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Lorico [155]
3 years ago
13

Determine whether the function is a linear transformation. T: P2 → P2, T(a0 + a1x + a2x2) = (a0 + a1 + a2) + (a1 + a2)x + a2x2.

Mathematics
1 answer:
Anvisha [2.4K]3 years ago
3 0

Answer with Step-by-step explanation:

We are given that a function

T:P_2\rightarrow P_2

T(a_0+a_1x+a_2x^2)=(a_0+a_1+a_2)+(a_1+a_2)x+a_2x^2

We have to determine the given function is a linear transformation.

If a function is linear transformation then it satisfied following properties

1.T(x+y)=T(x)+T(y)

2.T(ax)=aT(x)

T(a_0+a_1x+a_2x^2+b_0+b_1x+b_2x^2)=T((a_0+b_0)+(a_1+b_1)x+(a_2+b_2)x^2)=(a_0+b_0+a_1+b_1+a_2+b_2)+(a_1+b_1+a_2+b_2)x+(a_2+b_2)x^2

T(a_0+a_1x+a_2x^2+b_0+b-1x+b_2x^2)=(a_0+a_1+a_2)+(a_1+a_2)x+a_2x^2+(b_0+b_1+b_2)+(b_1+b_2)x+b_2x^2

T(a_0+a_1x+a_2x^2+b_0+b-1x+b_2x^2)=T(a_0+a_1x+a_2x^2)+T(b_0+b_1x+b_2x^2)

T(a(a_0+a_1x+a_2x^2))=T(aa_0+aa_1x+aa_2x^2)

T(a(a_0+a_1x+a_2x^2))=(aa_0+aa_1+aa_2)+(aa_1+aa_2)x+(aa_2)x^2

T(a(a_0+a_1x+a_2x^2))=a(a_0+a_1+a_2)+a(a_1+a_2)x+aa_2x^2=a((a_0+a_1+a_2)+(a_1+a_2)x+a_2x^2)=aT(a_0+a_1x+a_2x^2)

Hence, the function is a linear transformation because it satisfied both properties of linear transformation.

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Answer with steps!... Thanks
NNADVOKAT [17]

Answer:

A)  False

B)  True

Step-by-step explanation:

For A) First let's do x/11 = 15. If you solved it by now, you will know that the answer is 0.7333333. So now you just do 11x15. Because the answer is different then the answer we got when doing the division, the answer for A is false

For B) B is definitely true because equation are any mathematical sentences that have an equation symbol. So it's like saying 8x3=24 and 6x4=24. These two equations end up with the same answer so B is true.

I hope this helped you find out the rest :)

5 0
3 years ago
​ What are the zeros of the function?​ f(x)=x2+17x+72
Semmy [17]
In order to find zeroes of a function, we will probably want to use our quadratic formula.

-b±√b^2-4(a)(c)/2a 

If we know our values, we can plug it in. 

Our values:

A=1 (Since there is no number in front of x, it is an assumed 1)
B=17
C=72

Now, We can plug it into our formula.

BE SURE TO PUT PARENTHESIS AROUND ALL TERMS!

-(17)±√(17)^2-4(1)(72)/2(1)

Now we can type it into a calculator! 

When we plug it into the formula. It gives us two real solutions (or zeroes) which are represented as:

-8 & -9.

4 0
3 years ago
PLEASE HELP
Assoli18 [71]

Simplest form I can do

5 0
3 years ago
Which of the following is/are cube(s) of even number<br> pls help fast :( :(
Andre45 [30]

Answer:

write the question properly and <u>completely</u> then only you will get correct answer

5 0
3 years ago
Let f(x) = (5)^x+1 .Evaluate f (-3) without using a calculator
rewona [7]

Answer:

f\left(-3\right)=\frac{126}{125}

Step-by-step explanation:

Given

f\left(x\right)\:=\:\left(5\right)^x+1

Putting x = -3 to find f(-3)

f\left(-3\right)\:=\:\left(5\right)^{\left(-3\right)}+1

as

5^{-3}=\frac{1}{125}       ∵  \mathrm{Apply\:exponent\:rule}:\quad \:a^{-b}=\frac{1}{a^b}

so

f\left(-3\right)=\frac{1}{125}+1

\mathrm{Convert\:element\:to\:fraction}:\quad \:1=\frac{1\cdot \:125}{125}

f\left(-3\right)=\frac{1\cdot \:\:125}{125}+\frac{1}{125}

\mathrm{Since\:the\:denominators\:are\:equal,\:combine\:the\:fractions}:\quad \frac{a}{c}\pm \frac{b}{c}=\frac{a\pm \:b}{c}

f\left(-3\right)=\frac{1\cdot \:\:125+1}{125}

f\left(-3\right)=\frac{126}{125}      ∵ 1\cdot \:125+1=126

Thus,

f\left(-3\right)=\frac{126}{125}

6 0
3 years ago
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