All categories

3 years ago

If f(x)=2x squared -x, find f (3)

Mathematics

1 answer:

malfutka [58]3 years ago

6 0

Substitute the value of x = 3 to the equation of the function:

f(x) = 2x² - x → f(3) = 2(3²) - 3 = 2(9) - 3 = 18 - 3 = 15

Answer: f(3) = 15

You might be interested in

I need to find out x Q = 2x + 3 = 7

AlekseyPX

After subtracting 3 from 7, and dividing 4 by 2, you should end up with x=2.

3 0

3 years ago

Suppose lines EF and GH are reflected over the line Y equals X to form the lines JK and LM which statement would be true about t

katrin [286]

Answer:

A) JK and LM will be parallel to each other.

Step-by-step explanation:

On reflection on $y=x$ line the x co-ordinate changes with y co-ordinate and y co-ordinate changes with x co-ordinate

$(x,y)\rightarrow (y,x)$

Points on line EF

$(0,6) , (-5,-2)$

On reflection of this line on $y=x$ the new points we get for line JK are

$(6,0),(-2,-5)$

Points on line GH

$(-4,9),(-9,1)$

On reflection on y=x line the new points we get for line LM are

$(9,-4),(1,-9)$

Slope of line JK

$m=\frac{y_2-y_1}{x2-x1}\\m=\frac{(-5)-0}{(-2)-6} \\m=\frac{-5}{-8}=\frac{5}{8}$

Slope of line LM

$m=\frac{y_2-y_1}{x2-x1}\\m=\frac{(-9)-(-4)}{1-9} \\m=\frac{-9+4}{-8}=\frac{-5}{-8}\\m=\frac{5}{8}$

For two line to be parallel, their slopes will be same.

$m_{JK} =\frac{5}{8} , m_{LM}=\frac{5}{8}$

Since slopes of lines JK and LM are same therefore we can say that these are parallel to each other.

5 0

3 years ago

X^2-7×+10 solve for quadratic formula.

nordsb [41]

What does that even mean

5 0

3 years ago

Read 2 more answers

1. if x० and 70० makes an angle of complement, find the value of x.

stich3 [128]

Answer:

don't spend our time here

this is a spamm community

6 0

2 years ago

Find the area of the triangle ABC with the coordinates of A(10, 15) B(15, 15) C(30, 9).

lions [1.4K]

Check the picture below. so, that'd be the triangle's sides hmmm so let's use Heron's Area formula for it.

$~\hfill \stackrel{\textit{\large distance between 2 points}}{d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}}~\hfill~ \\\\[-0.35em] ~\dotfill\\\\ (\stackrel{x_1}{10}~,~\stackrel{y_1}{5})\qquad (\stackrel{x_2}{15}~,~\stackrel{y_2}{15}) ~\hfill a=\sqrt{[ 15- 10]^2 + [ 15- 5]^2} \\\\\\ ~\hfill \boxed{a=\sqrt{125}} \\\\\\ (\stackrel{x_1}{15}~,~\stackrel{y_1}{15})\qquad (\stackrel{x_2}{30}~,~\stackrel{y_2}{9}) ~\hfill b=\sqrt{[ 30- 15]^2 + [ 9- 15]^2} \\\\\\ ~\hfill \boxed{b=\sqrt{261}}$

$(\stackrel{x_1}{30}~,~\stackrel{y_1}{9})\qquad (\stackrel{x_2}{10}~,~\stackrel{y_2}{5}) ~\hfill c=\sqrt{[ 10- 30]^2 + [ 5- 9]^2} \\\\\\ ~\hfill \boxed{c=\sqrt{416}} \\\\[-0.35em] ~\dotfill$

$\qquad \textit{Heron's area formula} \\\\ A=\sqrt{s(s-a)(s-b)(s-c)}\qquad \begin{cases} s=\frac{a+b+c}{2}\\[-0.5em] \hrulefill\\ a=\sqrt{125}\\ b=\sqrt{261}\\ c=\sqrt{416}\\ s\approx 23.87 \end{cases} \\\\\\ A\approx\sqrt{23.87(23.87-\sqrt{125})(23.87-\sqrt{261})(23.87-\sqrt{416})}\implies \boxed{A\approx 90}$

6 0

2 years ago