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3 years ago

What is the completely factored form of

Mathematics

2 answers:

jekas [21]3 years ago

8 0

Answer:

(x - 3) (x + 3) (x + 4) or C

Step-by-step explanation:

x³ + 4x² - 9x - 36

factor out (x - 3) using synthetic division

3 ║ 1 4 -9 -36

1 7 12 0

(x - 3) (x² + 7x + 12)

factor x² + 7x + 12

x² + 7x + 12 = (x + 3) (x + 4)

kondor19780726 [428]3 years ago

3 0

Answer:

Step-by-step explanation:

i just did it

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Jasmine enlarged a drawing she made in art class,what is the value of x

malfutka [58]

Is there supposed to be an equation that follows? Can I see the entire question?

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3 years ago

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What is the inverse of the given equation y=7x^2-3?

NARA [144]

Answer:

The inverse of this equation would be y = $\sqrt{\frac{x - 3}{7} }$

Step-by-step explanation:

To find the inverse of any equation, start by switching the y and x values. Then solve for the new y value. That equation will be your inverse.

y = 7x^2 - 3

x = 7y^2 - 3

x + 3 = 7y^2

$\frac{x-3}{7}$ = y^2

$\sqrt{\frac{x - 3}{7} }$ = y

5 0

3 years ago

If the area of AABC is D, give the expressions that complete the equation for the measure of ZB?

Ira Lisetskai [31]

Given:

The figure of triangle ABC.

The area of the triangle ABC is D.

$m\angle B=\sin ^{-1}(\dfrac{m}{n})$

To find:

The value of m and n in the given expression.

Solution:

Let h be the height of the triangle ABC.

Area of a triangle is:

$Area=\dfrac{1}{2}\times base\times h$

Where, b is the base and h is the height of the triangle.

$Area=\dfrac{1}{2}\times a\times h$

The area of the triangle ABC is D.

$D=\dfrac{1}{2}\times a\times h$

$2D=ah$

$\dfrac{2D}{a}=h$ ...(i)

In a right angle triangle,

$\sin \theta =\dfrac{Perpendicular}{Hypotenuse}$

$\sin B =\dfrac{h}{c}$

$\sin B =\dfrac{1}{c}\times \dfrac{2D}{a}$ [Using (i)]

$\sin B =\dfrac{2D}{ac}$

$m\angle B =\sin ^{-1}\dfrac{2D}{ac}$ ...(ii)

We have,

$m\angle B=\sin ^{-1}(\dfrac{m}{n})$ ...(iii)

On comparing (ii) and (iii), we get

$m=2D$

$n=ac$

Therefore, the required values are $m=2D, n=ac$ .

6 0

3 years ago

Integrate e^x(sin(x) cos(x))

Karo-lina-s [1.5K]

$I=\int e^x(\sin(x)\cos(x))dx=\int e^x(\frac{1}{2}\sin(2x))dx=\frac{1}{2}\int e^x\sin(2x)dx$

$\text{If }u=\sin(2x)\to du=2\cos(2x)dx~\text{and}~dv=e^xdx\to v=e^x:\\\\
\text{Using }\int u\,dv=uv-\int v\,du:\\\\
I=\frac{1}{2}(e^x\sin(2x)-\int e^x(2\cos(2x))dx)\\\\
2I=e^x\sin(2x)-2\underbrace{\int e^x\cos(2x)dx}_{I_2}$

Looking for $I_2$ :

$\text{If}~u=\cos(2x)\to du=-2\sin(2x)dx~\text{and}~dv=e^xdx\to v=e^x:\\\\
I_2=e^x\cos(2x)-\int e^x(-2\sin(2x))dx\\\\ I_2=e^x\cos(2x)+2\int e^x(\sin(2x))dx\\\\ I_2=e^x\cos(2x)+2\int e^x(2\sin(x)\cos(x))dx\\\\ I_2=e^x\cos(2x)+4\int e^x(\sin(x)\cos(x))dx=e^x\cos(2x)+4I$

Replacing:

$2I=e^x\sin(2x)-2I_2\iff\\\\2I=e^x\sin(2x)-2(e^x\cos(2x)+4I)\iff\\\\
2I=e^x\sin(2x)-2e^x\cos(2x)-8I\iff\\\\
10I=e^x\sin(2x)-2e^x\cos(2x)\iff\\\\
I=\dfrac{e^x}{10}(\sin(2x)-2\cos(2x))\\\\
\boxed{\int e^x(\sin(x)\cos(x))dx=\dfrac{e^x}{10}(\sin(2x)-2\cos(2x))+C}$

6 0

3 years ago

ASAP!!!<br> Is the relation in the table below a function?

statuscvo [17]

Answer:

Step-by-step explanation:

A relation is a function if every x-value has exactly one y-value.

Because the x-value 1 has two y-values (-3 and 3), the relation is not a function.

Plotting (1, -3) and (1, 3) will produce a vertical line that intersects the function at two points.

Therefore, the relation is not a function.

Hope that helps.

4 0

3 years ago