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kupik [55]
2 years ago
11

A runner is 7/8 mile from the finish line. If she can travel 3/8 mile per minute, how long will it take her to finish the race?​

Mathematics
2 answers:
Ksivusya [100]2 years ago
8 0

Answer:

  • 2 1/3 minutes

Step-by-step explanation:

<u>Distance:</u>

  • d = 7/8 mile

<u>Speed:</u>

  • s = 3/8 mpm

<u>Time:</u>

  • t = d/s
  • t = 7/8 : 3/8 = 7/8 * 8/3 = 7/3 = 2 1/3 min
ra1l [238]2 years ago
6 0

Answer:

Step-by-step explanation:

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Step-by-step explanation:

16.80 - 20% of 16.80 = 13.44

you paid 13.44$

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What is the value of 0.363636... times 11/2? <br> PLEASE HELP
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.363636 * 5.5 = 1.999998
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Determine the area for the diagram above plzzzzz​
gulaghasi [49]

Answer:

see explanation

Step-by-step explanation:

The figure shown has one pair of parallel sides and is a trapezium

The area (A) of a trapezium is calculated using

A = \frac{1}{2} h(a + b)

where h is the height and a, b the length of the parallel sides

here h = 2w, a = w + 3 and b = 4w - 2, hence

A = \frac{1}{2} × 2w(w + 3 + 4w - 2)

   = w(5w + 1) = 5w² + w


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3 years ago
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F(x+h)-f(x)/h<br> f(x) = 2x 2 + 7x
Karolina [17]
174992774 there you go chicken
7 0
3 years ago
Find the dimensions of the open rectangular box of maximum volume that can be made from a sheet of cardboard 11 in. by 7 in. by
11111nata11111 [884]

Answer:

The dimension of the open rectangular box is 8.216\times 4.216\times 1.392.

The volume of the box is 8.217 cubic inches.

Step-by-step explanation:

Given : The open rectangular box of maximum volume that can be made from a sheet of cardboard 11 in. by 7 in. by cutting congruent squares from the corners and folding up the sides.

To find : The dimensions and the volume of the open rectangular box ?

Solution :

Let the height be 'x'.

The length of the box is '11-2x'.

The breadth of the box is '7-2x'.

The volume of the box is V=l\times b\times h

V=(11-2x)\times (7-2x)\times x

V=4x^3-36x^2+77x

Derivate w.r.t x,

V'(x)=4(3x^2)-2(36x)+77

V'(x)=12x^2-72x+77

The critical point when V'(x)=0

12x^2-72x+77=0

Solve by quadratic formula,

x=\frac{18+\sqrt{93}}{6},\frac{18-\sqrt{93}}{6}

x=4.607,1.392

Derivate again w.r.t x,

V''(x)=24x-72

Now, V''(4.607)=24(4.607)-72=38.568>0 (+ve)

V''(1.392)=24(1.392)-72=-38.592 (-ve)

So, there is maximum at x=1.392.

The length of the box is l=11-2x

l=11-2(1.392)=8.216

The breadth of the box is b=7-2x

b=7-2(1.392)=4.216

The height of the box is h=1.392.

The dimension of the open rectangular box is 8.216\times 4.216\times 1.392.

The volume of the box is V=l\times b\times h

V=8.216\times 4.216\times 1.392

V=48.217\ in.^3

The volume of the box is 8.217 cubic inches.

5 0
3 years ago
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