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3 years ago

If the height of this box was 10in, how many 2in cubes could fit inside this box

2 answers:

8 0

This question is incomplete because there is a missing variable. You would need the width or length for me to complete this.

Mashutka [201]3 years ago

8 0

Answer:

The number of 2 in cubes that could fit in a 10 in box (assuming that is a cube) are 125,

Step-by-step explanation:

Assuming that the box is a cube with 10 in heigh, then the volume of the cube correspond to:

$V=10^{3}=1000 in^{3}$

Now you should know the volume of the 2 in cube as follows:

$V=2^{3}=8 in^{3}$

If you want to know how many cubes with 2 in each side could fit inside the box of 10 in heigh, you should to divide the volume of the bigger cube over the smaller cube:

$2inCubes=\frac{1000ni^{3} }{8in^{3} }= 125$

The number of 2 in cubes that could fit in a 10 in box (assuming that is a cube) are 125.

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8) Find the endpoint Cif M is the midpoint of segment CD and M (2, 4) and D (5,7)

Elenna [48]

Answer:

8. c. (-1, -1)

9. a. (-6, -1)

b. True

Step-by-step Explanation:

8. Given the midpoint M(2, 4), and one endpoint D(5, 7) of segment CD, the coordinate pair of the other endpoint C, can be calculated as follows:

let $D(5, 7) = (x_2, y_2)$

$C(?, ?) = (x_1, y_1)$

$M(2, 4) = (\frac{x_1 + 5}{2}, \frac{y_1 + 7}{2})$

Rewrite the equation to find the coordinates of C

$2 = \frac{x_1 + 5}{2}$ and $4 = \frac{y_1 + 7}{2}$

Solve for each:

$2 = \frac{x_1 + 5}{2}$

$2*2 = \frac{x_1 + 5}{2}*2$

$4 = x_1 + 5$

$4 - 5 = x_1 + 5 - 5$

$-1 = x_1$

$x_1 = -1$

$4 = \frac{y_1 + 7}{2}$

$4*2 = \frac{y_1 + 7}{2}*2$

$8 = y_1 + 7$

$8 - 7 = y_1 + 7 - 7$

$1 = y_1$

$y_1 = 1$

Coordinates of endpoint C is (-1, 1)

9. a.Given segment AB, with midpoint M(-4, -5), and endpoint A(-2, -9), find endpoint B as follows:

let $A(-2, -9) = (x_2, y_2)$

$B(?, ?) = (x_1, y_1)$

$M(-4, -5) = (\frac{x_1 + (-2)}{2}, \frac{y_1 + (-9)}{2})$

$-4 = \frac{x_1 - 2}{2}$ and $-5 = \frac{y_1 - 9}{2}$

Solve for each:

$-4 = \frac{x_1 - 2}{2}$

$-4*2 = \frac{x_1 - 2}{2}*2$

$-8 = x_1 - 2$

$-8 + 2 = x_1 - 2 + 2$

$-6 = x_1$

$x_1 = -6$

$-5 = \frac{y_1 - 9}{2}$

$-5*2 = \frac{y_1 - 9}{2}*2$

$-10 = y_1 - 9$

$-10 + 9 = y_1 - 9 + 9$

$-1 = y_1$

$y_1 = -1$

Coordinates of endpoint B is (-6, -1)

b. The midpoint of a segment, is the middle of the segment. It divides the segment into two equal parts. The answer is TRUE.

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3 years ago

Anton Louis is a junior partner in an accounting firm. He is single

Sauron [17]

Answer:

Step-by-step explanation:

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2 years ago

Find the m∠SAQ. 44° 46° 57° 77°

dimulka [17.4K]

Answer:

X=57

Step-by-step explanation:

77+46+X=180

123+X=180

X=180-123

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7 0

3 years ago

Read 2 more answers

Josh is hiking Glacier National Park. He has now hiked a total of 17 \text{ km}17 km17, space, k, m and is 2 \text{ km}2 km2, sp

jeka94

Josh has hiked 17 km, which is 2 km away from being 1/2 done.
In a equation, this might look like: 17 + 2 = (1/2)h 
where h is the length of the hike. 
and solve for h, h = 38 km

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3 years ago

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The width of a rectangle is 6 2/3 inches. The length of the is twice it’s width. What so the perimeter of the rectangle?

Nookie1986 [14]

$\text {Width = } 6\dfrac{2}{3} \text { inches}
$

The length is twice its width:
$\text {Length = } 2 \times 6\dfrac{2}{3} \text { inches}$

Change to improper fraction:
$\text {Length = }2 \times \dfrac{20}{3} \text { inches}$

Combine into single fraction:
$\text {Length = } \dfrac{40}{3} \text { inches}$

Find Perimeter :
$\text {Perimeter = Length + Length + Width + Width}$

$\text {Perimeter = } \dfrac{40}{3} + \dfrac{40}{3} + \dfrac{20}{3} + \dfrac{20}{3} = \dfrac{120}{3} = 40 \text { inches}$

$\bf \text {Answer: Perimeter = 40 inches}$

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3 years ago

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