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nikklg [1K]
3 years ago
13

Find the value of the determinant. 4 8 10 4 0-1 3 - 2 4

Mathematics
1 answer:
VLD [36.1K]3 years ago
4 0

Answer:

- 240

Step-by-step explanation:

We have to find the value of the following determinant  

\left|\begin{array}{ccc}4&8&10\\4&0&-1\\3&-2&4\end{array}\right|

Now, we know that the value of a general determinant \left|\begin{array}{ccc}a&b&c\\d&e&f\\g&h&i\end{array}\right| is given by [a( ei - fh) + b(fg - di) + c(dh - eg)]

Therefore, the value of the given determinant is

= 4[0 × 4 - (-1)(-2)] + 8[3(-1) - 4 × 4] + 10[4(-2) - 3 × 0]

= - 8 - 152 - 80 = - 240 (Answer)

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3 years ago
Find an equation of a line that is tangent to the graph of f and parallel to the given line. Please see picture
ivanzaharov [21]

Answer:

y = 3x - 2 (smaller y-intercept)

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Step-by-step explanation:

First let's write the generic equation of a line:

y = ax + b

This line needs to be parallel to the line 3x - y + 5 = 0, so it needs to have the same slope of this line.

The line 3x - y + 5 = 0 has a slope of 3, so our line has a = 3:

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Now we need to find the values of b that make this line tangent to the function f(x) = x^3

Let's first find the derivative of f(x) in relation to x:

df(x)/dx = 3x^2

This derivative is the slope of the tangent line to the function for any value of x. We need a slope of 3, so:

3x^2 = 3

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Now, to find the y-values, we have:

f(1) = 1^3 = 1

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So, using the points (1,1) and (-1,-1) in our parallel line, we have:

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second line using (-1,-1) : -1 = -3*1 + b

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8 0
3 years ago
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Answer:

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Answer:

  A. (1, -2)

  B. the lines intersect at the solution point: (1, -2).

Step-by-step explanation:

A. The equations can be solve by substitution by using the y-expression provided by one of them to substitute for y in the other.

This gives ...

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