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Lorico [155]
4 years ago
8

Equivalent expression with fewer terms to -9x-3x+15+x-4

Mathematics
2 answers:
kykrilka [37]4 years ago
3 0

Answer:

−11x+11

Step-by-step explanation:

Nikolay [14]4 years ago
3 0
-11x+11 I think lol I’m just solving it
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According to the rational root theorem, the numbers below are some of the potential roots of f(x) = 10x3 29x2 – 66x 27. select a
exis [7]

The actual roots of the function f(x)=10x^{3}+29x^{2} -66x+27 are -9/2, 3/5 and 1.

Given  function f(x)=10x^{3}+29x^{2} -66x+27.

Function is a relationship between two or more variables expressed in equal to form.

The roots of a polynomial function are the zeroes of the polynomial function. A polynomial function is a function that involves only non negative integer powers in an equation.

The polynomial function is given as:

f(x)=10x^{3}+29x^{2} -66x+27

factorize the above function

f(x)=(2x+9)(5x-3)(x-1)

Now put the function f(x) equal to zero.

f(x)=(2x+9)(5x-3)(x-1)

split the function means put all the expressions equal to zero as under:

(2x+9)(5x-3)(x-1)=0

solve each for the value of x

x=-9/2,x=3/5,x=1

Hence the roots of the function f(x)=10x^{3} +29x^{2} -66x+27 are which are also the values of x are -9/2,3/5,1.

Learn more about function at brainly.com/question/10439235

#SPJ4

6 0
2 years ago
Just giving out points for no reason lol
Firlakuza [10]

Answer:

Hey! Thanks a bunch!

Step-by-step explanation:

Hope you have a great day!!

5 0
3 years ago
Read 2 more answers
Suppose that T : R3 → R2 is given by:
Ad libitum [116K]

Answer:  The required answers are

(a) T is proved to be a linear transformation.

(b) The matrix A such that T(x) = Ax is \begin{pmatrix}1 & 0 &0 \\ 0 & 1 &0 \end{pmatrix}

Step-by-step explanation:  We are given a linear transformation T : R³ → R² defined as follows :

T(a,b,c)=(a,b).

We are to

(a) prove that T is a linear transformation

and

(b) find a matrix A such that T(x) = Ax.

(a) Let s, t are any real numbers and (a, b, c), (a', b', c') ∈ R³.

Then, we have

T(s(a,b,c)+t(a',b',c'))\\\\=T(sa+ta',sb+tb',sc+tc')\\\\=(sa+ta',sb+tb')\\\\=(sa,sb)+(ta'+tb')\\\\=s(a,b)+t(a',b')\\\\=sT(a,b,c)+tT(a',b',c').

So, we get

T(s(a,b,c)+t(a',b',c'))=sT(a,b,c)+tT(a',b',c').

Therefore, T is a linear transformation.

(b) We know that B = {(1, 0, 0), (0, 1, 0), (0, 0, 1)} is a standard basis for R³ and B' = {(1, 0), (0, 1)} is a standard basis for R².

So, we have

T(1,0,0)=(1,0)=1(1,0)+0(0,1),\\\\T(0,1,0)=(0,1)=0(1,0)+1(0,1),\\\\T(0,0,1)=(0,0)=0(1,0)+0(0,1).

So, the matrix A such that T(x) = Ax will be given by

\begin{pmatrix}1 & 0 &0 \\ 0 & 1 &0 \end{pmatrix}

Thus,

(a) T is proved to be a linear transformation.

(b) The matrix A such that T(x) = Ax is  \begin{pmatrix}1 & 0 &0 \\ 0 & 1 &0 \end{pmatrix}

4 0
3 years ago
If I got 89 marks out of 121 what is my % average
Strike441 [17]
Is marks = questions you got wrong?
6 0
3 years ago
Nine students choose integers. Here are seven of them:
solmaris [256]

Answer:

-20, -12, -8, -1, 1, 5, 10

Step-by-step explanation:

6 0
3 years ago
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