One question please help a friend out thanks you guys :-)

Question 1<br> Let f(x) = 2x^2 - 3x - 4 and g(x) = x + 5. Find (fg)(x).

Scilla [17]

The function (fg)(x) is a composite function

The value of the function (fg)(x) is 2x^3 + 7x^2 - 19x - 20

<h3>How to determine the function (fg)(x)?</h3>

The functions are given as:

f(x) = 2x^2 - 3x - 4 and g(x) = x + 5.

To calculate (fg)(x), we make use of

(fg)(x) = f(x) * g(x)

So, we have:

(fg)(x) = (2x^2 - 3x - 4) * (x + 5)

Expand

(fg)(x) = 2x^3 - 3x^2 - 4x + 10x^2 - 15x - 20

Collect like terms

(fg)(x) = 2x^3 - 3x^2 + 10x^2 - 4x - 15x - 20

Evaluate

(fg)(x) = 2x^3 + 7x^2 - 19x - 20

Hence, the function (fg)(x) is 2x^3 + 7x^2 - 19x - 20

Read more about composite function at:

brainly.com/question/10687170

7 0

2 years ago

Please could you find the answers to the questions in the attachment.

Fudgin [204]

( $\frac{x1+x2}{2} , \frac{y1+y2}{2}$ )we need 3 equations
1. midpoint equation which is ( $\frac{x1+x2}{2} , \frac{y1+y2}{2}$ ) when you have 2 points

2. distance formula which is D= $\sqrt{(x2-x1)^{2}+(y2-y1)^{2}}$

3. area of trapezoid formula whhic is (b1+b2) times 1/2 times height

so

x is midpoint of B and C
B=11,10
c=19,6
x1=11
y1=10
x2=19
y2=6
midpoint=( $\frac{11+19}{2} , \frac{10+6}{2}$ )
midpoint=( $\frac{30}{2} , \frac{16}{2}$ )
midpoint= (15,8)

point x=(15,8)

y is midpoint of A and D
A=5,8
D=21,0
x1=5
y1=8
x2=21
y2=0
midpoint=( $\frac{5+21}{2} , \frac{8+0}{2}$ )
midpoint=( $\frac{26}{2} , \frac{8}{2}$ )
midpoint=(13,4)

Y=(13,4)

legnths of BC and XY
B=(11,10)
C=(19,6)
x1=11
y1=10
x2=19
y2=6
D= $\sqrt{(19-11)^{2}+(6-10)^{2}}$
D= $\sqrt{(8)^{2}+(-4)^{2}}$
D= $\sqrt{64+16}$
D= $\sqrt{80}$
D= $4 \sqrt{5}$
BC= $4 \sqrt{5}$

X=15,8
Y=(13,4)
x1=15
y1=8
x2=13
y2=4
D= $\sqrt{(13-15)^{2}+(4-8)^{2}}$
D= $\sqrt{(-2)^{2}+(-4)^{2}}$
D= $\sqrt{4+16}$
D= $\sqrt{20}$
D= $2 \sqrt{5}$
XY= $2 \sqrt{5}$

the thingummy is a trapezoid
we need to find AD and BC and XY
we already know that BC= $4 \sqrt{5}$ and XY= $2 \sqrt{5}$

AD distance
A=5,8
D=21,0
x1=5
y1=8
x2=21
y2=0
D= $\sqrt{(21-5)^{2}+(0-8)^{2}}$
D= $\sqrt{(16)^{2}+(-8)^{2}}$
D= $\sqrt{256+64}$
D= $\sqrt{320}$
D= $4 \sqrt{2}$
AD= $4 \sqrt{2}$

so we have
AD= $4 \sqrt{2}$
BC= $4 \sqrt{5}$
XY= $2 \sqrt{5}$

AD and BC are base1 and base 2
XY=height
so
(b1+b2) times 1/2 times height
( $4 \sqrt{2}+4 \sqrt{5}$ ) times 1/2 times $2 \sqrt{5}$ =
( $4 \sqrt{2}+4 \sqrt{5}$ ) times \sqrt{5} [/tex] =
$4 \sqrt{10}+4*5$ = $4 \sqrt{10}+20$ =80 $\sqrt{10}$ =252.982

X=(15,8)
Y=(13,4)
BC= $4 \sqrt{5}$
XY= $2 \sqrt{5}$
Area=80 $\sqrt{10}$ square unit or 252.982 square units

7 0

3 years ago

A parabola opens up and passes through (-4, 2) and (6, -3). How do you know that (-4, 2) is not the vertex

Ainat [17]

Answer:

Step-by-step explanation:

The minimum is at the vertex of this parabola because it opens up.

Now if (-4, 2) is the minimum then all the y values on the parabola must be > 2,

But we are given that y = -3 is on the graph ( the point (6,-3) - that is y < 2 here,

Therefore (-4, 2) cannot be the vertex .

4 0

1 year ago

Solve 2x2 - 6x = -8.

erma4kov [3.2K]

Answer:

x=2

Step-by-step explanation:

8 0

2 years ago

Write an equation of a line in slope intercept form with the given slope and the y-intercept.

defon

The way you put it is confusing. I'll answer but the way you put it was confusing to me.

7 0

3 years ago