Mathematics, 01.08.2019 21:50 kjn29

Points e, d, and h are the midpoints of the sides of triangle tuv. uv = 80, tv = 100 and hd = 80 2. find the lengths of segments he. 3. find the lengths of segments tu.

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Without the diagram, there are 6 possible arrangements for points E, D, H. Let's see if we can reduce that to something smaller. HD can not be parallel to either TV or UV since it would have to be half of either of those sides in length. So HD has to be parallel to side TU. This means that point E has to be the midpoint of side TU. However, it's indeterminate about which point is H, or which point is D. Therefore there's 2 possible answers to this question. Without the diagram, it's impossible to determine which of the two possibilities is correct, so you'll have to choose the correct one yourself. Option 1. D is midpoint of TV, H is midpoint of UV. Triangle TUV is similar to triangle DHV. Since the ratio of the corresponding sides is 2:1, the length of HE is half of the length of TV. So 100/2 = 50. And because of that 2:1 ratio, the length of TU is twice that of HD, so 2 * 80 = 160. In summary: HE = 50 TU = 160 Option 2. H is midpoint of TV, D is midpoint of UV. We still have a similar triangle, but H&D are swapped. The length of TU will be the same as in option 1 above. And the length of HE will be half that of UV instead of TV. So 80/2 = 40. So in summary: HE = 40 TU = 160

Answer from: Quest

just find the gcf and solve the gcf's until you get each answer i have a really good feeling somethings missing xd (like the dimensions of the fence) *cough *cough *hint *hint

step-by-step explanation:

Answer from: Quest

geometric series converge the common ratio r has magnitude less than one.

$|r| < 1$

which is equivalent to

$-1 < r < 1$

for the example we need

|2-x| < 1

-1 < 2-x and 2-x < 1

x < 3 and x > 1

we can write that

answer: 1 < x < 3

1a. common ratio r=250/25=10 > 1 diverges

1b. r=50/100=1/2 converges

when these converge they converge to

$s = \dfrac{a}{1 -r}$

where a is the first term.

$s = \dfrac{100}{1 - 1/2} = 200$

answer: 200

1c. r=-1 so it's not the case that |r| < 1. diverges

2a. x is the common ratio so we converge when |x|< 1

2b. r=-10x / -2 = 5x.

we converge when

|5x| < 1

|x| < 1/5

2c. r=3-2x

-1 < 3-2x and 3-2x < 1

-4 < -2x and 2 < 2x

x < 2 and 1 < x

1 < x < 2

2d. [/tex] r=-48x^3/12 = 4x^3[/tex]

|4x^3| < 1

|x^3| < \frac 1 4

|x| < 4^{-1/3}

$|x| < \frac 1 2 \sqrt[3]{2}$

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Points e, d, and h are the midpoints of the sides of triangle tuv. uv = 80, tv = 100 and hd = 80 2....

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