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jek_recluse [69]
3 years ago
15

A submarine elevation is -14 feet relative to sea level. A shark is swimming at elevation of -56 feet what is the difference in

evevation between the submarine and the shark
Mathematics
1 answer:
likoan [24]3 years ago
4 0

In order to solve this kind of problem, we must find the difference between two negative numbers describing elevation. Although when you have 2 negative numbers in a subtraction-elevation problem, the elevation difference won't be negative... It would be positive. -56 - (-14) would usually equal -42 feet but that would just describe a regular negative subtraction problem, so it becomes 42 feet. This answer is positive since it describe the distance between these integers.

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If a bird is flying at 30 meters above sea level and a fish is swimming 15 meters below sea, how many meters are between them? Y
barxatty [35]

Answer:

45 meters

Step-by-step explanation:

7 0
3 years ago
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The ratio of sixth grades to seventh grades in the club of science club is 4:7. If there are 20 sixth grades in the club, how ma
Vika [28.1K]

Answer: 35 seventh graders

Step-by-step explanation:

The simplified ratio is 4:7 sixth graders to seventh graders

The number of sixth-graders is 5 times the number shown in the ratio of attendees (20/4 = 5). Thus, the number of 7th graders must also be 5 times greater than the simplified ratio.

7*5 = 35, thus there are 35 seventh graders in the club

3 0
3 years ago
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I need help i will give 15 points and a Brainliest if its right
d1i1m1o1n [39]
A is the answer to the question
8 0
3 years ago
PLEASE HELP ASAP!!!!!
Akimi4 [234]

Answer:

Part A: 140

Step-by-step explanation:

Line ACE is a straight line meaning it equals 18 degrees. To find out what Angle CEF is you have to subtract 40 from 180. That gives you 140 degrees.

Hope its helpful!

3 0
3 years ago
If A = 1 2 1 1 and B= 0 -1 1 2 then show that (AB)^-1 = B^-1 A^-1<br><br><br> help meeeee plessss ​
Trava [24]

A = \begin{bmatrix}1&2\\1&1\end{bmatrix} \implies A^{-1} = \dfrac1{\det(A)}\begin{bmatrix}1&-1\\-2&1\end{bmatrix} = \begin{bmatrix}-1&1\\2&-1\end{bmatrix}

where det(<em>A</em>) = 1×1 - 2×1 = -1.

B = \begin{bmatrix}0&-1\\1&2\end{bmatrix} \implies B^{-1} = \dfrac1{\det(B)}\begin{bmatrix}2&1\\-1&0\end{bmatrix} = \begin{bmatrix}2&1\\-1&0\end{bmatrix}

where det(<em>B</em>) = 0×2 - (-1)×1 = 1. Then

B^{-1}A^{-1} = \begin{bmatrix}2&1\\-1&0\end{bmatrix} \begin{bmatrix}-1&1\\2&-1\end{bmatrix} = \begin{bmatrix}-1&3\\1&-2\end{bmatrix}

On the other side, we have

AB = \begin{bmatrix}1&2\\1&1\end{bmatrix} \begin{bmatrix}0&-1\\1&2\end{bmatrix} = \begin{bmatrix}2&3\\1&1\end{bmatrix}

and det(<em>AB</em>) = det(<em>A</em>) det(<em>B</em>) = (-1)×1 = -1. So

(AB)^{-1} = \dfrac1{\det(AB)}\begin{bmatrix}1&-3\\-1&2\end{bmatrix} = \begin{bmatrix}-1&3\\1&-2\end{bmatrix}

and both matrices are clearly the same.

More generally, we have by definition of inverse,

(AB)(AB)^{-1} = I

where I is the identity matrix. Multiply on the left by <em>A </em>⁻¹ to get

A^{-1}(AB)(AB)^{-1} = A^{-1}I = A^{-1}

Multiplication of matrices is associative, so we can regroup terms as

(A^{-1}A)B(AB)^{-1} = A^{-1} \\\\ B(AB)^{-1} = A^{-1}

Now multiply again on the left by <em>B</em> ⁻¹ and do the same thing:

B^{-1}\left(B(AB)^{-1}\right) = (B^{-1}B)(AB)^{-1} = B^{-1}A^{-1} \\\\ (AB)^{-1} = B^{-1}A^{-1}

7 0
3 years ago
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