All categories

3 years ago

Given that a, b, and c are non-zero real numbers and a + b ≠ 0, solve for x. ax + bx - c = 0

Mathematics

2 answers:

Stolb23 [73]3 years ago

4 0

$ax+bx-c=0\\
x(a+b)=c\\
x=\dfrac{c}{a+b}$

JulsSmile [24]3 years ago

3 0

$ax+bx-c=0 \ \ \ |+c \\
ax+bx=c \\
(a+b)x=c \ \ \ |\div (a+b), a+b \not= 0 \\
\boxed{x=\frac{c}{a+b}}$

You might be interested in

Which graph does NOT represent y as a function of x?

xxMikexx [17]

A
since it doesn’t pass the vertical line test, which proves if a graph is a function.

8 0

3 years ago

Mai's bank account is oversrawn by $60,which means her balance is -$60. She gets $85 for her birthday and deposits it into her a

Phoenix [80]

Answer:

Step-by-step explanation:

-60 plus 85 equals 25 so she has 25 dollars in the bank now.

6 0

3 years ago

Read 2 more answers

Please help, i'm horrible @ math :)

katovenus [111]

$\displaystyle\\
1 \frac{2}{5} +\Big(-5 \frac{1}{2} \Big) = ? \\ \\ 
1 \frac{2}{5} = \frac{1\times 5+2}{5}=\boxed{\frac{7}{5}} \\ \\ 
-5 \frac{1}{2} =-5 -\frac{1}{2} = \frac{-5\times2-1}{2} =\frac{-10-1}{2}=\frac{-11}{2}= \boxed{-\frac{11}{2} }\\ \\ \texttt{OR} \\ \\ 
-5 \frac{1}{2} = -\Big(5 \frac{1}{2} \Big)= -\Big( \frac{5\times2+1}{2} \Big)=-\Big( \frac{11}{2} \Big)= \boxed{-\frac{11}{2} }$

$\displaystyle\\
\Longrightarrow ~~1 \frac{2}{5} +\Big(-5 \frac{1}{2} \Big) =\frac{7}{5} -\frac{11}{2} = \frac{7\times 2}{5\times 2} -\frac{11\times 5}{2\times 5} = \\ \\ 
= \frac{14}{10} -\frac{55}{10} = \frac{14-55}{10} =\frac{-41}{10} = -\frac{41}{10}=-\frac{40+1}{10}=\boxed{\boxed{-4\frac{1}{10}}}$

7 0

3 years ago

X=A- B/A. How to find A? <br><br>

Airida [17]

Answer:

A = $\frac{ x+\sqrt{x^2+4b}}{2}, \frac{ x-\sqrt{x^2+4b}}{2}$

Step-by-step explanation:

First manipulate the equation as such,

$x=A-\frac{B}{A}$

$0=A - \frac{B}{A} - X$

$0=A^2 -AX - B$

Then use the quadratic formula,

$x= \frac{-b\pm\sqrt{b^2-4ac} }{2a}$ when, $0 = ax^2 +bx +c$ .

Which gives $A= \frac{ x+\sqrt{x^2+4b}}{2}, \frac{ x-\sqrt{x^2+4b}}{2}$ .

8 0

3 years ago

Serena drove 40 km on 3 L of gasoline. Which of the following would be an equation to help you find out how far she can travel o

velikii [3]

So here is the solution to the given problem above.
Given that Serena drove 40km on 3L of gasoline, we can say that she drove 13.33 kilometers per liter of gasoline. Since we want to find out how far she travelled with a full tank of 50L, we will only multiply 13.33 by 50 and we get 666.67 kilometers. So the answer would be 666.67 kilometers for 50 L of gasoline. Hope this helps.

8 0

3 years ago