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Valentin [98]
3 years ago
11

John was swimming for hour and a half hours at a speed of 2 miles per hour. How far did he swim?

Mathematics
2 answers:
alisha [4.7K]3 years ago
6 0

Answer: 1 mile.

Explanation: 1/2 of an hour is 1 mile because the speed was 2 miles an hour so if you do half of that it is 1 mile.

Dahasolnce [82]3 years ago
6 0

Answer:

Here's you're answer

I hope it's helpful

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c = the number of Nora's coworkers who attend the picnic v = the number of veggie burgers Nora prepares Which of the variables i
AveGali [126]

Answer: Independent variable = c

Dependent variable = v

Step-by-step explanation:

  • The independent variable does not depend on any other variable.
  • Dependent variable dependent on another variable.

Given variables:

c = the number of Nora's coworkers who attend the picnic

v = the number of veggie burgers

Here the number of burgers required is dependent on the number of Nora's coworkers who attend the picnic.

Therfeore, Independent variable = c

Dependent variable = v

3 0
3 years ago
Fill in the blank to make each equality true. x/(1/2x^3)=16x^6 y^4
inn [45]

Answer:

ur in rsm too? i have the same question sorry -_-

Step-by-step explanation:

7 0
3 years ago
1 + 1 x 346534565435 divided by 324395688392465 + 523984608796096299999999176 if u figure this out ill give brainliest, NO COPY
Ksju [112]

Answer:

Step-by-step explanation:

5.2398461e+26

6 0
3 years ago
Read 2 more answers
Simplify the expression. Show each step. -5 times 1 times 11 times 4
snow_lady [41]

STEP BY STEP:

-5 times 1 is -5

-5 times 11 which is positive makes it -55

-55 times 4 makes it -220

Hope this helps!


5 0
4 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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