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zhannawk [14.2K]

2 years ago

8

X2−14x−32=0 What are the roots of the polynomial equation?

1 answer:

stepladder [879]2 years ago

7 0

Answer:

x= - 8/3

Step-by-step explanation:

1- Multiply the monomials

2- Combine like terms

3- Rearrange variables to the left side of the equation

4- Divide both sides of the equation by the coefficient of the variable

5- Cross out the common factor

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Write an equation of the line

Deffense [45]

(3,0) and (0,9)

Slope= (9-0)/(0-3)= -3

Line equals Y= -3x+9

4 0

3 years ago

Which undefined geometric term is described as a location on a coordinated plane that designated by an ordered pair,(x,y)?

zepelin [54]

undefined slope = (16,3), ( 16, -80)

4 0

4 years ago

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What is the measure of ADC in quadrilateral ABCD?<br><br> 45°<br> 65°<br> 115°<br> 135°

ruslelena [56]

Answer:

115

Step-by-step explanation:

its the only that makes sense

8 0

3 years ago

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Help me with this pls

Dvinal [7]

Answer:

C

Step-by-step explanation:

The equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y- intercept )

Given

6x + 10y = 8 ( subtract 6x from both sides )

10y = - 6x + 8 ( divide each term by 10 )

y = - $\frac{6}{10}$ x + $\frac{8}{10}$ = - $\frac{3}{5}$ x 6 + $\frac{4}{5}$ ← in slope- intercept form

with slope m = - $\frac{3}{5}$ → C

7 0

3 years ago

The twice–differentiable function f is defined for all real numbers and satisfies the following conditions: f(0)=3 f′(0)=5 f″(0)

Vladimir79 [104]

$g(x)=e^{ax}+f(x)\implies g'(x)=ae^{ax}+f'(x)\implies g''(x)=a^2e^{ax}+f''(x)$

Given that $f'(0)=5$ and $f''(0)=7$ , it follows that

$g'(0)=a+5$

$g''(0)=a^2+7$

###

$h(x)=\cos(kx)f(x)+\sin x\implies h'(x)=-k\sin(kx)f(x)+\cos(kx)f'(x)+\cos x$

When $x=0$ , we have

$h(0)=\cos0f(0)+\sin0=f(0)=3$

The slope of the line tangent to $h(x)$ at (0, 3) has slope $h'(0)$ ,

$h'(0)=-k\sin0f(0)+\cos0f'(0)+\cos0=5+1=6$

Then the tangent line at this point has equation

$y-3=6(x-0)\implies y=6x+3$

###

Differentiating both sides of

$4x^2+y^2=48+2xy$

with respect to $x$ yields

$8x+2y\dfrac{\mathrm dy}{\mathrm dx}=2y+2x\dfrac{\mathrm dy}{\mathrm dx}$

$\implies(2y-2x)\dfrac{\mathrm dy}{\mathrm dx}=2y-8x$

$\implies\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{y-4x}{y-x}$

On this curve, when $x=2$ we have

$4(2)^2+y^2=48+2(2)y\implies y^2-4y-32=(y-8)(y+4)=0\implies y=8$

(ignoring the negative solution because we don't care about it)

The tangent to this curve at the point $(x,y)$ has slope $\dfrac{\mathrm dy}{\mathrm dx}$ . This tangent line is horizontal when its slope is 0. This happens for

$\dfrac{y-4x}{y-x}=0\implies y-4x=0\implies y=4x$

and when $x=2$ , there is a horizontal tangent line to the curve at the point (2, 8).

5 0

3 years ago