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2 years ago

Help me please ! 4,5 and 6

Mathematics

1 answer:

andreev551 [17]2 years ago

8 0

9514 1404 393

Answer:

4a. ∠V≅∠Y

4b. TU ≅ WX

5. No; no applicable postulate

6. see below

Step-by-step explanation:

a. When you use the ASA postulate, you are claiming you have shown two angles and the side between them to be congruent. Here, you're given side TV and angle T are congruent to their counterparts, sides WY and angle W. The angle at the other end of segment TV is angle V. Its counterpart is the other end of segment WY from angle W. In order to use ASA, we must show ...

∠V≅∠Y

b. When you use the SAS postulate, you are claiming you have shown two sides and the angle between them are congruent. The angle T is between sides TV and TU. The angle congruent to that, ∠W, is between sides WY and WX. Then the missing congruence that must be shown is ...

TU ≅ WX

The marked congruences are for two sides and a non-included angle. There is no SSA postulate for proving congruence. (In fact, there are two different possible triangles that have the given dimensions. This can be seen in the fact that the given angle is opposite the shortest of the given sides.)

"No, we cannot prove they are congruent because none of the five postulates or theorems can be used."

The first statement/reason is always the list of "given" statements.

1. ∠A≅∠D, AC≅DC . . . . given

2. . . . . vertical angles are congruent

3. . . . . ASA postulate

4. . . . . CPCTC

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2 years ago

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2 years ago

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wlad13 [49]

ANSWER

Eva is correct because 13/2 + 3/2 = 8

There is no mistake. Dakota may have thought that there was a mistake because Eva simplified when she got the answer for 4x which is correct but Dakota may have thought that it was incorrect.

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2 years ago

Read 2 more answers

Use the functions h(x) = 2x − 5 and t(x) = 6x + 4 to complete the function operations listed below.

Anika [276]

For this case we have the following functions:
h (x) = 2x - 5
t (x) = 6x + 4

Part A: (h + t) (x)
(h + t) (x) = h (x) + t (x)
(h + t) (x) = (2x - 5) + (6x + 4)
(h + t) (x) = 8x - 1

Part B: (h ⋅ t) (x)
(h ⋅ t) (x) = h (x) * t (x)
(h ⋅ t) (x) = (2x - 5) * (6x + 4)
(h ⋅ t) (x) = 12x ^ 2 + 8x - 30x - 20
(h ⋅ t) (x) = 12x ^ 2 - 22x - 20

Part C: h [t (x)]
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h [t (x)] = 12x + 8 - 5
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2 years ago

Identify the horizontal asymptote of f(x) = quantity 2 x minus 1 over quantity x squared minus 7 x plus 3.

Rina8888 [55]

We must recall that a horizontal asymptote is the value/s of y that the given function approaches to but never reaches. To find this in a rational function, we compare the expressions with highest degree in the numerator and denominator. There are three possible outcome when this happens.

1. if the highest degree (highest exponent) in the numerator is bigger than that of the denominator, then there won't be any horizontal asymptote.

2. if the highest degree in the denominator is bigger, then the horizontal symptote would be y = 0.

3. if they have the same highest degree, then we just get the quotient of their coefficient.

Now, going back to our function, we have

$f(x) = \frac{2x - 1}{x^{2} - 7x + 3}$

From this we can see that the highest degree in the numerator is 1 (from 2x) and 2 (from x²) for the denominator. Clearly, it shows that its denominator has a higher degree. And from our discussion, we can conclude that the horizontal asymptote would be y = 0.

Answer: y = 0

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