Mathematics, 27.08.2020 06:01 brae72

The map shows the route for a race. You are at X, 6000 ft from the first checkpoint C. The second checkpoint D is located at the midpoint between C and the end of the race Y. The total race is 3 miles how far apart are the 2 checkpoints? I know the answer is 4920 but i don’t know how to work it out

The map shows the route for a race.You are at X, 6000 ft from the first checkpoint C.The second che

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Answer from: mayalp

Distance between the two check point (C and D) is 4920 ft.

Step-by-step explanation:

From the question given above, we obtained the following information:

X is 6000 ft from C (1st check point)

D (2nd check point) is the midpoint point between C and Y ( the end of the race)

Total race = 3 miles

1 mile = 5280 ft

Next, we shall express the total race in feet (ft).

This is illustrated below:

1 mile = 5280 ft

Therefore,

3 miles = 3 × 5280 = 15840 ft

Next, we shall determine the distance between the 1st check point (C) and the end of the race (Y)

This can be obtained as follow:

Total distance = 15840 ft

Distance between X and C = 6000 ft

Distance between C and Y =?

Distance between C and Y = (Total distance) – (Distance between X and C)

Distance between C and Y = 15840 – 6000

Distance between C and Y = 9840 ft

Finally, we shall determine the distance between the two check point (C and D) as follow:

Since D is midpoint between C and Y, therefore, we shall we shall divide the distance between C and Y by 2 in order to obtain the distance between the two check point (C and D).

This is illustrated below:

Distance between C and Y = 9840 ft

Distance between the two check point (C and D) = 9840/2 = 4920 ft

Therefore, the distance between the two check point (C and D) is 4920 ft

Answer from: Quest

arithmetic sequence: a sequnce in which the diffrence between any two consecutive terms is the same

step-by-step explanation:

Answer from: Quest

< xcy ≅ < xby

step-by-step explanation:

in δcax and δbax,

ac ≅ ab (given)< cax ≅ < bax (given)ax ≅ ax (common side)

so, δcax ≅ δbax (side-angle-side or sas)

since δcax ≅ δbax, we can conclude < acx ≅ < abx

in δacy and δaby,

ac ≅ ab (given)< cay ≅ < bay (since < cax ≅ < bax is given and xy is the extension of line ax)ay ≅ ay (common side)

so, δacy ≅ δaby (side-angle-side or sas)

since δacy ≅ δaby, we can conclude < acy ≅ < aby

now,

< acy ≅ < aby

=> < acx + < xcy ≅ < abx + < xby

=> < acx + < xcy ≅ < acx + < xby (since < acx ≅ < abx already proved above)

subtracting < acx from both sides, we get

< acx + < xcy -< acx ≅ < acx + < xby -< acx

cancelling out < acx and -< acx from both sides, we get

< xcy ≅ < xby (proved)

Use a paragraph, flow chart, or two-column proof to prove the angle congruency. given: ∠cxy ≅ ∠bxy

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The map shows the route for a race. You are at X, 6000 ft from the first checkpoint C. The second ch...

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