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jekas [21]
2 years ago
8

A hot air balloon is dropping from a height above the ground at a steady rate of 8 feet per second. It starts from a height of 7

65 feet. After t-seconds it has hit a height of 421 feet. Set up and solve an equation for the value of t.
Mathematics
1 answer:
pickupchik [31]2 years ago
3 0
I believe the equation would be 765 -8t = 421. Then you would solve for t and get 43 seconds. I hope this helps!! :)
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Liz earns a salary of $2,200 per month, plus a commission of 3% of her sales. She wants to earn at least $2,800 this month. Ente
vagabundo [1.1K]

Answer:

The amount of sales in order to meet her goal must be greater than or equal to $20,000

Step-by-step explanation:

Let

x----> represent the amount of sales Liz will need

we know that

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so

The inequality that represent this situation is

0.03x+2,200\geq 2,800

solve for x

subtract 2,200 both sides

0.03x \geq 2,800-2,200

0.03x \geq 600

divide by 0.03 both sides

x \geq \$20,000

The amount of sales in order to meet her goal must be greater than or equal to $20,000

8 0
3 years ago
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Answer:C, B

Step-by-step explanation:

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3 years ago
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A science test, which is worth 100 points, consists of 24 questions. Each question is worth either 3 points or 5 points. If x is
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Find the six trigonometric function values for angle ∅ where its adjacent side is -9 and its hypotenuse is 41. (Theta is located
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Check the picture below.

\bf \textit{using the pythagorean theorem}
\\\\
c^2=a^2+b^2\implies \sqrt{c^2-a^2}=b
\qquad 
\begin{cases}
c=hypotenuse\\
a=adjacent\\
b=opposite\\
\end{cases}
\\\\\\
\sqrt{41^2-(-9)^2}=b\implies \sqrt{1681-81}=b\\\\\\ \sqrt{1600}=b\implies 40=b\\\\
-------------------------------

\bf sin(\theta )=\cfrac{\stackrel{opposite}{40}}{\stackrel{hypotenuse}{41}}\qquad~~  cos(\theta )=\cfrac{\stackrel{adjacent}{-9}}{\stackrel{hypotenuse}{41}}\qquad~~  tan(\theta )=\cfrac{\stackrel{opposite}{40}}{\stackrel{adjacent}{-9}}
\\\\\\
csc(\theta )=\cfrac{\stackrel{hypotenuse}{41}}{\stackrel{opposite}{40}}\qquad ~~sec(\theta )=\cfrac{\stackrel{hypotenuse}{41}}{\stackrel{adjacent}{-9}}\qquad ~~cot(\theta )=\cfrac{\stackrel{adjacent}{-9}}{\stackrel{opposite}{40}}

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3 years ago
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