Question

Can someone show me how to work these out please? I'll give a medal <3
1. -17y=da solve for d
2. 3a-p=r solve for a
3. -6d+5b=n solve for d
4. J=F(1+rt) solve for t
5. z=9+y, for y;z=5
y=
6.A certain rule for medication expresses a relationship between the recommended dosages for a child and for and for an adult. The rule states that a child's dosage C equals the product of the child's weight W in pounds divided by 140 and the adult dosage A. Write a formula for C in terms of W and A. Then solve this formula for A.
The formula for C in terms of W and A is
C=

Answer 1

yo

Answer 2

@zendaya-replay-guy can you help?

Answer 3

well....... i can probably do it on a calculator

Answer 4

hey @Darius25 and @johnweldon1993 can u help here?

Answer 5

I'll work out a couple and see if that helps you get going..

Well it will just be using algebra to isolate a variable...

1. -17y=da solve for d

So to isolate d on one side...you need to divide everything by 'a'

-17y
d = -----
a

2. 3a-p=r solve for a

Same thing here...add 'p' to both sides of the equation

3a = r + p

Now divide everything by 3 to isolate 'a'

a = r + p
------
3

Make sense?

Answer 6

@Savannah_Noelle

Answer 7

Okay so for number one the equation would be D/A=-17Y/A Does the "A" cancel out?

Answer 8

u got this? @johnweldon1993

Answer 9

\[\large \frac{D \cancel{A}}{ \cancel{A}} = \frac{-17y}{A} \]
\[\large D = \frac{-17y}{A}\]


So yes the A cancels

Answer 10

Okay thank you, is that the final answer?

Answer 11

That would be your final answer for 1) yes

Answer 12

@johnweldon1993 @zendaya-replay-guy Thank you guys so much! :) but I have a question. How do you know when you need to divide, multiply, add, etc for thee equations?

Answer 13

Also, the website i'm on says that a=r+p/3 was the wrong answer

Answer 14

Alright...so it's all about inverses of operations...Lets look again at 1)

We had

\[\large -17y = da\]

We want to isolate D....well we can see that D is being MULTIPLIED by A ...so in order to cancel that out...we need to DIVIDE by A....

Answer 15

Hint* a = (r + p)/3

Use parenthesis...it makes all the difference

Answer 16

Ohh okay, that makes sense

Answer 17

if that is how you put your answer...you entered

\[\large a = r + \frac{p}{3}\]

But the answer is

\[\large a = \frac{r + p}{3}\]

Answer 18

Okay so for number 3: -6d+5b=n solve for d, I should solve by dividing everything by n?

Answer 19

No...The goal would be to isolate d because that is what you are solving for...

-6d + 5b = n

You want this to look like

-6d = ...something...

Make sense? you want 'd' to be alone on 1 side of the equation...

So you would start by subtracting 5b from both sides of this equation

-6d + 5b = n
- 5b -5b
----------------
-6d = n - 5b

With me so far?

Answer 20

Yes

Answer 21

Alright....so now...we want to solve for 'd' not -6d....

So since 'd' is being MULTIPLIED by -6....to cancel that we...?

Answer 22

divide by -6

Answer 23

Right...but remember divide EVERYTHING by -6

-6d = n- 5b
-----------
-6 -6

d = n - 5b
------
-6

Answer 24

Okay thank you :) So for number 4: J=F(1+rt) solve for t, would I use the distributive property and get J=F1+FRT?

Answer 25

Right...now you want to isolate FRT first...so what would you do first?

Answer 26

subtract?

Answer 27

Correct...subtract what?

Answer 28

F1, so now I have J-F1=FRT

Answer 29

Just a short thing..

\[\large F1 = F\] it is just F being multiplied by 1 which is just F :)

But yes correct!

Answer 30

Oh okay :) and so now I need to divide?

Answer 31

Right! divide what?

Answer 32

everything by FR?

Answer 33

Perfect :)

so t = ?

Answer 34

T=J-F/FR

Answer 35

Right...just remember parenthesis

t = (J - F)/(FR)

Answer 36

Thank you so much! :)

Answer 37

Anytime :) think you got the last 2?

Answer 38

I thought number 5 would equal -4 since 5 substitutes for z, but it's saying that's wrong :/

Answer 39

@johnweldon1993

Answer 40

Hmm...well

z = 9 + y

First it says solve for 'y' so you would subtract 9 from both sides...

y = z - 9

When z = 5 y = ?

y = z - 9
y = 5 - 9
y = -4

You are correct...but maybe they just want the equation? y = z - 9?

Answer 41

Yes I figured that out, and I'm now on number 6. I figured out that the formula for C is C=(w/140)A
But now it's asking what is the formula when solved for A? I don't know how to do that

Answer 42

Well look how you have it...

You want to solve for 'A'

C = (w/140)A

Looks like A is just being MULTIPLIED by something right? so divide by w/140

HINT* when you divide by a fraction....you flip the fraction and turn it into multiplication...ie

\[\huge \frac{C}{\frac{w}{140}} = \frac{\frac{w}{140}}{\frac{w}{140}}A\]

Will become

\[\huge \frac{140}{w}C = A\]

Make sense? All I did was flip the fraction...and then turned it into multiplication

Answer 43

Oh okay I was typing in (W/140)C

Answer 44

Thank you for everything! I really appreciate it :)

Answer 45

Anytime :)