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3 years ago

Let g(x) = 2x and h(x) = x2 + 4. Evaluate (h ∘ g)(1).

Mathematics

1 answer:

anastassius [24]3 years ago

5 0

Answer:

$\large\boxed{(h\circ g)(1)=8}$

Step-by-step explanation:

$(f\circ g)(x)=f\bigg(g(x)\bigg)\\===================\\\\(h\circ g)(1)=h\bigg(g(1)\bigg)\\\\g(1)-\text{put}\ x=1\ \text{to the equation of the function}\ g(x)=2x:\\\\g(1)=2(1)=2\\\\h\bigg(g(1)\bigg)=h(2)-\text{put}\ x=2\ \text{to the equation of the function}\ h(x)=x^2+4:\\\\h(2)=2^2+4=4+4=8$

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A survey of an athletic shoe store’s customers showed that 8% of the customers run every day. The store has approximately 1900 c

dem82 [27]

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3 years ago

Simplify the following: a) [ -6 +22 – 6 + 8 ] ÷ ( -9 ) b) 400 ÷ { 40 – (-2) -3 – ( -1)} c) 40 x -23 + 40 x -17 d) 1673 x 99 – (-

oee [108]

Step-by-step explanation:

[ -6 +22 – 6 + 8 ] = 18

18 ÷ (-9) = -2

{40 – (-2) -3 – ( -1)} = { 40 + 2 -3 + 1} = 40

400 ÷ 40 = 10

(40 x 23) + (40 x -17)

920 + (-680)

= 240

(1673 x 99) + 1673

1673 x 100

= 167300

Hope this helps ^-^

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3 years ago

Consider the three points ( 1 , 3 ) , ( 2 , 3 ) and ( 3 , 6 ) . Let ¯ x be the average x-coordinate of these points, and let ¯ y

loris [4]

Answer:

$m=\dfrac{3}{2}$

Step-by-step explanation:

Given points are: ( 1 , 3 ) , ( 2 , 3 ) and ( 3 , 6 )

The average of x-coordinate will be:

$\overline{x} = \dfrac{x_1+x_2+x_3}{\text{number of points}}$

<u>1) Finding $(\overline{x},\overline{y})$ </u>

Average of the x coordinates:

$\overline{x} = \dfrac{1+2+3}{3}$

$\overline{x} = 2$

Average of the y coordinates:

similarly for y

$\overline{y} = \dfrac{3+3+6}{3}$

$\overline{y} = 4$

<u>2) Finding the line through $(\overline{x},\overline{y})$ with slope m.</u>

Given a point and a slope, the equation of a line can be found using:

$(y-y_1)=m(x-x_1)$

in our case this will be

$(y-\overline{y})=m(x-\overline{x})$

$(y-4)=m(x-2)$

$y=mx-2m+4$

this is our equation of the line!

<u>3) Find the squared vertical distances between this line and the three points.</u>

So what we up till now is a line, and three points. We need to find how much further away (only in the y direction) each point is from the line.

Distance from point (1,3)

We know that when x=1, y=3 for the point. But we need to find what does y equal when x=1 for the line?

we'll go back to our equation of the line and use x=1.

$y=m(1)-2m+4$

$y=-m+4$

now we know the two points at x=1: (1,3) and (1,-m+4)

to find the vertical distance we'll subtract the y-coordinates of each point.

$d_1=3-(-m+4)$

$d_1=m-1$

finally, as asked, we'll square the distance

$(d_1)^2=(m-1)^2$

Distance from point (2,3)

we'll do the same as above here:

$y=m(2)-2m+4$

$y=4$

vertical distance between the two points: (2,3) and (2,4)

$d_2=3-4$

$d_2=-1$

squaring:

$(d_2)^2=1$

Distance from point (3,6)

$y=m(3)-2m+4$

$y=m+4$

vertical distance between the two points: (3,6) and (3,m+4)

$d_3=6-(m+4)$

$d_3=2-m$

squaring:

$(d_3)^2=(2-m)^2$

3) Add up all the squared distances, we'll call this value R.

$R=(d_1)^2+(d_2)^2+(d_3)^2$

$R=(m-1)^2+4+(2-m)^2$

<u>4) Find the value of m that makes R minimum.</u>

Looking at the equation above, we can tell that R is a function of m:

$R(m)=(m-1)^2+4+(2-m)^2$

you can simplify this if you want to. What we're most concerned with is to find the minimum value of R at some value of m. To do that we'll need to derivate R with respect to m. (this is similar to finding the stationary point of a curve)

$\dfrac{d}{dm}\left(R(m)\right)=\dfrac{d}{dm}\left((m-1)^2+4+(2-m)^2\right)$