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3 years ago

Kevin claims he can draw a trapezoid with three right angles. Is this possible? Explain

2 answers:

5 0

No this is not possible because if you are drawing a trapezoid you cannot put it in three angles it is kind for like maybe doing a triangle because it has three sides but if i am wrong i am sorry hope that helps

Tanzania [10]3 years ago

4 0

No because a trapezoid only has 2 parrelagram

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Which of the following is not a geometric sequence?

Leviafan [203]

A.

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3 years ago

PLEASE HELP - Part A Find the value of x in the following figure: Part B Find the measure of ACD

Margarita [4]

Answer:

Part A = 90

Part B = 145

Step-by-step explanation:

For part A, we know that a triangle has a sum of 180 degrees. So we just add the angles we already know and then subtract them from 180. For part B, we know that angle ACB and angle ACD are supplementary angles, meaning they add up to 180 degrees. So just take 180 and subtract 35 to find the measure of angle ACD.

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3 years ago

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3 years ago

Read 2 more answers

Find the slope of the line that passes through the points (1,6) and (5,6)

Oksi-84 [34.3K]

Answer:

m=0 (m is the slope)

Step-by-step explanation:

m=y2-y1/x2-x1

m=6-6/1-5

m=0/4

m=0

*Note: If the zero is in the denominator (on the bottom) the slope is undefined (UND).

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3 years ago

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.

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2 years ago