Given the function h(x) = 3(5)x, Section A is from x = 0 to x = 1 and Section B is from x = 2 to x = 3. Part A: Find the average rate of change of each section. (4 points) Part B: How many times greater is the average rate of change of Section B than Section A? Explain why one rate of change is greater than the other. (6 points) i just need answer please. ill give modded cars on gta ps4 if help
HAHA MODDED CARS ON GTA
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Is this a test question?
no its my last question on my quiz
Quiz, ok.
need a screen shot ?
yes plz, send a screenshot instead
average rate of change is essentially the the slope formula. (y2 - y1)/(x2 - x1)
for section A, from x = 0 to x = 1 plug in 0 and 1 into the equation and solve for y. Then you will have your 2 coordinates
From there, plug into the formula and see what you get.
bro i got a f in math rn my dad is gonna beat me and my uncle is gonna jump in like its wwe smackdown but he likes to do some werid stuff
i just need the answer before my dad starts hitting me and my mom being his number one fan in the background
... can you do what i asked you to do?
lol ikr
okok how do i do that ?
Nope, you are not getting a direct answer, work with me lad
evaluate
?
so we have the equation 3(5)^x If 0 goes for x then 5^0 = 1 3 times 1 is 3 coordinate 1 = (0,3) If we plug in 1 for x 3(5)^1 = 3(5) =15 coordinate 2 = (1,15)
bro that one hurt the heck out my head but im listening...
now we plug it into the slope formula y2 =15 y1 = 3 x2 = 1 x1 = 0 so now (15-3)/(1-0)
and 15-3 = 12 1 - 0 = 1 12 is the average rate of change
Can you try to find the average rate of change for section B?
yea give me a second cause im trying to read this, understand it then look at the question for part b
yes ofc, if you need further clarification then ask away
ty
what number do i start with because you started with 3(5)^x but for part b says only x= 2 to x=3
you can decide, lets just assign x1 = 2 and x2 = 3 for the sake of understanding This means that the y value obtained when we plug in x = 2 into the equation will be assigned to y1. Because if x1 is plugged in, y1 is the output If x2 is plugged in y2 is the output
so we want to find from x = 2 to x = 3 Now lets say x1 = 2 and x2 = 3 Find y1 and y2 can you?
bro im kinda lost sorry. so do i use x1=2 and x2=3 to find it? if so am i supposed to put it in an equation ?
yes its all good if your lost. So x1 =2 and x2 = 3 So we have the equation 3(5)^x plug in the value for x1 into x in the equation right above this, and the y-value you get is y1 similarly for x2, plug into the equation 3(5)^x for x and the y value you get is y2. Can you tell me what y1 and y2 are now?
wait is y1 2 and y2 3 ??
@darkknight
no x1 is 2 and x2 is 3, you have to plug in 2 and 3 into the equations above to find what y1 and y2 are.
Okay, so x1 = 2 ill plug in 2 for x in this function 3(5)^x Now i have 3(5)^2 which is 75
ohh
so do the same for x2, and find what y2 is
so your telling me to do 3(5)^3
am i mutiplying orr ?
yes, that is what you are going to do.
you have to evaluate 3(5)^3
375 ?
yes, x2 = 3 and y2 = 375 x1 = 2 and y1 = 75. plug into this formula (y2-y1)/(x2-x1)
ok
like on a calculator ?
sure, you just have to plug in the values and find what the final value is
ok ok
still mutiplying ?
idk what you are saying, you have to evaluate
oh my bad
all good, type in what you get for the average rate of change
im putting (y2-y1)/(x2-x1) on the equation
?*
Have you done it yet? you don't need to ask for confirmation every second. Just plug in values and evaluate
lmao im trying to make sure cause i dont wanna do it wrong sorry
did you get your answer?
Did you get your answer?
@408juju The man is talkin to ya.....
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