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2 years ago

Question 3 of 10 Subtract these polynomials. (6x²-x+ 8) - (x² + 2) =

Mathematics

1 answer:

Julli [10]2 years ago

5 0

Answer:

$5 {x}^{2} - x + 6$

Step-by-step explanation:

The solution is in the image

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Plese help for brainlist !!!

Ipatiy [6.2K]

Answer:

{2, 4, 6}

Step-by-step explanation:

the domain is the interval or set of valid x (input) values.

the range is the interval or set of valid y (result) values.

all we need to do is use the given x values in the functional definition and collect the result values. that is the range.

x = -1

y = 2×-1 + 4 = 2

x = 0

y = 2×0 + 4 = 4

x = 1

y = 2×1 + 4 = 6

7 0

2 years ago

Given f(x)=x^2−2x−5. <br><br> Enter the quadratic function in vertex form.

FrozenT [24]

Answer:

$f(x) = (x-1)^2-6$

Step-by-step explanation:

Rewrite the quadratic using the method of completing the square:

$x^2-2x-5=(x^2-2x+1)-1-5=(x-1)^2-6\\f(x) = (x-1)^2-6$

From this form it is immediately clear that the vertex of the parabola is at x=1 and its y-intercept at -6.

8 0

3 years ago

The simplified answer of 6/35

Rzqust [24]

The simplified answer of 6/35 is 6/35. It is in simplest form.

7 0

3 years ago

Read 2 more answers

Calculate the total area of the shaded region.

LUCKY_DIMON [66]

so hmmm seemingly the graphs meet at -2 and +2 and 0, let's check

$\stackrel{f(x)}{2x^3-x^2-5x}~~ = ~~\stackrel{g(x)}{-x^2+3x}\implies 2x^3-5x=3x\implies 2x^3-8x=0 \\\\\\ 2x(x^2-4)=0\implies x^2=4\implies x=\pm\sqrt{4}\implies x= \begin{cases} 0\\ \pm 2 \end{cases}$

so f(x) = g(x) at those points, so let's take the integral of the top - bottom functions for both intervals, namely f(x) - g(x) from -2 to 0 and g(x) - f(x) from 0 to +2.

$\stackrel{f(x)}{2x^3-x^2-5x}~~ - ~~[\stackrel{g(x)}{-x^2+3x}]\implies 2x^3-x^2-5x+x^2-3x \\\\\\ 2x^3-8x\implies 2(x^3-4x)\implies \displaystyle 2\int\limits_{-2}^{0} (x^3-4x)dx \implies 2\left[ \cfrac{x^4}{4}-2x^2 \right]_{-2}^{0}\implies \boxed{8} \\\\[-0.35em] ~\dotfill$

$\stackrel{g(x)}{-x^2+3x}~~ - ~~[\stackrel{f(x)}{2x^3-x^2-5x}]\implies -x^2+3x-2x^3+x^2+5x \\\\\\ -2x^3+8x\implies 2(-x^3+4x) \\\\\\ \displaystyle 2\int\limits_{0}^{2} (-x^3+4x)dx \implies 2\left[ -\cfrac{x^4}{4}+2x^2 \right]_{0}^{2}\implies \boxed{8} ~\hfill \boxed{\stackrel{\textit{total area}}{8~~ + ~~8~~ = ~~16}}$

7 0

2 years ago

Shryia read a

Paraphin [41]

Answer:

Step-by-step explanation:

481 = 1.5h + 403

7 0

3 years ago