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densk [106]
2 years ago
15

The tire on a mountain bike measures 30 in. in diameter. A rider picks up a thorn in her tire as she passes through a patch of g

oathead thorns. If she is traveling at a rate of 10 MPH, find the equation that would best model the height above ground (inches) that the thorn is at any time (seconds) from the time it first got on the tire. Make sure to think about which equation would best model the situation ie. y = sinx, y = cosx, y = -cosx
Mathematics
1 answer:
mixas84 [53]2 years ago
5 0

Answer:

yrdjdhngessrhmsrymftfyhstdfkdtjyrarddtgjgbfffhr jganntfbeakye no tengo ru tkm re e

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The product 3/4 times 1/2 will be less than or greater than 3/4
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Answer:

less

Step-by-step explanation:

first you would multiply and get 4/8, which simplifies to 1/2, which would be 2/4, which is less than 3/4

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2 years ago
A building is 3 ft from an 8​-ft fence that surrounds the property. A worker wants to wash a window in the building 13 ft from t
zloy xaker [14]

Answer:

Length of the ladder used by worker = 17 feet

Step-by-step explanation:

Given:

Height of the window from the ground = 13 ft

Distance of fence from the building = 3 ft

Distance of ladder from the building = (3+8) = 11 ft

We have to find the length of the ladder.

Let the length of the ladder be 'x'

From the diagram we can also say that 'x' is the hypotenuse of the right angled triangle.

Using Pythagoras formula:

⇒ hypotenuse\ 'x' =\sqrt{(perpendicular)^2+(base)^2}

Here base length = 11 ft

Perpendicular = 13 ft

Plugging the values:

⇒ x=\sqrt{(13)^2+(11)^2}

⇒  x=\sqrt{(169+121)}

⇒ x= \sqrt{290}

⇒ x=17.02 feet

The length of the ladder = 17 feet to its nearest tenth.

6 0
3 years ago
Please help If using the method of completing the square to solve the quadratic equation
Luba_88 [7]
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PLEASE HELP! Find the missing part.
viktelen [127]

Answer:

Step-by-step explanation:

\frac{y}{a}= tan~60=\sqrt{3} \\y=a\sqrt{3} ~~~(1)\\\frac{y}{b} =tan~30\\y=\frac{b}{\sqrt{3} } ~~~(2)\\from~(1)~and~(2)\\a\sqrt{3} =\frac{b}{\sqrt{3} } \\b=a\sqrt{3} \times\sqrt{3} =3a\\a+b=15\\a+3a=15\\4a=15\\a=\frac{15}{4} \\b=3a=3 \times \frac{15}{4} =\frac{45}{4}\\y=a\sqrt{3} =\frac{15\sqrt{3}}{4} \\\frac{x}{15} =sin~30\\x=15 \times \frac{1}{2}=\frac{15}{2}\\\frac{z}{15}=cos~30\\z=15 \times \frac{\sqrt{3}}{2}=\frac{15\sqrt{3} }{2} \\

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2 years ago
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