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3 years ago

Are the points U, H, and L collinear? U E S H

Mathematics

1 answer:

butalik [34]3 years ago

4 0

Answer:

Yes

Step-by-step explanation:

Collinear means points are lying on the same straight line. U, H, and L are all lying on the line l.

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How to find the volume of a quarter-sphere?

Ganezh [65]

As the word suggests, a quarter-sphere is a quarter of a sphere, and thus the volume of a quarter-sphere is a quarter of the volume of the sphere.

The volume of a sphere is computed as

$\dfrac{4}{3} \pi r^3$

So, the volume of a quarter-sphere is

$\dfrac{1}{4}\cdot \dfrac{4}{3} \pi r^3 = \dfrac{1}{3} \pi r^3$

7 0

3 years ago

( -4 ) 3÷( -4 ) -12

bazaltina [42]

Answer:

(-4)3/(-4)-12

-12/48

-1/4

Step-by-step explanation:

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3 years ago

How many minutes are in 3 hours and 10 minutes

mamaluj [8]

190 min.................................

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3 years ago

Read 2 more answers

An airline claims that the no-show rate for passengers is less than 5%. In a sample of 420 randomly selected reservations, 19 we

aleksandr82 [10.1K]

Answer:

An airline claims that the no-show rate for passengers is less than 5%. In a sample of 420 randomly selected reservations, 19 were no-shows. At α=0.01, test the airline's claim. State the sample percentage and round it to three decimal places.

State the hypotheses.

State the critical value(s).

State the test statistics.

State the decision

State the conclusion.

3 0

2 years ago

Find limit as x approaches 2 from the left of the quotient of the absolute value of the quantity x minus 2, and the quantity x m

Molodets [167]

Answer:

$\lim_{x \to 2^-} \frac{|x-2|}{x-2}=-1$ .

Step-by-step explanation:

We want to find $\lim_{x \to 2^-} \frac{|x-2|}{x-2}$ .

By definition:

$|x-2|=\left \{ {{x-2,\:if\:x\:>\:2} \atop {-(x-2),\:if\:x\:$

Since we want to find the Left Hand Limit, we use f(x)=-(x-2)

$\implies \lim_{x \to 2^-} \frac{|x-2|}{x-2}=\lim_{x \to 2} \frac{-(x-2)}{x-2}$ .

$\implies \lim_{x \to 2^-} \frac{|x-2|}{x-2}=\lim_{x \to 2} (-1)$ .

The limit of a constant is the constant.

$\implies \lim_{x \to 2^-} \frac{|x-2|}{x-2}=-1$ .

8 0

3 years ago