What is the missing term? (10x-4x^2)-(7x+?)=3x-6x^2

Mathematics

2 answers:

Nana76 [90]4 years ago

7 0

The given problem is =

$( 10x-4x^{2} )-(7x+?)=3x-6x^{2}$

Lets suppose the missing term to be T

So equation becomes:

$10x-4x^{2} -7x-T=3x-6x^{2}$

$T=10x-4x^{2} -7x-3x+6x^{2}$

$T=2x^{2}$

Hence, the missing term is $2x^{2}$

mafiozo [28]4 years ago

6 0

The answer is 2x^2
Because everything on the right of the minus sign will turn negative

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What is the value of 11−3m , when m = 2?

user100 [1]

When a number is next to a letter like so: 3m You would multiply, so
3 ~ 2 = 6 then solve the problem like this

11-3m= ?
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so 11-3m=5
Hope this helps! :)

4 0

4 years ago

Read 2 more answers

<img src="https://tex.z-dn.net/?f=%20%5Cfrac%7B2%7D%7B3%7D%20k%20-%20%28k%20%2B%20%20%5Cfrac%7B1%7D%7B4%7D%29%20%3D%20%20%5Cfrac

VikaD [51]

Firstly, foil -(k + 1/4) (think of the minus sign as -1):

$\frac{2}{3}k-k-\frac{1}{4}=\frac{1}{12}k+\frac{4}{12}$

Next, combine like terms:

$-\frac{1}{3}k-\frac{1}{4}=\frac{1}{12}k+\frac{4}{12}$

Next, we have to add 1/3k on both sides, but first we have to find the LCD, or lowest common denominator, of 3 and 12. To do this, list the multiples of both and the lowest one they share is their LCD. In this case, the LCD is 12. Multiply both sides of -1/3 by 4/4 and 1/12 by 1/1:

$-\frac{1}{3}\times \frac{4}{4}=-\frac{4}{12}\\\\\frac{1}{12}\times \frac{1}{1}=\frac{1}{12}\\\\-\frac{4}{12}k-\frac{1}{4}=\frac{1}{12}k+\frac{4}{12}$

Now add 4/12k on both sides of the equation:

$-\frac{1}{4}=\frac{5}{12}k+\frac{4}{12}$

Next, to subtract 4/12 on both sides we need to find the LCD of 4 and 12. It's the similar process as we did with 12 and 3. This time the LCD is also 12. Multiply both sides of -1/4 by 3/3 and 4/12 by 1/1:

$-\frac{1}{4}\times \frac{3}{3}=-\frac{3}{12}\\\\\frac{4}{12}\times \frac{1}{1}=\frac{4}{12}\\\\-\frac{3}{12}=\frac{5}{12}k+\frac{4}{12}$