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DedPeter [7]
3 years ago
13

4) Sumalee had some candy to give to her

Mathematics
1 answer:
jonny [76]3 years ago
5 0

Answer:

31 pieces

Step-by-step explanation:

6 + 5 x 5 = 31

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Ava bought 1.7 lb of plums and 7.2 lb of peaches how many was her total. Plums:$0.23. per pound peaches:$1.22 per pound​
Rainbow [258]

Answer:

0.391+8.42=8.811

8.811 in dollars is 10.42 United States Dollars

Step-by-step explanation:

Your answer is 10.42 c:

6 0
2 years ago
Which of the following terms is a set of all points in a plane that are a given distance from a point?
Usimov [2.4K]

Answer: circle

Step-by-step explanation:

3 0
3 years ago
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A rocket is launched from a tower. The height of the rocket, y in feet, is related to the
IrinaK [193]

Answer:

1,041.9feet

Step-by-step explanation:

Given the height of the rocket expressed as

y = -16x² + 245x + 104

At maximum height, dy/dx = 0

dy/dx = -32x+245

0 = -32x+245

32x = 245

x = 245/32

x = 7.65625

Get the maximum height

Recall that;

y = -16x² + 245x + 104

Substitute the value of x;

y = -16(7.65625)² + 245(7.65625) + 104

y = -937.890625 + 1,875.78125 + 104

y = 1,041.890625feet

Hcne the maximum height to the nearest foot is 1,041.9feet

4 0
3 years ago
Jenna has 8 pairs of jeans. Three of them are blue. What percentage of her jeans is blue?
Bond [772]
A) blue/total = (3+2)/(8+2) = 5/10 = 50%

b) blue/total = (3)/(8 +2) = 3/10 = 30%
5 0
3 years ago
An open top box is to be built with a rectangular base whose length is twice its width and with a volume of 36 ft 3 . Find the d
denpristay [2]

Answer:

The dimensions of the box that minimize the materials used is 6\times 3\times 2\ ft

Step-by-step explanation:

Given : An open top box is to be built with a rectangular base whose length is twice its width and with a volume of 36 ft³.

To find : The dimensions of the box that minimize the materials used ?

Solution :

An open top box is to be built with a rectangular base whose length is twice its width.

Here, width = w

Length = 2w

Height = h

The volume of the box V=36 ft³

i.e. w\times 2w\times h=36

h=\frac{18}{w^2}

The equation form when top is open,

f(w)=2w^2+2wh+2(2w)h

Substitute the value of h,

f(w)=2w^2+2w(\frac{18}{w^2})+2(2w)(\frac{18}{w^2})

f(w)=2w^2+\frac{36}{w}+\frac{72}{w}

f(w)=2w^2+\frac{108}{w}

Derivate w.r.t 'w',

f'(w)=4w-\frac{108}{w^2}

For critical point put it to zero,

4w-\frac{108}{w^2}=0

4w=\frac{108}{w^2}

w^3=27

w^3=3^3

w=3

Derivate the function again w.r.t 'w',

f''(w)=4+\frac{216}{w^3}

For w=3, f''(3)=4+\frac{216}{3^3}=12 >0

So, it is minimum at w=3.

Now, the dimensions of the box is

Width = 3 ft.

Length = 2(3)= 6 ft

Height = \frac{18}{3^2}=2\ ft

Therefore, the dimensions of the box that minimize the materials used is 6\times 3\times 2\ ft

4 0
3 years ago
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