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3 years ago

2x-3 divided by x-4=2divided by 3

Mathematics

1 answer:

Marrrta [24]3 years ago

7 0

Given:

$\dfrac{2x-3}{x-4}=\dfrac{2}{3}$

To find:

The value of x.

Solution:

We have,

$\dfrac{2x-3}{x-4}=\dfrac{2}{3}$

By cross multiplication, we get

$3(2x-3)=2(x-4)$

$6x-9=2x-8$

Isolating variable terms, we get

$6x-2x=9-8$

$4x=1$

Divide both sides by 4.

$x=\dfrac{1}{4}$

$x=0.25$

Therefore, the solution of the given equation is x = 0.25.

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What is equivalent to 4 5/3

storchak [24]

Answer: 17/3

Step-by-step explanation:

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3 years ago

In △ABC, point M is the midpoint of AB , point D∈ AC so that AD:DC=2:5. If AABC=56 yd2, find ABMC, AAMD, and ACMD.

Komok [63]

Since point M is the midpoint of AB, then AM=MB.

Consider the area of the triangles ABC and BMC:

$A_{ABC}=\dfrac{1}{2}\cdot AB\cdot h_c=56\ yd^2,$

where $h_c$ is the height drawn from the vertex C to the side AB.

So, $AB\cdot h_c=112\ yd^2.$

Now

$A_{BMC}=\dfrac{1}{2}\cdot BM\cdot h_c=\dfrac{1}{2}\cdot \dfrac{AB}{2}\cdot h_c=\dfrac{1}{4}\cdot AB\cdot h_c=\dfrac{1}{4}\cdot 112=28\ yd^2.$

Also

$A_{AMC}=A_{ABC}-A_{BMC}=56-28=28\ yd^2.$

Now consider the area of the triangles AMD and CMD. Let $h_M$ be the height drawn from the point M to the side AC.

$A_{AMD}=\dfrac{1}{2}\cdot AD\cdot h_M=\dfrac{1}{2}\cdot \dfrac{2AC}{7}\cdot h_M=\dfrac{2}{7}\cdot \left(\dfrac{1}{2}\cdot AC\cdot h_M\right)=\dfrac{2}{7}\cdot A_{AMC}=\dfrac{2}{7}\cdot 28=8\ yd^2.$

Therefore,

$A_{MDC}=A_{AMC}-A_{AMD}=28-8=20\ yd^2.$

Answer: $A_{MBC}=28\ yd^2, A_{AMD}=8\ yd^2, A_{MDC}=20\ yd^2.$

5 0

3 years ago

Read 2 more answers

The sum of two numbers is x. if one of the numbers is 12, what is the other? 12 - x x - 12 12 x 12 · x user: one side of a squar

I am Lyosha [343]

I don't get your question, can u type it again more clearly? thanks

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4 years ago

Which graph COULD represent y = x2 - 3? A) B) C) D)

Naya [18.7K]

I believe the answer should be A.

6 0

3 years ago

Read 2 more answers

How do I solve systems of equations? I was absent and the test is next week!

MrRa [10]

Well, there are three different ways to solve systems of equations:
a) Graphing
b) Substitution
c) Elimination

Graphing:
If your system of equations are written in slope-intercept form (y=mx+b "m" being the slope and "b" being the y-intercept) its quite easy.
The y-intercept is pretty much where your line touches the y-axis so if your first equation is: y=3x-2 then you will put your first point on (0,-2). Then from that point you follow the slope. if the slope is 3 then you'll rise 3 and run 1 (go three times up and one to the right and put your point right there) till you feel you have enough points to make a line. You do the same process with your second equation.

Substitution:
This is best when you have a system of equations in which an equation tells you what x or y are equal to. For example: if I have "4x+2=y" and "2y+3x= 26" you can "replace" your "2y" with the equation "4x+2=y" which gives you a brand new equation: " 2(4x+2)+3x=26". To solve this you do the distributive property and link like terms which gives you the equation "11x+4=26". Then you solve like a normal equation, isolating x and then you'll get the value of x (in this case x=2). Now you go to the first equation "4x+2=y" and replace "x" to its value (in this case it's 2) and you get y (in this example, y=10).

Elimination:
I think this one is one of the easiest methods. Let's say my systems of equations is "5x-6y=-32" & "3x+6y=48". If you take a look at y's coefficients, you see they are the same. So you can "cancel them out" like this:
5x-6y=-32
+ 3x+6y=48
-----------------------
8x=16 <-------- when you get this, you can solve like a normal equation
x=2

Also, if you have, for example:
2x+6y=190 & 2x+3y=130
You can solve this by: canceling x or multiplying the second equation by 2 and canceling y.
I'm gonna show you how to do the second way (multiplying and canceling y)
My second equation is "2x+3y=130". To get an equivalent equation, I just have to multiply the whole equation by any number. If I want to cancel y, then I have to multiply by 2.
So "2x+3y=130" turns into "4x+6y=260"
Now you follow the same procedure as before:
2x+6y=190
- 4x+6y=260
-------------------------------
-2x= -70
2x=70--------> x=35
Then you pretty much follow the substitution method by replacing "x" to "35" in one of the equations.

Hope this covers all you missed in class. If you have any questions you can comment them here or message me. :D Good luck in your exam!!!

5 0

3 years ago