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Neko [114]
3 years ago
7

Which of the following statements best explains the difference between literal and figurative language? (4 points)

Mathematics
1 answer:
SOVA2 [1]3 years ago
4 0
2nd answer is the correct one
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Help me solve this problem and explain how you got the answer plz
Vitek1552 [10]
I hope this helps you

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- 1/5[y+x]<br> whats the answer
stiv31 [10]

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i aint never seen 2 pretty best friends always one of them gotta be ugly....

Step-by-step explanation:

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3 years ago
3. What is the probability of choosing a gray button, replacing it, and choosing gray again
Vikentia [17]

Answer:

36%

Please Mark Brainliest If This Helped!

8 0
2 years ago
Read 2 more answers
Please help will mark Brainly
SVETLANKA909090 [29]

Answer: C

Step-by-step explanation:

We are given the information that:

  1. The airplane is at 50,000 feet
  2. It descends 2,500 feet in 1 minute
  3. x = minutes

For  ease of solving this question, let's set the number of minutes (x) to 1.

  • In one minute, we know the plane will drop to 47,500 feet. (50,000 - 2,500(x))

There are multiple ways to write this equation bu shifting around negative and positive values. Another way to write this equation would be (-2,500(x) + 50,000). That would be choice C.

Try it out~

(-2,500(x) + 50,000)\\ (-2,500(1) + 50,000)\\ (-2,500 + 50,000)\\47,500

It works, so our answer is C.

5 0
3 years ago
A rock is thrown upward with an initial velocity of 16 ft/s from an initial height of 5 ft. How much time does it take for the r
Bogdan [553]
Check the picture below.

so it hits the ground when y = 0, thus

\bf ~~~~~~\textit{initial velocity}\\\\&#10;\begin{array}{llll}&#10;~~~~~~\textit{in feet}\\\\&#10;h(t) = -16t^2+v_ot+h_o&#10;\end{array} &#10;\quad &#10;\begin{cases}&#10;v_o=\stackrel{16}{\textit{initial velocity of the object}}\\\\&#10;h_o=\stackrel{5}{\textit{initial height of the object}}\\\\&#10;h=\stackrel{}{\textit{height of the object at "t" seconds}}&#10;\end{cases}

\bf h(t)=-16t^2+16t+5\implies \stackrel{h(t)}{0}=-16t^2+16t+5&#10;\\\\\\&#10;16t^2-16t-5=0\implies (4t+1)(4t-5)=0\\\\&#10;-------------------------------\\\\&#10;4t+1=0\implies 4t=-1\implies t=-\cfrac{1}{4}\\\\&#10;-------------------------------\\\\&#10;4t-5=0\implies 4t=5\implies t=\cfrac{5}{4}\implies \boxed{t=1\frac{1}{4}}

since "t" is seconds it took, it can't be a negative amount.

5 0
3 years ago
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