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sineoko [7]
3 years ago
8

. If F, G, and H are the midpoints of the sides of ∆CDE, FG = 9, GH = 7, and CD = 24, find each measure.

Mathematics
1 answer:
Lady_Fox [76]3 years ago
7 0

WHERE'S THE LEAK MA'AM

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If AB = 8 and BC = 24, then AC =<br><br> If AB = 17 and AC = 68, then BC =
Kryger [21]

Answer:

(i) The length of AC is 32 units, (ii) The length of BC is 51 units.

Step-by-step explanation:

(i) Let suppose that AB and BC are collinear to each other, that is, that both segments are contained in the same line. Algebraically, it can be translated into this identity:

AC = AB + BC

If we know that AB = 8 and BC = 24, then:

AC = 8 + 24

AC = 32

The length of AC is 32 units.

(ii) Let suppose that AB and AC are collinear to each other, that is, that both segments are contained in the same line. Algebraically, it can be translated into this identity:

AC = AB + BC

BC = AC - AB

If we know that AC = 68 and AB = 17, then:

BC = 68-17

BC = 51

The length of BC is 51 units.

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Anna11 [10]

Answer:

Step-by-step explanation:

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3 years ago
Show all work and reasoning
Natalija [7]
Split up the interval [2, 5] into n equally spaced subintervals, then consider the value of f(x) at the right endpoint of each subinterval.

The length of the interval is 5-2=3, so the length of each subinterval would be \dfrac3n. This means the first rectangle's height would be taken to be x^2 when x=2+\dfrac3n, so that the height is \left(2+\dfrac3n\right)^2, and its base would have length \dfrac{3k}n. So the area under x^2 over the first subinterval is \left(2+\dfrac3n\right)^2\dfrac3n.

Continuing in this fashion, the area under x^2 over the kth subinterval is approximated by \left(2+\dfrac{3k}n\right)^2\dfrac{3k}n, and so the Riemann approximation to the definite integral is

\displaystyle\sum_{k=1}^n\left(2+\frac{3k}n\right)^2\frac{3k}n

and its value is given exactly by taking n\to\infty. So the answer is D (and the value of the integral is exactly 39).
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