Consider the function f(x) = |x|. Let g(x) = |–4(x – 7)|. Which shows the graphs of f(x) and g(x)?

A.) On a coordinate plane, y = g (x) opens up and goes through (negative 3, 4), has a vertex at (negative 2, 0) and goes through (0, 7). Y = f (x) opens up and goes through (negative 4, 4), has a vertex at (0, 0), and goes through (4, 4).
B.) On a coordinate plane, y = g (x) opens up and goes through (6, 4), has a vertex at (7, 0) and goes through (8, 4). Y = f (x) opens up and goes through (negative 4, 4), has a vertex at (0, 0), and goes through (4, 4).
C.) On a coordinate plane, y = g (x) opens down and goes through (6, negative 4), has a vertex at (7, 0) and goes through (8, negative 4). Y = f (x) opens up and goes through (negative 4, 4), has a vertex at (0, 0), and goes through (4, 4).
D.) On a coordinate plane, y = g (x) opens up and goes through (negative 8, 4), has a vertex at (negative 7, 0) and goes through (negative 6, 4). Y = f (x) opens up and goes through (negative 4, 4), has a vertex at (0, 0), and goes through (4, 4).

option a, c,d are correct.

step-by-step explanation:

from the given figure, it is given that z is equidistant from the sides of the triangle rst, then from triangle tzb and triangle szb, we have

tz=sz(given)

bz=zb(common)

therefore, by rhs rule,δtzb ≅δszb

by cpctc, sz≅tz

also, from δctz and δasz,

tz=sz(given)

∠tcz=∠saz(90°)

by rhs rule, δctz ≅ δasz, therefore by cpctc, ∠ctz≅∠asz

also,from δasz and δzsb,

zs=sz(common)

∠zbs=∠saz=90°

by rhs rule, δasz ≅δzsb, therefore, by cpctc, ∠asz≅∠zsb

hence, option a, c,d are correct.

false

step-by-step explanation:

false : it can not go to 9 the highest that part can go is 7

here is your proof

your exact question is kind of difficult to understand, so maybe an example will explain what you're supposed to do? i'll work through the first problem on the worksheet.

suppose you have a tray of 652 brownies, and you have 5 friends that you want to feed without picking favorites. how many brownies can you dole out to each of your friends in equal amounts? how many will be left over for you to eat?

what you're doing here is finding a way to arrange the number 652 as the sum of integer multiple of 5 and some remainder:

here ("q" for quotient) is the number of brownies each of your friends get, and is the remainder. that's the basic idea behind division.

we break up the number 652 into parts so that the leading digit stands on its own:

652 = 600 + 52

first we check if it's possible to hand out exactly 100 brownies to each friend. with 5 friends, this means we want to know if we can take out 500 brownies from the tray. we can do this 1 time because we have 600 brownies to take from, and when we do that we're left with 100 + 52 brownies to divvy up. so we know that

652 = 1*500 + 152

next we want to know if it's possible to hand out exactly 10 brownies per friend. if we can, we have to remove 50 brownies from the tray each time until we can't. we have 152 brownies left, and we know that 3(50) = 150, so

152 = 3*50 + 2

now we have 2 brownies left in the tray. is it possible to hand 1 out to each of your 5 friends? no, you'd need a minimum of 5 brownies to do that. so the last 2 are yours.

putting everything together, we found that

652 = 1*500 + 3*50 + 2

and we can rearrange this to write

652 = 5*100 + 5*30 + 2

652 = 5(100 + 30) + 2

652 = 5*130 + 2

so and .

you might be more accustomed to the table depicting the algorithm. that's just a compact way of organizing what we've done above. you can consult the image i've attached below.