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katrin [286]
3 years ago
9

Bobby bought 3 5/12 fraction numerator 5 1/12 star times text end text over denominator $1.20 star times text end text end fract

ion ft of red ribbon and 5 fraction numerator 1 star times text end text over denominator 12 star times text end text end fraction ft of yellow ribbon. Ribbon costs Syntax error from line 1 column 87 to line 1 column 107. Unexpected '1'. a foot. How much did Bobby spend on ribbon?
Mathematics
1 answer:
TEA [102]3 years ago
6 0
Hey Ender, I'm pretty sure you got some text things wrong, or just copied and pasted something, because what you wrote there makes 0 sense :)

You might be interested in
What is the greatest common factor and the least common multiple of 45,75 and 90?
nata0808 [166]
3, 3, 2 is your answer
6 0
3 years ago
What is the slope of the line equation 7x-3y=21
kramer
Equation: 7x - 3y = 21

3y = 7x - 21

Divide both sides by 3, 

y = 7/3x - 7

Compare it with principle equation, y = mx + c

Here, slope (m) = 7/3

In short, Your Answer would be: 7/3

Hope this helps!
8 0
3 years ago
A circle has a diameter of 42 mm. What is the circumference of this circle? (use 22/7 for pi; show your work in numbers)
stellarik [79]

Answer:

132

Step-by-step explanation:

Find Radius (r)

42/2 = 21

Plug 21 into Circumference Equation

2πr (In this case, π is 22/7)

Simplify

42 (22/7)

<u>132</u>

8 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
spencer surveyed five of his friends to find out how many pets they have his results are shown in the table lara has 3 cody has
Phantasy [73]
The mean is the average.
Add all the number of pets and divide by how many people there were.
3+5+2+4+1
15/number of people
15/5
=3

Hope this helps :)
4 0
3 years ago
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