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-Dominant- [34]
3 years ago
15

PLS HELP!!! YOU'LL BE A GRADE SAVER. I'LL IVE BRAINLIEST IF CORRECT. Triangle XYZ was dilated by a scale factor of 2 to create t

riangle ACB and tan ∠X = 5 over 2 and 5 tenths. Part A: Use complete sentences to explain the special relationship between the trigonometric ratios of triangles XYZ and ABC. You must show all work and calculations to receive full credit. (5 points) Part B: Explain how to find the measures of segments AC and CB. You must show all work and calculations to receive full credit. (5 points) Triangles XYZ and ACB; angles Y and C both measure 90 degrees, angles A and X are congruent.
Mathematics
1 answer:
tankabanditka [31]3 years ago
6 0

Answer:

Step-by-step explanation:

As for the angles of both triangles; they’re the same. The sides are 1:2.

I’m giving you formulas that are labeled: side a shortest

” b mid length

” c hypotenuse

angle α(alpha) opposite side a

” β(beta) ” ” b

” γ(gamma) ” ” c

A major formula for rt triangles is: a^2+b^2=c^2.

*Another is: a/sinα=b/sinβ=c/sinγ.

Remember α+β+γ=180°.

As for sides a&b use the above formula.

As for <ACB; the angle is γ which is a rt <.

Given: tan<x=5/2+1/2=6/2=3atan=71.565……….°=β. So α=18.44…….°. γ= rt angle.

To get the sides use the formulas at *.

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Discuss the continuity of the function on the closed interval.Function Intervalf(x) = 9 − x, x ≤ 09 + 12x, x &gt; 0 [−4, 5]The f
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Answer:

It is continuous since \lim_{x\to 0^{-}} = f(0) = \lim_{x \to 0^{+} f(x)

Step-by-step explanation:

We are given that the function is defined as follows f(x) = 9-x, x\leq 0 and f(x) = 9+12x, x>0 and we want to check the continuity in the interval [-4,5]. Note that this a piecewise function whose only critical point (that might be a candidate of a discontinuity)  x=0 since at this point is where the function "changes" of definition. Note that 9-x and 9+12x are polynomials that are continous over all \mathbb{R}. So F is continous in the intervals [-4,0) and (0,5]. To check if f(x) is continuous at 0, we must check that

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Note that if x=0, then f(x) = 9-x. So, f(0)=9. On the same time, note that

\lim_{x\to 0^{-}} f(x) = \lim_{x\to 0^{-}} 9-x = 9. This result is because the function 9-x is continous at x=0, so the left-hand limit is equal to the value of the function at 0.

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\lim_{x\to 0^{+}} f(x) = \lim_{x\to 0^{+}} 9+12x = 9. As before, this result is because the function 9+12x is continous at x=0, so the right-hand limit is equal to the value of the function at 0.

Thus, \lim_{x\to 0^{-}} = f(0) = \lim_{x \to 0^{+} f(x)=9, so by definition, f is continuous at x=0, hence continuous over the interval [-4,5].

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