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3 years ago

What is the approximate length of the hypotenuse of a right triangle with vertices (-6, 2),

Mathematics

1 answer:

muminat3 years ago

4 0

Answer:

Step-by-step explanation:

A(-6, 2), B(6,-3) and C(-6, -3)

AB² = (x₂ - x₁)² + (y₂ - y₁)²

= ( 6 -[-6])² + ( -3 -2)²

= ( 6 + 6)² + ( -3 -2)² = 12² + (-5)² = 144 + 25 =169

AB = √169 = 13 units

BC² = ( -6 -6)² + ( -3 - [-3])² = (-6-6)² + (-3 +3)²

= (-12)² + 0 = 144

BC = √144 = 12 unis

CA² = (-6 - [-6])² +(-3-2)² = (-6 + 6)² + ( -3-2)²

= 0 + (-5)² = 25

CA =√25 = 5 units

length of the hypotenuse of a right triangle = 13units

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3 years ago

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morpeh [17]

Answer:

245 days per year ; 1715 hours per year

Step-by-step explanation:

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3 years ago

Find the distance between (1, 4) and (-2, 0)

Vanyuwa [196]

Answer:

$\boxed {d = 5}$

Step-by-step explanation:

Use the Distance Formula to help you determine the distance between two given points:

$d = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}}$

First point: $(x_{1}, y_{1})$

Second Point: $(x_{2}, y_{2})$

-Apply the two given points onto the formula:

First point: $(1, 4)$

Second point: $(-2, 0)$

$d = \sqrt{(-2 - 1)^{2} + (0 - 4)^{2}}$

-Solve for the distance:

$d = \sqrt{(-2 - 1)^{2} + (0 - 4)^{2}}$

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3 years ago

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3 years ago

You always need some time to get up after the alarm has rung. You get up from 10 to 20 minutes later, with any time in that inte

Mama L [17]

Answer:

a) P(x<5)=0.

b) E(X)=15.

c) P(8<x<13)=0.3.

d) P=0.216.

e) P=1.

Step-by-step explanation:

We have the function:

$f(x)=\left \{ {{\frac{1}{10},\, \, \, 10\leq x\leq 20 } \atop {0, \, \, \, \, \, \, otherwise }} \right.$

a) We calculate the probability that you need less than 5 minutes to get up:

$P(x$

Therefore, the probability is P(x<5)=0.

b) It takes us between 10 and 20 minutes to get up. The expected value is to get up in 15 minutes.

E(X)=15.

c) We calculate the probability that you will need between 8 and 13 minutes:

$P(8\leq x\leq 13)=P(10\leqx\leq 13)\\\\P(8\leq x\leq 13)=\int_{10}^{13} f(x)\, dx\\\\P(8\leq x\leq 13)=\int_{10}^{13} \frac{1}{10} \, dx\\\\P(8\leq x\leq 13)=\frac{1}{10} \cdot [x]_{10}^{13}\\\\P(8\leq x\leq 13)=\frac{1}{10} \cdot (13-10)\\\\P(8\leq x\leq 13)=\frac{3}{10}\\\\P(8\leq x\leq 13)=0.3$

Therefore, the probability is P(8<x<13)=0.3.

d) We calculate the probability that you will be late to each of the 9:30am classes next week:

$P(x>14)=\int_{14}^{20} f(x)\, dx\\\\P(x>14)=\int_{14}^{20} \frac{1}{10} \, dx\\\\P(x>14)=\frac{1}{10} [x]_{14}^{20}\\\\P(x>14)=\frac{6}{10}\\\\P(x>14)=0.6$

You have 9:30am classes three times a week. So, we get:

$P=0.6^3=0.216$

Therefore, the probability is P=0.216.

e) We calculate the probability that you are late to at least one 9am class next week:

$P(x>9.5)=\int_{10}^{20} f(x)\, dx\\\\P(x>9.5)=\int_{10}^{20} \frac{1}{10} \, dx\\\\P(x>9.5)=\frac{1}{10} [x]_{10}^{20}\\\\P(x>9.5)=1$

Therefore, the probability is P=1.

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3 years ago