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tankabanditka [31]
2 years ago
12

Two buses leave a station at the same time and travel in opposite directions. One bus travels12m/hrslower than the other. If the

two buses are 765 miles apart after 6 hours, what is the rate of each bus?
Mathematics
1 answer:
Alex Ar [27]2 years ago
6 0

distance = speed * time

the first bus speed = x

second bus = x - 12

the sum of distance = 765 miles

first distance = 6x

second distance = 6 (x -12)

6x + 6(x-12) = 765

x + x -12 = 127.5

2x = 139.5

x = 69.75 mi/h which is the faster bus

the second bus speed = 57.75 mi/h

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Y = (x+3)^2 - 5 here's the answer hope it helps
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4 years ago
Sabiendo que 16x10=160¿como resolverías, sin hacer la cuenta escrita los siguientes calculos?A_16x20=
-Dominant- [34]

Answer:

A_16x20=320

Step-by-step explanation:

Esto se puede hacer fácilmente de 2 formas.

1) 20 se puede dividir en un múltiplo de 10 y luego multiplicar

16 x 20 =16 x 10 x 2 = 160 x 2 = 320

2) Uso de cero.

Cualquier cosa multiplicada por cero da un cero y luego multiplicar el otro número da el resultado.

16 x 20 = (16 x 2) * 10 = (32) 10 = 320

Así que separamos el cero añadiéndole un 1, lo que lo convierte en 10 y obtenemos el mismo resultado.

This can be easily done in 2 ways .

1) 20 can be broken into a multiple of 10 and then multiplied

16 x 20 =16 x10 x 2=160x 2= 320

2) Use of zero.

Anything multiplied with zero gives a zero and then multiplying the other number gives the result.

16 x 20 = (16x 2) *10= (32)10= 320

So we separated the zero by adding a 1 to it thus making it 10 and we get the same result.

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3 years ago
The vertex of this angle is​
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4 years ago
Read 2 more answers
Please explain, thanks! :)<br><br> Multiply DE
Anit [1.1K]
First, we have to make sure that the number of columns in the first matrix is equal to the number of rows in the second matrix.

\left[\begin{array}{cc}1&-3&2&0\\\end{array}\right] *   \left[\begin{array}{ccc}2&3&4\\1&2&3\end{array}\right]

Since this is true, we can continue to solve the problem.
To multiply two matrices, multiply each row element in the first matrix by each column element in the second matrix. For example:
1*2 = 2
-3*1=-3
Then we add them to get our new matrix element.
-3+2=-1
Then we move to the next column of the second matrix.
1*3=3
-3*2=-6
-6+3=-3
Then the final column of the second matrix.
1*4=4
-3*3=-9
-9+4=-5
Our matrix so far:
\left[\begin{array}{ccc}-1&-3&-5\\x&x&x\end{array}\right]
We do the same for the bottom row of the first matrix.
<em>First Column</em>
2*2=4
0*1=0
4+0=4
<em>Second Column
</em>2*3=6
0*2=0
6+0=6
<em>Third Column</em>
2*4=8
0*3=0
8+0=8
Our final matrix is:
\left[\begin{array}{ccc}-1&-3&-5\\4&6&8\end{array}\right]

:)

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3 years ago
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Answer:

Is it 12?Yes,I think it's 12.

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