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3 years ago

Choose ASA, SAA, or neither to describe this figure. A B ASA SAA neither

Mathematics

1 answer:

ad-work [718]3 years ago

6 0

Answer:

SAA

Step-by-step explanation:

there is a side, and vertical angles equal eachother so it would be SAA

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This is a math question if you are good this is good for you this is hard

lubasha [3.4K]

Answer:

Hope this helps!

Step-by-step explanation:

1- -24

2- -23

3- -2

4- 11

5- 5

6- -3

7- 28

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7 0

3 years ago

Find the product 4.51 x 3.4

kifflom [539]

Multiply 4.51 by 3.4
That's all it is asking you

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3 years ago

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Integrate this two questions

tankabanditka [31]

Simplify the integrands by polynomial division.

$\dfrac{t^2}{1 - 3t} = -\dfrac19 \left(3t + 1 - \dfrac1{1 - 3t}\right)$

$\dfrac t{1 + 4t} = \dfrac14 \left(1 - \dfrac1{1 + 4t}\right)$

Now computing the integrals is trivial.

$\displaystyle \int \frac{t^2}{1 - 3t} \, dt = -\frac19 \int \left(3t + 1 - \frac1{1-3t}\right) \, dt \\\\ = -\frac19 \left(\frac32 t^2 + t + \frac13 \ln|1 - 3t|\right) + C \\\\ = \boxed{-\frac{t^2}6 - \frac t9 - \frac{\ln|1-3t|}{27} + C}$

where we use the power rule,

$\displaystyle \int x^n \, dx = \frac{x^{n+1}}{n+1} + C ~~~~ (n\neq-1)$

and a substitution to integrate the last term,

$\displaystyle \int \frac{dt}{1-3t} = -\frac13 \int \frac{du}u \\\\ = -\frac13 \ln|u| + C \\\\ = -\frac13 \ln|1-3t| + C ~~~ (u=1-3t \text{ and } du = -3\,dt)$

$\displaystyle \int \frac t{1+4t} \, dt = \frac14 \int \left(1 - \frac1{1+4t}\right) \, dt \\\\ = \frac14 \left(t - \frac14 \ln|1 + 4t|\right) + C \\\\ = \boxed{\frac t4 - \frac{\ln|1+4t|}{16} + C}$

using the same approach as above.

5 0

2 years ago

Just a few questions that I need help with.

zvonat [6]

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2: I just used a calculator: 2.25 * 5 = 11.25 kilometers

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3 years ago

Which two triangles are congruent by the ASA Theorem? Complete the congruence<br><br> statement

astraxan [27]

It would sure help a lot if I could take a peek at the drawings of the triangles.

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3 years ago

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