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2 years ago

PLEASE ANSWER QUICKLY AND HELP ME!! THANK YOU SO MUCH!!

Mathematics

1 answer:

coldgirl [10]2 years ago

5 0

Hello there!

To start off this problem, let's replace f(x) with x in the equation. This gives us the following.

$x= \dfrac{3}{4} x+12$

Subtract 3/4*x from both sides.

$x- \dfrac{3}{4} x=12$

$\dfrac{4}{4} x-\dfrac{3}{4}x=12$

$\dfrac{1}{4} x=12$

Now, multiply both sides by 4 to eliminate the fraction on the left side.

$x=12*4=48$

Your answer is B. Hope this helps! :)

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Corrected work and description of error

hichkok12 [17]

Answer:

hmm i have no idea can u send a clear pic of question?

8 0

2 years ago

20 Points! Which ordered pair is the best estimate for the solution of the system of equations?

yan [13]

Answer:

The best estimate for the solution is the ordered pair $(-6.5,-3.5)$

Step-by-step explanation:

we have

$y=\frac{3}{2}x+6$ ------> equation A

$y=\frac{1}{4}x-2$ ------> equation B

we know that

using a graphing tool, the solution of the system of equations is the intersection point both graphs

The intersection point is $(-6.4,-3.6)$

therefore

The best estimate for the solution is the ordered pair $(-6.5,-3.5)$

3 0

2 years ago

What number is 20% of 60

KonstantinChe [14]

We know that 10% of 60 is 6.

60*.1= 6

So,

6+6= 12 or 20% of 60.

12 is the number.

I hope this helps!
~kaikers

5 0

2 years ago

A horizontal trough is 16 m long, and its end are isosceles trapezoids with an altitude of 4 m, a lower base of 4 m, and an uppe

Ganezh [65]

Answer:

0.28cm/min

Step-by-step explanation:

Given the horizontal trough whose ends are isosceles trapezoid

Volume of the Trough =Base Area X Height

=Area of the Trapezoid X Height of the Trough (H)

The length of the base of the trough is constant but as water leaves the trough, the length of the top of the trough at any height h is 4+2x (See the Diagram)

The Volume of water in the trough at any time

$Volume=\frac{1}{2} (b_{1}+4+2x)h X H$