Calculate the value of x, given line segment FM = 3x and line segment MN = x + 40.
A. 13.3
B. 10
C. 40.
D. 20

- anonymous

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- anonymous

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- ash2326

@alainabbyboo22 Given that the arcs NM and FM are equal

- ash2326

so chords will be equal too
NM=FM
Can you solve it from here ?

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## More answers

- anonymous

i'm confused on this . what do i need to do ?

- ash2326

We have the arcs equal as it's indicated in the diagram, isn't it?

- anonymous

ya

- ash2326

so the chords ( the straight lines ) NM and FM will be equal
NM=FM

- anonymous

then what do i do ?

- ash2326

You are given the values of NM and FM in the question. Plug them here :
NM=FM

- ash2326

NM=x+40
FM=3x
NM=FM
so
\[x+40=3x\]
Can you solve this @alainabbyboo22 ?

- anonymous

how do i solve that though ? what do i do ?

- anonymous

i am sorry if i am irritating you with this but i just don't quite understand this stuff

- ash2326

No problem. I 'll help you figure this out :)

- anonymous

thanks so much ! (: i really appreciate it ! thank you for taking your time out of your day to try and help me out on this !

- ash2326

We are all here to help. You're welcome

- ash2326

Can you solve this equation for x ?
\[x+4=6\]

- anonymous

would i take the numbers from the answer choices and plug them into x or would i add 4 + 6 ?

- ash2326

It's just an example, unrelated to the question. Do you know how to solve these type of equations?

- anonymous

no /:

- ash2326

Ok. I'll explain you

- ash2326

Her we have
x+4=6
it means that x+4 is equal to 6
x is a variable, we need to find its value which will make this equation true. There is only one value of x which will make this equation true. It's easy to find that :)
We need to modify it in such a way that we have x separate and no or no.s separate.
\[x+\underline 4=6\]
I can see that 4 is on left side. Want to remove it !!
So I subtract both sides by 4, it's important to remember that whatever operation I do be it add, subtract, multiply or divide. I have to do on both sides not one side
\[x+4-4=6-4\]
Now I'll simplify
\[x+\cancel 4- \cancel 4=2\]
\[x+0=2\]
or
\[x=2\]
Do you get this?

- anonymous

alright i got that part now !

- ash2326

good, similarly we have to find the value of x here.
We know
FM=MN
so
\[3x=x+40\]
here x on the right side is to be removed, what can you do remove x from right side?
remember operations on both sides !!

- anonymous

okay now i'm lost again ..

- ash2326

I'll give you a hint, subtract both sides by x. what would you get?

- anonymous

the right side would 38 and the left side would be 1 .

- ash2326

What did you subtract?

- anonymous

am i right ?

- anonymous

i subtracted 40-2
and 3-2

- ash2326

You made a mistake somewhere. No worries, I'll guide you
\[3x=x+40\]
subtract both sides by x, because we want to remove x from the right side
\[3x-x=x+40-x\]
Now what would you get?

- anonymous

3x=40 ?

- ash2326

we have 3x-x, on left side
or
3x-1x, what would you get on the left side?

- anonymous

x + 38

- ash2326

we have
\[3x-1x=40\]
or
\[x(3-1)=40\]
What would you get next?

- ash2326

what's 3-1 ?

- anonymous

2

- ash2326

@alainabbyboo22 ?

- anonymous

i got 2

- anonymous

where do i go from there now ?

- anonymous

@ash2326

- ash2326

so you have
\[x \times 2=40\]
If you subtract variables, like 4x-2x
just subtract 4-2=2 and put the 2 before x
but if you have x-x or 9x-9x it'll become 0
do you get this?

- anonymous

yes i see now . okay what do i do to get my answer ?

- ash2326

now what do we have?

- anonymous

2x = 40

- ash2326

divide both sides by 2, what would you get?

- anonymous

2 and 20

- ash2326

\[\frac{2\times x}{2}=\frac{40}{2}\]
\[\frac{\cancel 2 \times x}{\cancel 2}=20\]
what would you get for x?

- anonymous

2

- ash2326

how you would get 2???

- anonymous

i'm lost .. again .

- anonymous

would it be 10 ?

- anonymous

because 20/2

- anonymous

?

- ash2326

we have
\[2x=40\]
divide both sides by 2 only once

- anonymous

1x = 20

- ash2326

yeah so x =???

- anonymous

20 !

- ash2326

good, did you understand this?

- anonymous

yes now i do ! thank you very much ! (:

- ash2326

Welcome and feel free to contact me here, Anytime for help !!

- anonymous

thanks !! (: have a great day !

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