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Iteru [2.4K]
3 years ago
15

Just need help with these 2 questions

Mathematics
1 answer:
liberstina [14]3 years ago
5 0

Answer:

I wanna help but I can't see it properly

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An airplane is currently at an elevation of 8500 feet above sea level. The airplane is losing elevation at a rate of 150 feet pe
solniwko [45]

Answer:the answer is y=8500-150m

Step-by-step explanation:I say this because 150 is the amount of feet that it drops per minute m stands for minute so you would be multiplying 150 by whatever m would be.

7 0
4 years ago
NEED HELP FAST WILL GIVE 50 PTS AND BRAINLIEST FOR BEST AND CORRECT ANSWER OTHERS WILL BE DELEATED
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C, 1,020. There are 67 out of 100 students who do at least 2 hrs of chores a day, or 67%. Converting that to a decimal, you get .68, which you multiply by the number of students at the school to get your answer.
4 0
3 years ago
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Henry needs a minimum of $20 to buy a toy truck that he wants. If he had $10 and then spent $2, how much more does he need to sa
Zigmanuir [339]
If he needs $20, for a truck only has 10 and spends $2 already. He only has $8 left which means he needs a minimum of $12. So I'd say best guess. C or D.
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4 years ago
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1. Where will the line y = -1/2 -3 cross the y-axis?
soldier1979 [14.2K]

1) The y- axis is where x = 0, so the line "y = -1/2x - 3" will cross the y-axis when x = 0. You plug in "x = 0" into the equation, and get y = 0 -3, so y = -3. That means the line " y = -1/2x - 3" will cross the y-axis ar (0, -3)

2) If there are 2 parallel lines, then there is no solution, because the definition of parallel lines is, " 2 lines in the same plane, that never meet." That makes the solution inconsistent.

7 0
3 years ago
Solve the system of equations:x + 3y - z = -4 2x - y + 2z = 13 3x - 2y - z = -9
tatiyna

Answer:

The solution to the system of equations is

\begin{gathered} x=\frac{179}{13} \\  \\ y=-\frac{279}{39} \\  \\ z=-\frac{48}{13} \end{gathered}

Explanation:

Giving the system of equations:

\begin{gathered} x+3y-z=-4\ldots\ldots\ldots\ldots\ldots\ldots..........\ldots\ldots\ldots\ldots.\ldots\text{.}\mathrm{}(1) \\ 2x-y+2z=13\ldots\ldots...\ldots\ldots\ldots\ldots..\ldots..\ldots\ldots\ldots\ldots\ldots.(2) \\ 3x-2y-z=-9\ldots\ldots\ldots.\ldots\ldots\ldots\ldots....\ldots\ldots.\ldots\ldots\ldots\text{.}\mathrm{}(3) \end{gathered}

To solve this, we need to first of all eliminate one variable from any two of the equations.

Subtracting (2) from twice of (1), we have:

5y-4z=-21\ldots\ldots\ldots\ldots\ldots.\ldots.\ldots..\ldots..\ldots\ldots.\ldots..\ldots\text{...}\mathrm{}(4)

Subtracting (3) from 3 times (1), we have

3y-5z=-3\ldots\ldots...\ldots\ldots..\ldots\ldots\ldots\ldots\ldots.\ldots\ldots\ldots\ldots\ldots..\ldots\ldots(5)

From (4) and (5), we can solve for y and z.

Subtract 5 times (5) from 3 times (4)

\begin{gathered} 13z=-48 \\  \\ z=-\frac{48}{13} \end{gathered}

Using the value of z obtained in (5), we have

\begin{gathered} 3y-5(-\frac{48}{13})=-3 \\  \\ 3y+\frac{240}{13}=-3 \\  \\ 3y=-3-\frac{240}{13} \\  \\ 3y=-\frac{279}{13} \\  \\ y=-\frac{279}{39} \end{gathered}

Using the values obtained for y and z in (1), we have

\begin{gathered} x+3(-\frac{279}{39})-(-\frac{48}{13})=-4 \\  \\ x-\frac{279}{13}+\frac{48}{13}=-4 \\  \\ x-\frac{231}{13}=-4 \\  \\ x=-4+\frac{231}{13} \\  \\ x=\frac{179}{13} \end{gathered}

8 0
1 year ago
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