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2 years ago

Idk how to do this can you help me

Mathematics

1 answer:

slavikrds [6]2 years ago

4 0

Equilateral triangle are also equiangular which means all angles are 60 degrees. thus our needed altitude is the leg of a right triangle with other leg 4 and hypotenuse 8. 4² +a² = 8²

a² = 48
a=√48 = 4√3

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PLZ HELP I BEG DUE IN 30 MIN!!! 30 POINTS!!!!

katrin [286]

The total distance from A to B is 5 ( -3 to 2 = 5).

Using the ratio 1/5, split the distance in to 1/5th's, point K would be at -2.

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3 years ago

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soldier1979 [14.2K]

Answer:

1.12/30 which is equals to 2/5

2.9 more students

3.171*1.50 =$25.65

Step-by-step explanation:

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2 years ago

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Carpenters use a tool called a speed square to help them mark right angles . A speed square is a right triangle

Ronch [10]

Answer:

A. $Area = \frac{1}{2}(2x^2 - 2x - 24)$

B. $Area = 3.75 in^2$

Step-by-step explanation:

A. $Area = \frac{1}{2}bh$

Where,

$b = 2x + 6$

$h = x - 4$

Plug in the values to get a polynomial that represents the area of the tool

$Area = \frac{1}{2}(2x + 6)(x - 4)$

$Area = \frac{1}{2}(2x(x - 4) + 6(x - 4)$

$Area = \frac{1}{2}(2x^2 - 8x + 6x - 24)$

$Area = \frac{1}{2}(2x^2 - 2x - 24)$

B. To find area, when x = 4.5 in, plug in the value of x into the equation for the area of the tool.

$Area = \frac{1}{2}(2(4.5)2 - 2(4.5) - 24)$

$Area = \frac{1}{2}(2(20.25) - 9 - 24)$

$Area = \frac{1}{2}(40.5 - 9 - 24)$

$Area = \frac{1}{2}(7.5)$

$Area = 3.75 in^2$

8 0

3 years ago

Please explain and help me with this question.

k0ka [10]

it is so easy

Step-by-step explanation:

can i help you

5 0

2 years ago

A coin, having probability p of landing heads, is continually flipped until at least one head and one tail have been flipped. (a

Natali [406]

Answer:

(a)

The probability that you stop at the fifth flip would be

$p^4 (1-p) + (1-p)^4 p$

(b)

The expected numbers of flips needed would be

$\sum\limits_{n=1}^{\infty} n p(1-p)^{n-1} = 1/p$

Therefore, suppose that $p = 0.5$ , then the expected number of flips needed would be 1/0.5 = 2.

Step-by-step explanation:

(a)

Case 1

Imagine that you throw your coin and you get only heads, then you would stop when you get the first tail. So the probability that you stop at the fifth flip would be

$p^4 (1-p)$

Case 2

Imagine that you throw your coin and you get only tails, then you would stop when you get the first head. So the probability that you stop at the fifth flip would be

$(1-p)^4p$

Therefore the probability that you stop at the fifth flip would be

$p^4 (1-p) + (1-p)^4 p$

(b)

The expected numbers of flips needed would be

$\sum\limits_{n=1}^{\infty} n p(1-p)^{n-1} = 1/p$

Therefore, suppose that $p = 0.5$ , then the expected number of flips needed would be 1/0.5 = 2.

7 0

2 years ago