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3 years ago

5

What are the zeros of the polynomial function f(x)=x^2+5x+6

2 answers:

Molodets [167]3 years ago

8 0

Find 2 numbers that multiply to 6 and add to 5.

3*2 = 6
3+2 = 5

x^2 + 5x+6 = (x+3)(x+2) = 0

Set each factor equal to zero.
x+3 = 0 ----> x = -3
x+2 = 0 -----> x = -2

Answer:
There are two zeros at -2 and -3.

Alika [10]3 years ago

5 0

Answer:

It’s -1 and 6 the likes above threw me off why are u liking when its wrong?

Step-by-step explanation:

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Use the algebraic procedure explained in section 8.9 in your book to find the derivative of f(x)=1/x. Use h for the small number

Triss [41]

Answer:

By definition, the derivative of f(x) is

$lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$

Let's use the definition for $f(x)=\frac{1}{x}$

$lim_{h\rightarrow 0} \frac{\frac{1}{x+h}-\frac{1}{x}}{h}=\\lim_{h\rightarrow 0} \frac{\frac{x-(x+h)}{x(x+h)}}{h}=\\lim_{h\rightarrow 0} \frac{\frac{(-1)h}{x^2+xh}}{h}=\\lim_{h\rightarrow 0} \frac{(-1)h}{h(x^2+xh)}=\\lim_{h\rightarrow 0} \frac{-1}{x^2+xh)}=\frac{-1}{x^2+x*0}=\frac{-1}{x^2}$

Then, $f'(x)=\frac{-1}{x^2}$

7 0

3 years ago

Given that f(x) = x2 + 3x + 6 and g(x) = the quantity of x plus two, over three, solve for f(g(x)) when x = 1

MaRussiya [10]

Easy
f(g(1))
evaluate g(1) then plug thatin for x in f(x)
g(1)=(x+2)/3
g(1)=(1+2)/3
g(1)=1

f(g(1))=
f(1)=(1)^2+3(1)+6
f(1)=1+3+6
f(1)=10
f(g(1))=10

3 0

3 years ago

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The following triangle ABC is rotated clockwise about the origin the new coordinates are A'(1,-2) B'(4,6) and C'(-2,3). Which of

Yanka [14]

B.90 because 90 degrees is above the original angel

5 0

3 years ago

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How do you simplify this and write as a polynomial in standar form? (x-9)3x-7)+(3x^2-5x+2)

nasty-shy [4]

Answer:

Standard form of $(x-9)(3x-7) + (3x^{2} - 5x+2)$ $=6x^{2} - 39x + 65$

Step-by-step explanation:

Here, the given expression is $(x-9)(3x-7) + (3x^{2} - 5x+2)$

Now, simplifying the above expression in parts, we get

$(x-9)(3x-7) = 3x^{2} - 7x -27x + 63 = 3x^{2} - 34x + 63$

hence, combining both parts:

$(x-9)(3x-7) + (3x^{2} - 5x+2)$ $=(3x^{2} -34x +63) + (3x^{2} - 5x+2)$

= $6x^{2} - 39x + 65$

The above expression is of the STANDARD FORM: $ax^{2} +bx + c$

Hence, the standard form of $(x-9)(3x-7) + (3x^{2} - 5x+2)$ $=6x^{2} - 39x + 65$

6 0

3 years ago

Csc A - sin A = <br><br> A. (cos^2A) / (sinA)<br> B. (1 - sinA) / (sinA) <br> C. (1 - sinA) / (cosA)

julsineya [31]

Answer:

A) $\frac{cos^2A}{sinA}$

Step-by-step explanation:

$cscA-sinA\\\\\frac{1}{sinA}-sinA\\ \\\frac{1}{sinA}-\frac{sin^2A}{sinA}\\ \\ \frac{1-sin^2A}{sinA}\\ \\\frac{cos^2A}{sinA}$

4 0

3 years ago

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