

2(4\cdot -7+3)=a(-7+7)+b(-7-3)

2(-28+3)=a(0)+b(-10)

2(-25)=0-10b

-50=-10b

Divide both sides by -10:

\frac{-50}{-10}=b

5=b

Now we have:

2(4x+3)=a(x+7)+b(x-3) with b=5

I notice that x-3 is 0 when x=3. So I'm going to replace x with 3.

2(4\cdot 3+3)=a(3+7)+b(3-3)

2(12+3)=a(10)+b(0)

2(15)=10a+0

30=10a

Divide both sides by 10:

\frac{30}{10}=a

3=a

So a=3 and b=5.

4 0
3 years ago
Read 2 more answers
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