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tatyana61 [14]
3 years ago
12

The distance traveled (d) varies directly as the time (t) of

Mathematics
1 answer:
Helen [10]3 years ago
6 0

Answer:

2.70 hours to travel 180 mi.

Step-by-step explanation:

The distance traveled (d) varies directly as the time (t) of travel, by the following formula:

s = \frac{d}{t}

In which s is the velocity.

It takes 63 min to travel 70 mi.

So i am going to find the velocity in miles per minute

s = \frac{70}{63} = 1.11

how many hours would it take to travel 180 mi?

I am first going to find the time in minutes, then divide by 60 to pass to hours.

s = \frac{d}{t}

st = d

t = \frac{d}{s}

t = \frac{180}{1.11}

t = 162.16

It takes 162.16 minutes = 162.16/60 = 2.70 hours to travel 180 mi.

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What times 5 equals 62?
Anni [7]
12.4 because 62 divided by 5 = 12.4
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3 years ago
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Evaluate the expression for = −3 and = 5.<br><br> r^{-4} s^2
Butoxors [25]
If r = -3 and s = 5:
(r^-4)(s^2) = (-3^-4)(5^2) = (1/81)(25) = 25/81

If r = 5 and s = -3:
(r^-4)(s^2) = (5^-4)(3^2) = (1/625)(9) = 9/625
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HAY IN THE MIDDLE OF THE BARN: There are one thousand and forty-one bales of hay in the barn. Farmer John is getting ready for F
snow_tiger [21]

Answer:

Kindly check explanation

Step-by-step explanation:

Given that:

Initial number of bales of hay = 1041 bales

After stacking more bales in the barn:

Total number of bales after stacking = 2,358 bales

Equation for scenario b

Initial bales + added bales = total number of bales after stacking

Let added bales = b

1,041 + b = 2,357

Added bales = Newly stacked bales:

1041 + b = 2,357

b = 2357 - 1041

b = 1,316

The stacked bales of hay stacked to join the initial is. 1,316 bales

3 0
3 years ago
Please help. I only need parts a and b.
Alex

Answer:

oI know that for a they are pretty much the same shape im not sure how else to desribe it sorry

Step-by-step explanation:

7 0
2 years ago
The graph of the function P(x) = −0.34x2 + 12x + 62 is shown. The function models the profits, P, in thousands of dollars for a
STatiana [176]

Answer:

0 ≤ x < 1.12 and 34.18 < x ≤ 39.87

Step-by-step explanation:

<u><em>The options of the question are</em></u>

−4.57 ≤ x ≤ 39.87

1.12 ≤ x ≤ 34.18

−4.57 ≤ x ≤ 1.12 and 34.18 ≤ x ≤ 39.87

0 ≤ x < 1.12 and 34.18 < x ≤ 39.87

Let

x ----> is the number of tires produced, in thousands

C(x) --->  the production cost, in thousands of dollars

we have

C(x)=-0.34x^{2} +12x+62

This is a vertical parabola open downward (the leading coefficient is negative)

The vertex represent a maximum

The graph in the attached figure

we know that

Looking at the graph

For the interval [0,1.12) ----->  0\leq x

The value of C(x) ----> C(x) < 75

That means ----> The production cost is under $75,000

For the interval (34.18,39.87] -----> 34.18 < x\leq 39.87

The value of C(x) ----> C(x) < 75

That means ----> The production cost is under $75,000

Remember that the variable x (number of tires) cannot be a negative number

therefore

If the company wants to keep its production costs under $75,000 a reasonable domain for the constraint x is

0 ≤ x < 1.12 and 34.18 < x ≤ 39.87

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3 years ago
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