What is the distance between the points (7,3) and (-8, O) on a coordinate grid?

Mathematics

1 answer:

Georgia [21]3 years ago

5 0

Answer:15.29 or sqrt of 234

Step-by-step explanation:

Distance Formula

Sqrt of (-8-7)^2 + (0-3)^2

Sqrt of -15^2 + -3^2

Sqrt of 225 +9

Sqrt of 234 or 15.297

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125 pounds decreased by 4% what is the new amount

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By driving 8 mph faster than Bob, John can make a 230 mile trip in one half hour less. How fast does Bob drive on the trip? Roun

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Answer: The speed of Bob is 56.8 km/hr.

Step-by-step explanation:

Let the speed of Bob be 'x'.

Let the speed of John be 'x-8'.

Distance covered = 230 miles

time = $1\dfrac{1}{2}=\dfrac{3}{2}\ hr$

According to question, we get that

$\dfrac{230}{x}-\dfrac{230}{x+8}=\dfrac{3}{2}\\\\230\dfrac{x+8-x}{x(x+8)}=\dfrac{3}{2}\\\\\dfrac{230\times 8}{x^2+8x}=\dfrac{3}{2}\\\\\dfrac{1840}{x^2+8x}=\dfrac{3}{2}\\\\1840\times 2=3x^2+24x\\\\3680=3x^2+24x\\\\3x^2+24x-3680=0\\\\x\approx 56.8\ km/hr$

Hence, the speed of Bob is 56.8 km/hr.

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3 years ago

A ball is dropped from a height of 192 inches onto a level floor. After the third bounce it is still 3 inches off the ground. Pr

ANEK [815]

Answer: The required fraction = $\dfrac18$

Step-by-step explanation:

Let the required fraction = $\dfrac{p}{q}$

Given: Initial height = 192 inches

Height of ball after second bounce = $\dfrac{p}{q}\times192$

Height of ball after third bounce = $\dfrac{p}{q}\times\dfrac{p}{q}\times192=192\dfrac{p^2}{q^2}$

After the third bounce it is 3 inches off the ground.

So,

$(\dfrac{p}{q})^2192=3\\\\\\(\dfrac{p}{q})^2=\dfrac{3}{192}\\\\(\dfrac{p}{q})^2=\dfrac{1}{64}\\\\(\dfrac{p}{q})^2=(\dfrac{1}{8})^2\\\\ \dfrac{p}{q}=\dfrac{1}{8}$