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Ne4ueva [31]
3 years ago
6

How can the Angle-Angle Similarity Postulate be used to prove the two triangles below are similar? Explain your answer using com

plete sentences, and provide evidence to support your claims

Mathematics
1 answer:
zloy xaker [14]3 years ago
5 0

To use the AA postulate directly, you need to show that two corresponding angles are congruent. In order to show that here, you must calculate the value of one of the missing angle measures. Either of the missing angles can be found by invoking the fact that the sum of angles in a triangle is 180°.

After finding either missing angle, you can show that the measures of two angles in one triangle are identical to the measures of two angles in the other triangle, hence the triangles are similar by the AA postulate.

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zzz [600]

Answer: z=9

Step-by-step explanation:

If you meant to solve for "z", then refer to the attachment below.

3 0
3 years ago
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A)<br> Identify all x-values not in the domain of<br> y=(x² – 4x²) / (3x² - 6x – 24)
True [87]

Answer:

x ≠ 4 or -2

Step-by-step explanation:

the denominator cannot be zero, so factor the bottom equation to get the zeros and those are the domain restrictions.

3x^2 - 6x - 24 ≠ 0

3(x^2 - 2x - 8) ≠ 0          (factor out a 3)

3(x - 4)(x + 2) ≠ 0            (factor equation)

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3 0
3 years ago
Determine (without solving the problem) an interval in which the solution of the given initial value problem is certain to exist
yuradex [85]

Answer:

0 < t < 5 is the required interval for the differential equation (t - 5)y' + (ln t)y = 6t to have a solution.

Step-by-step explanation:

Given the differential equation

(t - 5)y' + (ln t)y = 6t

and the condition y(1) = 6

We can rewrite the differential equation by dividing it by (t - 5) as

y' + [(ln t)/(t - 5)]y = 6t/(t - 5)

(ln t)/(t - 5) is continuous on the interval (0, 5) and (5, +infinity).

6t/(t - 5) is continuous on (-infinity, 5) and (5, +infinity)

We see that for these expressions, we have continuity at the intervals (0, 5) and (5, +infinity).

But the initial condition is y = 6, when t = 1.

The solution to differential equation is certain to exist at (0, 5)

Which implies that

0 < t < 5

is the required interval.

3 0
3 years ago
What is the value of the expression 30n when n=2
iris [78.8K]
Hey there!
Since it's 30n, 30 is the coefficient and it's being multiplied by n.
We plug in n.
30(2) =
60
Hope this helps!
8 0
3 years ago
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Simplify Expression <br> 3/5b - 2 + 7b - 12 - 2/5b + 14
Vinvika [58]

Answer:

The answer is 7 1/5 b

Step-by-step explanation:

8 0
2 years ago
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