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-Dominant- [34]

3 years ago

15

janet makes the conjecture that line p is also parallel to line r. what must the value of y be for her conjecture to be correct?

2 answers:

Llana [10]3 years ago

4 0

Answer:

Step-by-step explanation:

:)

solmaris [256]3 years ago

3 0

Answer:

The answer is 6.75

Step-by-step explanation:

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Problem: a circle is inscribed in a square. The squares perimeter is 16 times the square root of 6. What is the probability of s

Oksi-84 [34.3K]

I got 21% at selecting a point outside.
Explanation: Since the perimeter is 16sqrt (6), each side must be 4sqrt (6)
This also means that the diameter is 4sqrt (6).

To find the area of a circle, we need to find the radius (2sqrt (6)).
Area of circle is pi × (2sqrt (6))^2 = 24pi or approximately 75.4 units^2
Area of square is 4sqrt (6)^2 = 96

Thus, let's take the area of the circle and subtract that from the area of our square to yield approx. 20.6 units^2
and now divide through by 96 to yield 21%

8 0

4 years ago

Not sure if any of this is correct, but it’s what I got so far

Irina18 [472]

Problem 1 is correct. You use the pythagorean theorem to find the hypotenuse.

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Problem 2 has the correct answer, but one part of the steps is a bit strange. I agree with the 132 ft/sec portion; however, I'm not sure why you wrote $\frac{1 \text{ sec}}{132 \text{ ft}}=\frac{0.59\overline{09}}{78 \text{ ft}}*127 \text{ ft}$

I would write it as $\frac{1\text{ sec}}{132 \text{ ft}}*127 \text{ ft} = \frac{127}{132} \text{ sec} \approx 0.96 \text{ sec}$

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For problem 3, we first need to convert the runner's speed from mph to feet per second.

$17.5 \text{ mph} = \frac{17.5 \text{ mi}}{1 \text{ hr}}*\frac{1 \text{ hr}}{60 \text{ min}}*\frac{1 \text{ min}}{60 \text{ sec}}*\frac{5280 \text{ ft}}{1 \text{ mi}} \approx 25.667 \text{ ft per sec}$

Since the runner needs to travel 90-12 = 78 ft, this means $\text{time} = \frac{\text{distance}}{\text{speed}} \approx \frac{78 \text{ ft}}{25.667 \text{ ft per sec}} \approx 3.039 \text{ sec}$

So the runner needs about 3.039 seconds. In problem 2, you calculated that it takes about 0.96 seconds for the ball to go from home to second base. The runner will not beat the throw. The ball gets where it needs to go well before the runner arrives there too.

-------------

The question is now: how much of a lead does the runner need in order to beat the throw?

Well the runner needs to get to second base in under 0.96 seconds.

Let's calculate the distance based on that, and based on the speed we calculated earlier above.

$\text{distance} = \text{rate}*\text{time} \approx (25.667 \text{ ft per sec})*(0.96 \text{ sec}) \approx 24.64032 \text{ ft}$

This is the distance the runner can travel if the runner only has 0.96 seconds. So the lead needed is 90-24.64032 = 65.35968 feet

This is probably not reasonable considering it's well over halfway (because 65.35968/90 = 0.726 = 72.6%). If the runner is leading over halfway, then the runner is probably already in the running motion and not being stationary.

As you can see, the runner is very unlikely to steal second base. Though of course such events do happen in real life. What may explain this is the reaction time of the catcher may add on just enough time for the runner to steal second base. For this problem however, we aren't considering the reaction time. Also, not all catchers can throw the ball at 90 mph which is quite fast. According to quick research, the MLB says the average catcher speed is about 81.8 mph. This slower throwing speed may account for why stealing second base isn't literally impossible, although it's still fairly difficult.

5 0

3 years ago

WHAT IS X³-27 SIMPLIFIED

Eduardwww [97]

Answer:

<u>It</u><u> </u><u>is</u><u> </u><u>(</u><u>x</u><u> </u><u>-</u><u> </u><u>3</u><u>)</u><u>³</u><u> </u><u>-</u><u> </u><u>9</u><u>x</u><u>(</u><u>3</u><u> </u><u>-</u><u> </u><u>x</u><u>)</u>

Step-by-step explanation:

Express 27 in terms of cubes, 27 = 3³:

$= {x}^{3} - {3}^{3}$

From trinomial expansion:

${(x - y)}^{3} = (x - y)(x - y)(x - y) \\$

open first two brackets to get a quadratic equation:

${(x - y)}^{3} = ( {x}^{2} - 2xy + {y}^{2} )(x - y)$

expand further:

${(x - y)}^{3} = {x}^{3} - y {x}^{2} - 2y {x}^{2} + 2x {y}^{2} + x {y}^{2} - {y}^{3} \\ {(x - y)}^{3} = {x}^{3} - {y}^{3} + 3x {y}^{2} - 3y {x}^{2} \\ {(x - y)}^{3} = {x}^{3} - {y}^{3} + 3xy(y - x) \\ \\ { \boxed{( {x}^{3} - {y}^{3} ) = {(x - y)}^{3} - 3xy(y - x)}}$

take y to be 3, then substitute:

$( {x}^{3} - 3^3) = {(x - 3)}^{3} - 9x(3 - x)$

5 0

3 years ago

Sonja's house is 4 blocks west and 1 block south of the center of town. Her school is 3 blocks east and 2 blocks north of the ce

adoni [48]

Answer:

The answer is √58 = 7.62

Step-by-step explanation:

So, if you draw this down, you will see that the direct distance is hypotenuse c of a right triangle which sides are 3 blocks (1 south and 2 north) and 7 blocks (4 west and 3 east).

Use the Pythagorean theorem:

c² = a² + b²

a = 3

b = 7

c² = 3² + 7²

c² = 9 + 49

c² = 58

c = √58

c = 7.62

8 0

4 years ago

Read 2 more answers

How do you solve this

olchik [2.2K]

The answer would have to be 4x^2y

4 0

4 years ago