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Softa [21]
3 years ago
6

A child has a piece of flat cardboard with dimensions of 40 cm by 44 cm. She cuts out small congruent squares from each corner o

f the cardboard. She then folds up the sides of the cardboard creating a box with a height that is the same as the side length of the removed squares. The box does not have a top.
The side length of each of the removed squares is 4x. Which of the following equations represents the volume of the box?

None of these.

V(x)=(40−8x)(44−8x)(4x)

V(x)=(40−4x)(44−4x)(4x)

V(x)=(40)(44)(4x)

V(x)=(40−2x)(44−2x)(2x)
Mathematics
1 answer:
Rom4ik [11]3 years ago
4 0

Answer:

someone please answer this

Step-by-step explanation:

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Given that ∠STV = 108° and ∠RST = 50°, determine the missing angles:
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A. 108 degrees
b. 72 degrees
c. 108 degrees

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Can someone help please
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Top is 18, Bottom goes 6 for 12, then 7 for 14, then 10 for 20

Step-by-step explanation:

2 guests = 1 table

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Last week 24,000 fans attended a football match. this week three times as many bought tickets, but one-sixth of them cancelled t
lukranit [14]

Answer:

C. 60,000

Step-by-step explanation:

<u>Given</u>

  • 24,000 fans attended a football match
  • three times as many bought tickets
  • one-sixth of them cancelled their tickets

<u>Three times as many bought tickets</u>

24,000 * 3 = 72,000

<u>one-sixth of them cancelled their tickets</u>

72,000 * 1/6 = 12,000

<u>Subtract</u>

72,000 - 12,000 = 60,000

<u>Answer</u>

60,000 people are attending this week

6 0
3 years ago
Find+the+positive+value+for+α+if+the+radius+of+the+circle+3x^2+3y-6αx+12y-3α=0+is+4
Alik [6]

Answer:

\alpha=3

Step-by-step explanation:

<u>Equation of a Circle</u>

A circle of radius r and centered on the point (h,k) can be expressed by the equation

(x-h)^2+(y-k)^2=r^2

We are given the equation of a circle as

3x^2+3y^2-6\alpha x+12y-3\alpha=0

Note we have corrected it by adding the square to the y. Simplify by 3

x^2+y^2-2\alpha x+4y-\alpha=0

Complete squares and rearrange:

x^2-2\alpha x+y^2+4y=\alpha

x^2-2\alpha x+\alpha^2+y^2+4y+4=\alpha+\alpha^2+4

(x-\alpha)^2+(y+2)^2=r^2

We can see that, if r=4, then

\alpha+\alpha^2+4=16

Or, equivalently

\alpha^2+\alpha-12=0

There are two solutions for \alpha:

\alpha=-4,\ \alpha=3

Keeping the positive solution, as required:

\boxed{\alpha=3}

8 0
3 years ago
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