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3 years ago

FIND THE AREA OF THE SHADED REGION. SHOW ALL YOUR WORK.

Mathematics

2 answers:

RideAnS [48]3 years ago

6 0

The teachers this is commonly known as a doughnut problem. Basically you find the area of the large trays and subtract the area of the small shape.

Lg. Rectangle. Sm. Rectangle
A = lw. A=lw
A=(3x - 4)(2x + 2) A=(x - 3)(x - 6). Use FOIL on each
A = 6x^2 + 6x - 8x -8 A=x^2 - 6x - 3x + 18. Combine like terms
A = 6x^2 - 2x - 8. A=x^2 - 9x + 18. Subtract rectangles to find area of shaded region
A = 6x^2 -2x - 8 - (x^2 - 9x + 18) Distribute the negative into the parenthesis (multiply each term by -1)
A = 6x^2 - 2x - 8 - x^2 + 9x -18. Combine like terms
A = 5x^2 + 7x - 26. <— answer

erik [133]3 years ago

3 0

Answer:

he area that is un-shaded is (x-3)(x-6).

The entire area is (2x+2)(3x-4).

Area of the shaded region is:

(2x+2)(3x-4) - (x-3)(x-6) =

(6x2-2x-8) - (x2-9x+18) =

6x2-2x-8-x2+9x-18 =

5x2+7x-26 hopes this help btw (;

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2/2 simplify the fraction into its simplest terms

Illusion [34]

Answer: 1

Step-by-step explanation:

2/2= 1

3 0

3 years ago

What is the solution of the equation (x – 5)2 3(x – 5) 9 = 0? use u substitution and the quadratic formula to solve.

Ahat [919]

The value of x in the equation $(x-5)^{2} +3(x-5)+9=0$ is x=(7+3 $\sqrt{3}i$ )/2,

(7-3 $\sqrt{3}i$ )/2.

Given equation $(x-5)^{2} +3(x-5)+9=0$ .

We have to find the solution of the equation.

Equation is solved to find the value of variables. The number of values of the variable depends on the highest power of variables.

let (x-5)=u-------------------1

equation becomes $u^{2} +3u+9=0$

solution of a quadratic formula

x=-b± $\sqrt{b^{2} -4ac}$ /2a

on comparing with general form a=1,b=3, c=9

u=-3± $\sqrt{3^{2} -4*1*9} /2$

=-3± $\sqrt{9-36}/2$

=-3± $3\sqrt{-27}/2$

=-3± $3\sqrt{3} i/2$

$u_{1} =-3$ +3 $\sqrt{3}/2$ , $u_{2}$ =-3-3 $\sqrt{3}/2$

put the values of $u_{1}$ in equation 1

x-5= $u_{1}$

x-5=-3+ $3\sqrt{3} /2$

x=-3+3 $\sqrt{3}i/2+5$

x=7+3 $\sqrt{3} i/2$

put the values of $u_{2}$ in equation2

x-5= $u_{2}$

x-5=-3-3 $\sqrt{3} /2$

x=7-3 $\sqrt{3}/2$

Hence the values of x are 7+3 $\sqrt{3}i/2$ , 7-3 $\sqrt{3}i/2$ .

Learn more about equation at brainly.com/question/2972832

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5 0

2 years ago

What is the average rate of change of f over the interval -1 on a separate sheet of paper and upload the work as a photo. Show a

Anarel [89]

Answer:

Average rate of change for the function $f(x)= x^2-x-1$ over the interval -1<x<1 is -1

Step-by-step explanation:

We need to find average rate of change of f over the interval -1 < x < 1

The function given is: $f(x)= x^2-x-1$

The formula used to find average rate of change is:

$Average\:rate\:of\:change=\frac{f(b)-f(a)}{b-a}$

We have, a = -1 and b = 1

Finding f(b) when b=1

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

Now, finding f(a), when a= -1

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

Now, putting values and finding average rate of change

$Average\:rate\:of\:change=\frac{f(b)-f(a)}{b-a}\\Average\:rate\:of\:change=\frac{f(1)-f(-1)}{1-(-1)}\\Average\:rate\:of\:change=\frac{-1-(1)}{1-(-1)}\\Average\:rate\:of\:change=\frac{-1-1}{1+1}\\Average\:rate\:of\:change=\frac{-2}{2}\\Average\:rate\:of\:change=-1$

So, average rate of change for the function $f(x)= x^2-x-1$ over the interval -1<x<1 is -1

4 0

3 years ago

Help meeeeeeeeeeeeee

blagie [28]

Answer:

perimeter= 2(l+b)

=2(36+23)

=118m

area =l×b

=36×23

=828m²

8 0

3 years ago

GIVING BRAINLIEST+30 POINTS TO BEST ANSWER!! REMEMBER, FOLLOW DIRECTIONS!!

vitfil [10]

Answer:

Step-by-step explanation:

First, you would multiply 3x4 =12

next you would keep the denominator the same

last you would divide 8 by 12 10 get 1 3/8 for your answer

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2 years ago