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3 years ago

ILL MARK BRAINLIST

Mathematics

2 answers:

Alik [6]3 years ago

7 0

Answer:

vertical angles

Step-by-step explanation:

hope this helps

Mkey [24]3 years ago

5 0

Answer:

vertical angles

Step-by-step explanation:

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Let f(x)=⌊x/2⌋. We learned that the floor and the ceiling functions are NOT invertible, but we also learned about the set of pre

romanna [79]

Answer:

Value of $f^{-1}(\{4\})$ is 8.

Step-by-step explanation:

Given function,

$f(x)=\big[\frac{x}{2}\big]$

we have to find : $f^{-1}(\{4\})$

It is known that,

$[x]$ =the greatest integer <=x

Then,

$f(x)=\big[\frac{x}{2}\big]$

$\implies x=f^{-1}(\big[\frac{x}{2}\big])$

Taking x=8 we get,

$f^{-1}(\big[\frac{x}{2}\big])=x$

$\implies f^{-1}(\big[\frac{8}{2}\big])=8$

$\implies f^{-1}([4])=8$

$\implies f^{-1}(\{4\})=8$ ( Since [4]=4 )

Hence the value of $f^{-1}(\{4\})$ is 8.

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3 years ago

Whats on the dba 08.08

Paul [167]

Answer:

Step-by-step explanation:

You haven't went into the assignment enough to say. Its asking for the date you spoke to the teacher and for you to submit that with the assignment.

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3 years ago

What Does C equal in

mina [271]

Square the entire thing which gets ride of the root so (8x+4)^2 now solve. 8x^2+4^2. 64x+16

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3 years ago

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Precalc help needed<br> convert y = x^2 − x −6 to vertex form

Nadya [2.5K]

I think it would it be -6,-4

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3 years ago

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T(v) = Av represents the linear transformation T. Find a basis for the kernel of T and the range of T.

lawyer [7]

Answer:

Ker(T) has basis $\emptyset$

Range(T) has basis $\left\{\left(\begin{array}{c}2&3\end{array}\right), \left(\begin{array}{c}1&4\end{array}\right) \right\}.$

Step-by-step explanation:

If the matrix you wanted to represent is $A=\left(\begin{array}{cc}2&1\\3&4\end{array}\right)$ , then:

$ker(T)=\left\{{\bf x}\in \mathbb{R}^{2}: A{\bf x}=0\right\}$ .

So, to find ker(T) you must solve the homogenous equation

$A{\bf x}=\left(\begin{array}{cc}2&1\\3&4\end{array}\right) \left(\begin{array}{c} x&y \end{array}\right)=\left(\begin{array}{c}0&0\end{array}\right)$

Using Gauss elimination we obtain the simpler equivalent system

$\left(\begin{array}{cc} -6&3\\0&5\end{array}\right) \left(\begin{array}{c} x&y\end{array}\right)=\left(\begin{array}{c}0&0\end{array}\right)$

Then, we have that

$x=0,y=0$ .

We have that $ker(T)=\{\left(\begin{array}{c}0&0\end{array}\right)\}$ . On this case we say that the basis is the empty set $\emptyset$ .

The range of $T$ is the set of vectors of the form

$\left(\begin{array}{c} \alpha &\beta\end{array}\right)=\left(\begin{array}{cc}2&1\\3&4\end{array}\right)\left(\begin{array}{c}x&y\end{array}\right)=x\left(\begin{array}{c}2&3\end{array}\right)+y\left(\begin{array}{c}1&4\end{array}\right)$

So,

$Range(T)=\langle \left(\begin{array}{c}2&3\end{array}\right), \left(\begin{array}{c}1&4\end{array}\right) \rangle.$ Where the angle brakets denotes the span.

5 0

3 years ago