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2 years ago

Graph: f(x) = (1/4)x

1 answer:

7 0

Answer: The initial value of a function is its value when equals zero. This is the same as the -intercept. We can see from the graph that our line crosses the -axis at the point zero, eight. Therefore, the initial value is eight.

Step-by-step explanation:

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Which fraction is not equivalent to the shaded area of the circle?

LuckyWell [14K]

Did this come with multiple choice answers?

4 0

3 years ago

Read 2 more answers

Suppose T is a transformation from ℝ2 to ℝ2. Find the matrix A that induces T if T is reflection over the line y=1/2x

trapecia [35]

Answer:

$A = \left[\begin{array}{cc}1&\frac{4}{5}\\\frac{4}{3}&-\frac{3}{5}\end{array}\right]$

Step-by-step explanation:

We have to see how the canonical vectors are transformed throught T. Lets first define T in any basis.

Since T is a reflection, then any element of the line y = x/2 if fixed by T. Therefore T(2,1) = (2,1).

On the other hand, any vector perpendicular to the line direction should be sent to its opposite value. We can take, for example, (-1,2) (note that the scalar product (2,1) * (-1,2) = -2+2 = 0). As a consecuence T(-1,2) = (1,-2). We have

T(2,1) = (2,1)
T(-1,2) = (1,-2)

By summing the first vector with the double of the second one we get, using linearity

T(0,5) = T( (2,1) + 2(-1,2)) = T(2,1) + 2T(-1,2) = (2,1) + 2(1,-2) = (4,-3)

Hence, T(0,1) = (4/5,-3/5)

Now, we take the second vector and substract it the double of the first one (to kill the second variable)

T(-3,0) = T( (-1,2) - 2*(2,1) ) = T(-1,2) -2T(2,1) = (1,-2)-2(2,1) = (-3,-4)

Therefore, T(1,0) = (1,4/3)

The matrix A induced by T has in its first column T(1,0) and in its second column T(0,1). We conclude that

$A = \left[\begin{array}{cc}1&\frac{4}{5}\\\frac{4}{3}&-\frac{3}{5}\end{array}\right]$

3 0

3 years ago

Perform the indicated operation and express the result as a simplified complex number. I need help with 35

skelet666 [1.2K]

Given:-

$\frac{3+4i}{2-i}$

To find:-

The simplified form.

At first we take conjucate and multiply below and above.

The conjucate is,

$2+i$

So now we multiply. we get,

$\frac{3+4i}{2-i}\times\frac{2+i}{2+i}$

Now we simplify. so we get,

$\frac{3+4i}{2-i}\times\frac{2+i}{2+i}=\frac{6+3i+8i+4(i)^2}{2^2-i^2}$

We know the value of,

$i^2=-1$

Substituting the value -1. we get,

$\begin{gathered} \frac{6+3i+8i+4(i)^2}{2^2-i^2}=\frac{6+3i+8i+4(-1)_{}^{}}{2^2-(-1)^{}} \\ \text{ =}\frac{6-4+11i}{2+1} \\ \text{ =}\frac{2+11i}{3} \end{gathered}$

So now we split the term to bring it into the form a+ib. so we get,