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WITCHER [35]
3 years ago
8

Sara was on a cliff 25 feet above sea level. she jumped into the sea and descended a total of 37 feet. how many feet below sea l

evel did she travel? justify your reasoning
Mathematics
2 answers:
USPshnik [31]3 years ago
7 0
25 feet above sea level means she travelled 25 feet to the surface of the sea. She would have to descend (37-25=12) 12 more feet to achieve 37 total feet of travel, so she descended 12 feet below sea level.
aksik [14]3 years ago
5 0
 (37-25=12) 12  12 feet below sea level
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is it a multiple choice?

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3 years ago
There are 6 Programmers and 8 Analysts working on a team at CIA. What is the unsimplified ratio of Programmers to Analysts?
Rzqust [24]

Answer:  \dfrac{6}{8}

Step-by-step explanation:

Given

There are 6 Programmers and 8 Analysts working on a team at CIA.

Ratio of Programmers to analysts is

\Rightarrow \dfrac{6}{8}

Un-simplified ratio is \dfrac{6}{8}

Simplified ratio is \dfrac{3}{4}

6 0
2 years ago
1/8 plus 3/4: plz tell me I need help
xenn [34]
The answer should be 7/8
6 0
3 years ago
Read 2 more answers
Solve the initial value problem where y′′+4y′−21 y=0, y(1)=1, y′(1)=0 . Use t as the independent variable.
igor_vitrenko [27]

Answer:

y = \frac{7}{10} e^{3(t - 1)} + \frac{3}{10}e^{-7(t - 1)}

Step-by-step explanation:

y′′ + 4y′ − 21y = 0

The auxiliary equation is given by

m² + 4m - 21 = 0

We solve this using the quadratic formula. So

m = \frac{-4 +/- \sqrt{4^{2} - 4 X 1 X (-21))} }{2 X 1}\\ = \frac{-4 +/- \sqrt{16 + 84} }{2}\\= \frac{-4 +/- \sqrt{100} }{2}\\= \frac{-4 +/- 10 }{2}\\= -2 +/- 5\\= -2 + 5 or -2 -5\\= 3 or -7

So, the solution of the equation is

y = Ae^{m_{1} t} + Be^{m_{2} t}

where m₁ = 3 and m₂ = -7.

So,

y = Ae^{3t} + Be^{-7t}

Also,

y' = 3Ae^{3t} - 7e^{-7t}

Since y(1) = 1 and y'(1) = 0, we substitute them into the equations above. So,

y(1) = Ae^{3X1} + Be^{-7X1}\\1 = Ae^{3} + Be^{-7}\\Ae^{3} + Be^{-7} = 1      (1)

y'(1) = 3Ae^{3X1} - 7Be^{-7X1}\\0 = 3Ae^{3} - 7Be^{-7}\\3Ae^{3} - 7Be^{-7} = 0 \\3Ae^{3} = 7Be^{-7}\\A = \frac{7}{3} Be^{-10}

Substituting A into (1) above, we have

\frac{7}{3}B e^{-10}e^{3} + Be^{-7} = 1      \\\frac{7}{3}B e^{-7} + Be^{-7} = 1\\\frac{10}{3}B e^{-7} = 1\\B = \frac{3}{10} e^{7}

Substituting B into A, we have

A = \frac{7}{3} \frac{3}{10} e^{7}e^{-10}\\A = \frac{7}{10} e^{-3}

Substituting A and B into y, we have

y = Ae^{3t} + Be^{-7t}\\y = \frac{7}{10} e^{-3}e^{3t} + \frac{3}{10} e^{7}e^{-7t}\\y = \frac{7}{10} e^{3(t - 1)} + \frac{3}{10}e^{-7(t - 1)}

So the solution to the differential equation is

y = \frac{7}{10} e^{3(t - 1)} + \frac{3}{10}e^{-7(t - 1)}

6 0
3 years ago
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tresset_1 [31]

Answer:

x is smaller than 10in so x<5

Step-by-step explanation:

4 0
3 years ago
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