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3 years ago

What are the coefficients in the polynomial 4x2 + 3x-3?

Mathematics

1 answer:

Ksenya-84 [330]3 years ago

4 0

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$\diamond\large\blue\textsf{\textbf{\underline{\underline{Question:-}}}}$

What are the coefficients in the polynomial 4x²+3x-3?

$\diamond\large\blue\textsf{\textbf{\underline{\underline{Answer and How to solve:-}}}}\diamond$

<u>Coefficient:-</u> A number before a variable

<u>Coefficients in this polynomial:-</u>

4 (the number before x²) and 3 (the number before x)

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So we conclude that the right option is:-

$\bigcirc\!\!\!\!\!\large\checkmark$ 4, 3 (Option B)

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What is the domain and range of h(t)=cot t

NeTakaya

Domain = R
Range = R .............

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4 years ago

Factorise 9w² - 100<br><br><br>

ohaa [14]

Answer:

$9\, w^{2} - 100 = (3\, w - 10) \, (3\, w + 10)$ .

Step-by-step explanation:

Fact:

$\begin{aligned} & (a - b)\, (a + b)\\ =\; & a^{2} + a\, b - a\, b - b^{2} \\ =\; & a^{2} - b^{2} \end{aligned}$ .

In other words, $(a^{2} - b^{2})$ , the difference of two squares in the form $a^{2}$ and $b^{2}$ , could be factorized into $(a - b)\, (a + b)$ .

In this question, the expression $(9\, w^{2} - 100)$ is the difference between two terms: $9\, w^{2}$ and $100$ .

$9\, w^{2}$ is the square of $3\, w$ . That is: $(3\, w)^{2} = 9\, w^{2}$ .
On the other hand, $10^{2} = 100$ .

Hence:

$9\, w^{2} - 100 = (3\, w)^{2} - (10)^{2}$ .

Apply the fact that $a^{2} - b^{2} = (a - b) \, (a + b)$ to factorize this expression. (In this case, $a = 3\, w$ whereas $b = 10$ .)

$\begin{aligned}& 9\, w^{2} - 100 \\ =\; & (3\, w)^{2} - (10)^{2} \\ = \; & (3\, w - 10)\, (3\, w + 10)\end{aligned}$ .

8 0

3 years ago

What's the square root of 2 rounded to the nearest tenth ?

vichka [17]

It is 1.4 when rounded

8 0

3 years ago

Read 2 more answers

Can anyone please help with these 2 questions. If you can, please show step-by-step.

valentinak56 [21]

Answer:

1. x = 21

2. m∡ABC = 51°

Step-by-step explanation:

First problem, solve for x

the sum of inside angles of a triangle is 180

also the supplementary angle for L = 180 - 100 is 80°

now you can add all angles

80 + 2x - 11 + 2x + 27 = 180

4x + 96 = 180

4x = 84

x = 21

Second problem, solve for m∡ABC

the sum of inside angles of a triangle is 180

also the supplementary angle for C = 180 - 148 is 32°

now you can add all angles

31 + 2x - 15 + x - 5 = 180

3x + 12 = 180

3x = 168

x= 56,

now solve for m∡ABC = (x - 5)° = (56 - 5)° = 51°

8 0

3 years ago

Read 2 more answers

Given that the expression 2x^3 + mx^2 + nx + c leaves the same remainder when divided by x -2 or by x+1 I prove that m+n =-6

Alla [95]

Given:

The expression is:

$2x^3+mx^2+nx+c$

It leaves the same remainder when divided by x -2 or by x+1.

To prove:

$m+n=-6$

Solution:

Remainder theorem: If a polynomial P(x) is divided by (x-c), thent he remainder is P(c).

Let the given polynomial is:

$P(x)=2x^3+mx^2+nx+c$

It leaves the same remainder when divided by x -2 or by x+1. By using remainder theorem, we can say that

$P(2)=P(-1)$ ...(i)

Substituting $x=-1$ in the given polynomial.

$P(-1)=2(-1)^3+m(-1)^2+n(-1)+c$

$P(-1)=-2+m-n+c$

Substituting $x=2$ in the given polynomial.

$P(2)=2(2)^3+m(2)^2+n(2)+c$

$P(2)=2(8)+m(4)+2n+c$

$P(2)=16+4m+2n+c$

Now, substitute the values of P(2) and P(-1) in (i), we get

$16+4m+2n+c=-2+m-n+c$

$16+4m+2n+c+2-m+n-c=0$

$18+3m+3n=0$

$3m+3n=-18$

Divide both sides by 3.

$\dfrac{3m+3n}{3}=\dfrac{-18}{3}$

$m+n=-6$

Hence proved.

7 0

3 years ago