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klemol [59]
3 years ago
12

Find the value of b that makes the system of equations have the solution (3 ,5).

Mathematics
1 answer:
dexar [7]3 years ago
4 0
5=b\cdot3+2\\
3b=3\\
b=1 \Rightarrow A
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Find an equation for the line that passes through the points (1. - 4) and (-5.4).
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Answer:

y=-4/3x-8/3

Step-by-step explanation:

Use the slope formula y2-y1/x2-x1

Finding the slope, 4-(-4)/-5-1 = 8/-6 = -4/3

y=-4/3x + b

Substituting,

-4=-4/3+b

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b=-8/3

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Theo buys 5 packages of paper. There are 500 sheets of paper in each package. How many sheets of paper does Theo buy?
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10 POINTS!!! FULL ANSWER WITH FULL STEP BY STEP SOLUTION PLEASE. DO BOTH PARTS OF 1 AND ALL OF 2.
diamong [38]
One A
y = e^x
dy/dx = e^x The f(x) = the differentiated function. Any value that e^x can have, the derivative has the same value. x is contained in all the reals.
One B
y = x*e^x
y' = e^x + xe^x Using the multiplication rule.
You want the slope and the value of the of y to be the same. The slope is y' of the tangent line
xe^x = e^x + xe^x
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This happens only when x is very "small" like x = - 4444444

y = e^x * ln(x) Using the multiplication rule again, we need the slope of the line with is y'
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y1' = e^x
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y' = e^x*ln(x) + e^x/x So at x = 1 the slope of the line = 
y' = e^1*ln(1) + e^1/1
y' = e*0+e  = e
y = mx + b
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3 years ago
These tables of value represent continuous functions. In which table do the values represent an exponential function?
oksano4ka [1.4K]

Answer:

B.

Step-by-step explanation:

A goes up by 1

C goes up by 8

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5 0
3 years ago
Find the inverse of the given​ matrix, if it exists.Aequals=left bracket Start 3 By 3 Matrix 1st Row 1st Column 1 2nd Column 0 3
BabaBlast [244]

Answer:

A^{-1}=\left[ \begin{array}{ccc} \frac{1}{9} & \frac{4}{27} & - \frac{2}{27} \\\\ \frac{8}{9} & \frac{5}{27} & \frac{11}{27} \\\\ - \frac{4}{9} & \frac{2}{27} & - \frac{1}{27} \end{array} \right]

Step-by-step explanation:

We want to find the inverse of A=\left[ \begin{array}{ccc} 1 & 0 & -2 \\\\ 4 & 1 & 3 \\\\ -4 & 2 & 3 \end{array} \right]

To find the inverse matrix, augment it with the identity matrix and perform row operations trying to make the identity matrix to the left. Then to the right will be inverse matrix.

So, augment the matrix with identity matrix:

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

  • Subtract row 1 multiplied by 4 from row 2

\left[ \begin{array}{ccc|ccc}1&0&-2&1&0&0 \\\\ 0&1&11&-4&1&0 \\\\ -4&2&3&0&0&1\end{array}\right]

  • Add row 1 multiplied by 4 to row 3

\left[ \begin{array}{ccc|ccc}1&0&-2&1&0&0 \\\\ 0&1&11&-4&1&0 \\\\ 0&2&-5&4&0&1\end{array}\right]

  • Subtract row 2 multiplied by 2 from row 3

\left[ \begin{array}{ccc|ccc}1&0&-2&1&0&0 \\\\ 0&1&11&-4&1&0 \\\\ 0&0&-27&12&-2&1\end{array}\right]

  • Divide row 3 by −27

\left[ \begin{array}{ccc|ccc}1&0&-2&1&0&0 \\\\ 0&1&11&-4&1&0 \\\\ 0&0&1&- \frac{4}{9}&\frac{2}{27}&- \frac{1}{27}\end{array}\right]

  • Add row 3 multiplied by 2 to row 1

\left[ \begin{array}{ccc|ccc}1&0&0&\frac{1}{9}&\frac{4}{27}&- \frac{2}{27} \\\\ 0&1&11&-4&1&0 \\\\ 0&0&1&- \frac{4}{9}&\frac{2}{27}&- \frac{1}{27}\end{array}\right]

  • Subtract row 3 multiplied by 11 from row 2

\left[ \begin{array}{ccc|ccc}1&0&0&\frac{1}{9}&\frac{4}{27}&- \frac{2}{27} \\\\ 0&1&0&\frac{8}{9}&\frac{5}{27}&\frac{11}{27} \\\\ 0&0&1&- \frac{4}{9}&\frac{2}{27}&- \frac{1}{27}\end{array}\right]

As can be seen, we have obtained the identity matrix to the left. So, we are done.

6 0
3 years ago
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