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Advocard [28]
3 years ago
13

Pls help fast its urgent

Mathematics
1 answer:
telo118 [61]3 years ago
6 0

Answer: -8

Step-by-step explanation:

3 - (-8) = 3+8

3+ 8=11

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Express the edge length of a cube as a function of the cube's diagonal length d. Then express the surface area & volume of t
Lorico [155]
D = sqrt(3s^2) where s is the length of the side. Solving for s, 

<span>3s^2 = d^2 iff </span>
<span>s^2 = d^2 / 3 iff </span>
<span>s = sqrt(d^2 / 3) </span>
<span>= d / sqrt(3) or d sqrt(3) / 3 </span>

<span>Surface area of the cube = 6 s^2. Thus, </span>
<span>A = 6 (d / sqrt(3))^2 </span>
<span>= 6d^2 / 3 </span>
<span>= 2d^2 </span>

<span>Volume = s^3. Thus, </span>
<span>V = (d / sqrt(3))^3 </span>
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<span>= d^3 sqrt(3) / 9</span>
8 0
3 years ago
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Leo reads 14 pages in 1/3 hourWhat is the unit rate for pages per hour
pogonyaev

To determine the unit rate of pages, that is, the number of pages Leo reads in one hour you can use cross multiplication

1/3h _____14 pages

1h______x pages

\begin{gathered} \frac{14}{\frac{1}{3}}=\frac{x}{1} \\ 14\cdot3=x \\ x=42 \end{gathered}

Leo reads 42pages/hour

7 0
1 year ago
I NEED THE ANSWER FOR NUMBER 1
garik1379 [7]

Answer:

Answer would be x=1.

Step-by-step explanation:

7 0
3 years ago
A rectangle has its vertices at (-4, -3), (-4,7), (1,7), (1, -3). What part, in percent, of the rectangle is located in Quadrant
Rasek [7]

Consider the attached figure. The whole rectangle is ABCD, while AEGF is the part located in the third quadrant. In fact, this quadrant is composed by all the points with both coordinates negative.

To answer the question, let's compute the area of the two rectangles and see what part of ABCD is AEGF.

A and B have the same x coordinate, so the length of AB is given by the absolute difference of their y coordinates:

\overline{AB} = |A_y-B_y| = |-3-7| = |-10| = 10

Similarly, but exchanging the role of x and y, we compute the length of BC:

\overline{BC} = |B_x-C_x| = |-4-1| = |-5| = 5

So, the area of the rectangle is \overline{AB} \cdot \overline{BC} = 10\cdot 5 = 50

The same procedure allows us to compute width and height of the sub-rectangle in the third quadrant:

\overline{AE} = |A_y-E_y| = |-3-0| = |-3| = 3

\overline{EG} = |E_x-G_x| = |-4-0| = |-4| = 4

So, the area of the portion located in the third quadrant is \overline{AE} \cdot \overline{EG} = 3\cdot 4 = 12

This means that the ratio between the two area is

\cfrac{\text{area }AEGF}{\text{area }ABCD} = \cfrac{12}{50}

If we want this ratio to be a percentage, just make sure that the denominator is 100:

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3 0
3 years ago
ASAP PLEASE please please
Aleksandr [31]

{ \qquad\qquad\huge\underline{{\sf Answer}}}

Let's solve ~

\qquad \sf  \dashrightarrow \:  \cfrac{1}{b}  + 10 =  \cfrac{9}{b}  + 7

\qquad \sf  \dashrightarrow \:  \cfrac{9}{b}  -   \cfrac{1}{b} =  10 -   7

\qquad \sf  \dashrightarrow \:  \cfrac{8}{b}   =  3

\qquad \sf  \dashrightarrow \: b =  \cfrac{8}{3}

5 0
2 years ago
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