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Bumek [7]
3 years ago
10

Determine if the lines are parallel, perpendicular, or neither 10x+5y=-5 and y=-2x+6

Mathematics
1 answer:
Ksenya-84 [330]3 years ago
7 0
If they are parallel they will have the same slope , m

So in y = mx + c, if there are two equations which both have the same m value they will be parallel.
If the lines are perpendicular they'll have slopes like this: 1/2 to -2/1 - where they flip and a negative gets added.

In the equations: 10x + 5y = -5 , and y = -2x + 6
We can rearrange 10x + 5y = -5 to be in the form y = mx + c
10x + 5y = -5
5y = -5 - 10x
y = -1 - 2x
y = -2x - 1

Since y = -2x - 1 and y = -2x + 6 both have the same slope of -2 they are parallel!

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To make concrete, the ratio of cement to sand is 1 : 3. If cement and sand are sold in bags of equal mass, how many bags of ceme
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Answer:

5 bags of cement are required.

Step-by-step explanation:

Since to make concrete, the ratio of cement to sand is 1: 3, if cement and sand are sold in bags of equal mass, to determine how many bags of cement are required to make concrete using 15 bags of sand the following calculation must be done:

Cement = 1

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Therefore, 5 bags of cement are required.

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dimulka [17.4K]

Answer:

1 and 2.

Midpoints calculated, plotted and connected to make the triangle DEF, see the attached.

  • D= (-2, 2), E = (-1, -2), F = (-4, -1)

3.

As per definition, midsegment is parallel to a side.

Parallel lines have same slope.

<u>Find slopes of FD and CB and compare. </u>

  • m(FD) = (2 - (-1))/(-2 -(-4)) = 3/2
  • m(CB) = (1 - (-5))/(1 - (-3)) = 6/4 = 3/2
  • As we see the slopes are same

<u>Find the slopes of FE and AB and compare.</u>

  • m(FE) = (-2 - (- 1))/(-1 - (-4)) = -1/3
  • m(AB) = (1 - 3)/(1 - (-5)) = -2/6 = -1/3
  • Slopes are same

<u>Find the slopes of DE and AC and compare.</u>

  • m(DE) = (-2 - 2)/(-1 - (-2)) = -4/1 = -4
  • m(AC) = (-5 - 3)/(-3 - (-5)) = -8/2 = -4
  • Slopes are same

4.

As per definition, midsegment is half the parallel side.

<u>We'll show that FD = 1/2CB</u>

  • FD = \sqrt{(2+1)^2+(-2+4)^2} = \sqrt{3^2+2^2} = \sqrt{13}
  • CB = \sqrt{(1 + 5)^2+(1+3)^2} = \sqrt{6^2+4^2} = 2\sqrt{13}
  • As we see FD = 1/2CB

<u>FE = 1/2AB</u>

  • FE = \sqrt{(-4+1)^2+(-1+2)^2} = \sqrt{3^2+1^2} = \sqrt{10}
  • AB = \sqrt{(-5 -1)^2+(3-1)^2} = \sqrt{6^2+2^2} = 2\sqrt{10}
  • As we see FE = 1/2AB

<u>DE = 1/2AC</u>

  • DE = \sqrt{(-2+1)^2+(2+2)^2} = \sqrt{1^2+4^2} = \sqrt{17}
  • AC = \sqrt{(-5 +3)^2+(3+5)^2} = \sqrt{2^2+8^2} = 2\sqrt{17}
  • As we see DE = 1/2AC

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Answer:

5th term: 20

6th term: 23

Step-by-step explanation:

T5=3n+5

T5=3(5)+5

T5=15+5

T5=20.

T6=3n+5

T6=3(6)+5

T6=18+5

T6=23

T5= 20

T6= 23

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