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2 years ago

100 points! Look at the picture attached. Random answers and links will be reported

Mathematics

1 answer:

____ [38]2 years ago

7 0

Answer and Step-by-step explanation:

4. We are finding line AC.

AB is congruent to BC because the angles at the top are both 30 because of angle bisector. BD is congruent to BD, so triangles are congruent by SAS theorem. That means AB is congruent to BC by CPCTC (Congruent Parts of Congruent Triangles are Congruent).

With that being said, we can equal the expressions together.

-x + 25 = 3(2x - 8)

-x + 25 = 6x - 24

49 = 7x

7 = x

-(7) + 25 = 18

3(2(7) - 8) = 3(14 - 8) = 3(6) = 18

18 + 18 = 36

AC = 36

6. We are finding measure of angle UTW.

We have another angle bisector, so that side is equal to itself (TV ≅ TV)

We have UV and VW having the same length, so those are congruent.

Since this is a right triangle, these triangles are congruent by the HL (Hypotenuse-Leg) Theorem.

So, now we can equal the angles together.

2x + 3 = 5x - 24

27 = 3x

9 = x

2(9) + 3 = 18 + 3 = 21

5(9) - 24 = 45 - 24 = 21

21 + 21 = 42

m∠UTW = 42

#teamtrees #PAW (Plant And Water)

(Sorry for the late response, this took some time).

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3 years ago

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Step-by-step explanation:

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3 years ago

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Mai is flying a plane at an altitude of 1600 ft. She sights a stadium at an angle of

LekaFEV [45]

Answer:

2,789ft

Step-by-step explanation:

The set up will give a right angled triangle.

The altitude will be the opposite side = 1600 ft

The angle of depression = 35°

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3 years ago

Evaluate the expression

babunello [35]

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Step-by-step explanation:

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2 years ago

Consider a discrete random variable x with pmf px (1) = c 3 ; px (2) = c 6 ; px (5) = c 3 and 0 otherwise, where c is a positive

PtichkaEL [24]

Looks like the PMF is supposed to be

$\mathbb P(X=x)=\begin{cases}\dfrac c3&\text{for }x\in\{1,5\}\\\\\dfrac c6&\text{for }x=2\\\\0&\text{otherwise}\end{cases}$

which is kinda weird, but it's not entirely clear what you meant...

Anyway, assuming the PMF above, for this to be a valid PMF, we need the probabilities of all events to sum to 1:

$\displaystyle\sum_{x\in\{1,2,5\}}\mathbb P(X=x)=\dfrac c3+\dfrac c6+\dfrac c3=\dfrac{5c}6=1\implies c=\dfrac65$

Next,

$\mathbb P(X>2)=\mathbb P(X=5)=\dfrac c3=\dfrac25$

$\mathbb E(X)=\displaystyle\sum_{x\in\{1,2,5\}}x\,\mathbb P(X=x)=\dfrac c3+\dfrac{2c}6+\dfrac{5c}3=\dfrac{7c}3=\dfrac{14}5$

$\mathbb V(X)=\mathbb E\bigg((X-\mathbb E(X))^2\bigg)=\mathbb E(X^2)-\mathbb E(X)^2$
$\mathbb E(X^2)=\displaystyle\sum_{x\in\{1,2,5\}}x^2\,\mathbb P(X=x)=\dfrac c3+\dfrac{4c}6+\dfrac{25c}3=\dfrac{28c}3=\dfrac{56}5$
$\implies\mathbb V(X)=\dfrac{56}5-\left(\dfrac{14}5\right)^2=\dfrac{84}{25}$

If $Y=X^2+1$ , then $X^2=Y-1\implies X=\sqrt{Y-1}$ , where we take the positive root because we know $X$ can only take on positive values, namely 1, 2, and 5. Correspondingly, we know that $Y$ can take on the values $1^2+1=2$ , $2^2+1=5$ , and $5^2+1=26$ . At these values of $Y$ , we would have the same probability as we did for the respective value of $X$ . That is,

$\mathbb P(Y=y)=\begin{cases}\dfrac c3&\text{for }y=2\\\\\dfrac c6&\text{for }y=5\\\\\dfrac c3&\text{for }y=26\\\\0&\text{otherwise}\end{cases}$

Part (5) is incomplete, so I'll stop here.

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3 years ago