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zzz [600]
3 years ago
9

I can’t answer this can someone please help me

Mathematics
2 answers:
KATRIN_1 [288]3 years ago
8 0

Answer:

5,400.3 * 10^3

ella [17]3 years ago
7 0

5,004,300 i think is the answer

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Find the parametric equations for the line that passes through the points P (1,1,0) and Q (0,2,2).
sp2606 [1]

Answer:

The parametric equations for the given line are x=1-t, y=1+t and z=2t.

Step-by-step explanation:

Given information: P (1,1,0) and Q (0,2,2).

The parametric equation of line are

x=x_0+at

y=y_0+bt

z=z_0+ct

where, (x_0,y_0,z_0) is point on line and <a,b,c> is direction vector.

The line passes through the points P (1,1,0) and Q (0,2,2). So, the direction vector is

\overrightarrow{v}=

\overrightarrow{v}=

\overrightarrow{v}=

The direction vector is <-1,1,2>. So, a=-1, b=1 and c=2. The parametric equation of line are

x=1+(-1)t=1-t

y=1+(1)t=1+t

z=0+(2)t=2t

Therefore the parametric equations for the given line are x=1-t, y=1+t and z=2t.

5 0
3 years ago
How many 1-inch cubes do you need to create a cube with an edge length of 6 inches?
Ray Of Light [21]
36, so that way each edge would be 6 in.
4 0
3 years ago
Read 2 more answers
Solve for -4[x+9]=-8
Ivahew [28]

Answer:

x= -7

Step-by-step explanation:

-4x - 36 = -8

-4x = - 8 + 36

-4x= 28

x= 28/-4

x= - 7

3 0
3 years ago
HELP THIS IS DUE IN AN HOUR!!!​
umka2103 [35]

Answer:

The perimeter of the first question is 62

If you draw it, you'll you'll see that it's a rectangle. The length of each side is calculated by adding the widths and heights as in the photo ( red ), and then adding them up. 13 + 18 + 13 + 18 = 62

You do the same with question 2, but this time the rectangle is smaller. (green in the photo)

So the perimeter is:

13 + 14 + 13 + 14 = 54

The answer to question 3 is (-15, -5) as drawn in the photo.

8 0
3 years ago
What would you get if you add all odd numbers 1 through 7 inclusive
alexdok [17]
Well, it would be 1+3 which is 4. Then 4+5 which is 9. Then 9+7 which is 16 so your final answer would be 16
8 0
3 years ago
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