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Luda [366]
2 years ago
5

Solve for t. 4 (t + 1/4) = 3

Mathematics
1 answer:
tankabanditka [31]2 years ago
3 0

Answer:

t=1/2

Step-by-step explanation:

You might be interested in
1. if x and y are both even whole numbers. what is a possible solution for x + y?
vivado [14]

Answer:

1. A possible solution for x +y is 9

Step-by-step explanation:

A possible solution for m-n is 0

6 0
3 years ago
Write an equation perpendicular to the line y=3/2x-2 that goes through (-4,3)
Goryan [66]

Answer: y=-2/3x-2/3

Step-by-step explanation:

concept to know: two lines that are perpendicular has opposite reciprocal slopes.

y=-2/3x+b

in order to find b or the y-intercept, we need to plug in a point

3=-2/3(-4)+b

3=8/3+b

b=-2/3

y=-2/3x-2/3

Hope this helps!! :)

4 0
3 years ago
What is sum of two mixed numbers of 5.815+6.021
Julli [10]
5.815+6.021= 11.836


good luck!
3 0
3 years ago
If f(x)=2+abs(x-3) for all x, what's the value of the derivative f'(x) at x=3?
ruslelena [56]
No derivation at x=3 .
only at x=3
8 0
3 years ago
Use matrices and elementary row to solve the following system:
LiRa [457]

I assume the first equation is supposed to be

5x-3y+2z=13

and not

5x-3x+2x=4x=13

As an augmented matrix, this system is given by

\left[\begin{array}{ccc|c}5&-3&2&13\\2&-1&-3&1\\4&-2&4&12\end{array}\right]

Multiply through row 3 by 1/2:

\left[\begin{array}{ccc|c}5&-3&2&13\\2&-1&-3&1\\2&-1&2&6\end{array}\right]

Add -1(row 2) to row 3:

\left[\begin{array}{ccc|c}5&-3&2&13\\2&-1&-3&1\\0&0&5&5\end{array}\right]

Multiply through row 3 by 1/5:

\left[\begin{array}{ccc|c}5&-3&2&13\\2&-1&-3&1\\0&0&1&1\end{array}\right]

Add -2(row 3) to row 1, and add 3(row 3) to row 2:

\left[\begin{array}{ccc|c}5&-3&0&11\\2&-1&0&4\\0&0&1&1\end{array}\right]

Add -3(row 2) to row 1:

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

Multiply through row 1 by -1:

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

Add -2(row 1) to row 2:

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

Multipy through row 2 by -1:

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

The solution to the system is then

\boxed{x=1,y=-2,z=1}

5 0
3 years ago
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