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2 years ago

Can you pls help thanks

Mathematics

2 answers:

olga55 [171]2 years ago

7 0

Okay so if the question is asking for the area the answer is 13 times 5 is 65 and 65 divided by 2 is 32.5 so 32.5 is the area of this triangle. If the question is what is x then the answer is 5
I hope this helps! :)

Andrew [12]2 years ago

6 0

I think the answer you are looking for is the number 5. Hope I helped!

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Find the magnitude of AB. A(-2, 6), B(1, 10) O A 2 OB. ✔️15 O C.5 OD ✔️2

natita [175]

Answer:

C. 5

Step-by-step explanation:

Use the Distance Formula.

Substitute the values of x1 , y1 , x2 , and y2 .

|AB|² =|(1--2)²+(10-6)²|

|AB|² = |9+16|

|AB| = √ 25

|AB| =5

6 0

3 years ago

A student answers a multiple-choice examination question that offers four possible answers. Suppose the probability that the stu

Nitella [24]

Answer: 0.9730

Step-by-step explanation:

Let A be the event of the answer being correct and B be the event of the knew the answer.

Given: $P(A)=0.9$

$P(A^c)=0.1$

$P(B|A^{C})=0.25$

If it is given that the answer is correct , then the probability that he guess the answer $P(B|A)= 1$

By Bayes theorem , we have

$P(A|B)=\dfrac{P(B|A)P(A)}{P(B|A)P(A)+P(C|A^c)P(A^c)}$

$=\dfrac{(1)(0.9)}{(1))(0.9)+(0.25)(0.1)}\\\\=0.972972972973\approx0.9730$

Hence, the student correctly answers a question, the probability that the student really knew the correct answer is 0.9730.

7 0

3 years ago

I need help finishing a question. I used the explanation function on the program, but it didn't specify what to do to get the fi

leva [86]

I would compute sqrt(1 + 160pi^2) first to get approximately 39.75093337

Add this to 1 and we have 40.75093337

Then divide over 2pi to get a final approximate result of 6.48571248

So x = 6.48571248 is one approximate solution

In short, I computed $\frac{1+\sqrt{1+160\pi^2}}{2\pi}$ only focusing on the plus for now.

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If you were to compute $\frac{1-\sqrt{1+160\pi^2}}{2\pi}$ you should get roughly -6.167402596 as your other solution. Each solution can then be plugged into the original equation to check if you get 0 or not. You likely won't land exactly on 0 but you'll get close enough.