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____ [38]
3 years ago
7

Read the following adage.

Mathematics
1 answer:
Natasha_Volkova [10]3 years ago
4 0

Answer:

1

Step-by-step explanation:

Everyone sees beuty as a diiferent thing, aka, eye of the beholder,

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Help me please !<br> 4,5 and 6
andreev551 [17]

9514 1404 393

Answer:

  4a. ∠V≅∠Y

  4b. TU ≅ WX

  5. No; no applicable postulate

  6. see below

Step-by-step explanation:

<h3>4.</h3>

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

__

<h3>5.</h3>

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."

__

<h3>6.</h3>

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

8 0
3 years ago
Please help!!! I will mark brainliest :)))
zubka84 [21]

Answer:

its the first top one the 3rd one and the bottom one

Step-by-step explanation:

4 0
3 years ago
<img src="https://tex.z-dn.net/?f=%20%7C%20%7Bx%7D%5E%7B2%7D%20%2B%203x%20%2B2%20%7C%20%3C%202x%20%2B%204" id="TexFormula1" titl
Snezhnost [94]

\\ \tt\hookrightarrow |x^2+3x+2|

Either

\\ \tt\hookrightarrow x^2+3x+2

\\ \tt\hookrightarrow x^2+3x+2x

\\ \tt\hookrightarrow x^2+5x+6

\\ \tt\hookrightarrow (x+2)(x+3)

\\ \tt\hookrightarrow x

Or

\\ \tt\hookrightarrow x^2+3x+2

\\ \tt\hookrightarrow x^2+x-2

\\ \tt\hookrightarrow x^2+2x-x-2

\\ \tt\hookrightarrow (x+2)(x-1)

\\ \tt\hookrightarrow x

So

\\ \tt\hookrightarrow x\in (-3,1)

7 0
2 years ago
What is the ratio of the area of the triangle to the area of the rectangle?
Slav-nsk [51]
C

We can see that both triangles occupy exactly half of the rectangle each, so the area of one triangle is half of the area of the rectangle
7 0
3 years ago
Read 2 more answers
Exercise is mixed —some require integration by parts, while others can be integrated by using techniques discussed in the chapte
wlad13 [49]

Answer:

\frac{2x^{\frac{7}{2}}}{7}+4x+c

Step-by-step explanation:

We are asked to find the value of the indefinite integral \int \:x^2\sqrt{x}+4\:dx.

Let us solve our given problem applying sum rule as:

\int \:x^2\sqrt{x}\:dx+\int \:4\:dx

\int \:x^2*x^{\frac{1}{2}}\:dx+\int \:4\:dx

\int \:x^{\frac{4}{2}}*x^{\frac{1}{2}}\:dx+\int \:4\:dx

\int \:x^{\frac{5}{2}}\:dx+\int \:4\:dx

Now, we will use power rule.

=\frac{x^{\frac{5}{2}+1}}{\frac{5}{2}+1}}+4x+c

=\frac{x^{\frac{7}{2}}}{\frac{7}{2}}+4x+c

=\frac{2x^{\frac{7}{2}}}{7}+4x+c

Therefore, our required integral would be \frac{2x^{\frac{7}{2}}}{7}+4x+c.  

7 0
3 years ago
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