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gtnhenbr [62]
3 years ago
12

The area of a 2/3 cube

Mathematics
1 answer:
algol133 years ago
3 0

Answer:

2 that's what soup calculated says

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Select the equation of the line that passes through the point (3, 5) and is
ohaa [14]

Answer:

<h2>1) y = 5</h2>

Step-by-step explanation:

x = 4 it's a vertical line.

Perpendicular line to a vertical line is a horizontal line.

A horizontal line has equation y = a.

A horizontal line passes through the point (3, 5) → x = 3 and y = 5.

Therefore the equation is y = 5

3 0
3 years ago
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Which number rounds to 0.32 rounded to the nearest hundredth
Naddik [55]
0.32 to the nearest hundredth would be 0.3
5 0
3 years ago
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You throw three darts in a numbered region of the dart board that has the scores of 2, 5, and 8. How many different sums are pos
givi [52]

Answer:

12 different sums

Step-by-step explanation:

there's three darts and three numbers.

if you use the factorial 3! for both the number of darts and the score numbers and add the two together, you would get twelve :

( 3! = 3*2*1 = 6 ) * 2

6 + 6 = 12

or you could just write out all the combinations like this :

2+2+2

2+2+5

2+5+5

2+5+8

2+8+8

5+5+2

5+5+5

5+5+8

5+8+8

8+8+2

8+8+5

8+8+8

6 0
4 years ago
For the equation y = 2x2 − 5x + 18, choose the correct application of the quadratic formula.
Mazyrski [523]
X equals five plus or minus the square root of negative five squared minus four times two times eighteen, all divided by two times two
6 0
3 years ago
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A rectangular swimming pool has a length twice as long as its width. The pool has a sidewalk around it that is 2 feet wide. Writ
xxMikexx [17]

Answer:

A_p=2w^2

A_s=12w+16

Step-by-step explanation:

Functions and Geometry

Sometimes we need to express areas, volumes, and distances in terms of unknown quantities (or variables). When we can find such expressions, they can be referred to as functions. Let's recall the area of a rectangle is A=wl, being w its wide and l its length

Please refer to the image below. The length of the swimming pool is l and its width is w. Its area is

A_p=lw

We know the length is twice as long as its width, l=2w, so

A_p=(2w)w=2w^2

A_p=2w^2

The sidewalk is 2 feet wide. The length of the pool and its sidewalk is

l'=l+4

Similarly, the width of the pool+sidewalk is

w'=w+4

The area of the pool+sidewalk is

A'=l'w'=(l+4)(w+4)=lw+4l+4w+16

We know that l=2w, so

A'=(2w)w+4(2w)+4w+16

A'=2w^2+12w+16

The area of the sidewalk alone is the subtraction of both areas

A_s=A'-A_p=2w^2+12w+16-2w^2

Simplifying

A_s=12w+16

7 0
3 years ago
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