3 years ago

Solve ax^2=bx for x

Mathematics

1 answer:

Vladimir79 [104]3 years ago

3 0

$ax^2=bx\\
ax^2-bx=0\\
x(ax-b)=0\\
x=0 \vee x=\dfrac{b}{a}$

You might be interested in

Solve and graph 4x<40

allochka39001 [22]

I am sorry , i no have graph

7 0

3 years ago

The rate change of (3,0) and (0,9)

ddd [48]

Answer:

rate of change is -3

A = (9 - 0) / (0 - 3) = 9 / -3 = -3

Step-by-step explanation:

8 0

1 year ago

Read 2 more answers

5/100+8/10=?......help

lukranit [14]

85/100
17/20 when simplified

8 0

2 years ago

Read 2 more answers

-194=-6x+2(-7x+13)<br> ?????

lara31 [8.8K]

Answer:l

Step-by-step explanation:

8 0

2 years ago

It takes machine A x hours to manufacture a deck of cards that machine B can manufacture in y hours. If machine A operates alone

nlexa [21]

Answer:

The expression that shows how long the two machines will operate simultaneously is:

$\frac{y(100x-z)}{x+y}$

Step-by-step explanation:

We know that:

x: hours to manufacture a deck of cards for machine A

y: hours to manufacture a deck of cards for machine B

z: hours that machine A operates alone

The number of decks manufactured only by machine A is:

$\frac{z}{x}$

So, the remaining decks are given by:

$100-\frac{z}{x}=\frac{100x-z}{x}$

Then, the combined rate of machines A and B would be:

$\frac{1}{x} +\frac{1}{y} =\frac{x+y}{xy}$

The work-rate formula is:

$Amount= Rate \times Time$

Hence, the time that the two machines work simultaneously is:

$Time=\frac{Amount}{Rate}$

$Time=\frac{Amount}{Rate} =\frac{\frac{100x-z}{x} }{\frac{x+y}{xy} } ={\frac{100x-z}{x} \times \frac{xy}{x+y}=\frac{y(100x-z)}{x+y}$

3 0

3 years ago