Put your answer in standard notation to 4 decimal places.)

Mrs.smells math class has 20 questions worth 100 points. if the multiple choice are worth 3 points and the short answer is worth

andreyandreev [35.5K]

Let, Number of multiple choice = x
Number of short answer question = y

Now, x + y = 20
& 3x + 11y = 100

Multiply 1st equation by 3, and subtract from eq,2,
3x + 3y = 60
After subtraction,
8y = 40
y = 40/8
y = 5

Now, substitute in eq (1),
x + 5 = 20
x = 15

In short, There are 15 Multiple choice & 5 short answer questions

Hope this helps!

4 0

3 years ago

The table below represents a linear function f(x) and the equation represents a function g(x):

notsponge [240]

(0,0)(1,3) = (3 - 0) / (1 - 0) = 3

slope = 3
equation for this line is y = 3x...and since u have (0,0) as a point on this line, ur y int is (0,0)

g(x) = 7x + 2...slope here is 7 and y int is (0,2)
==========================
The slope of f(x) = 3. I found this using the slope formula (y2 - y1) / (x2 - x1).
The slope of g(x) = 7. Because in y = mx + b form, the slope is in the m position.
g(x) has a steeper slope.
=========================
f(x) has a y int of (0,0)...the line cuts thru the origin.

g(x) has a y int of (0,2)...because in y = mx + b form, the y int will be in the b position.

So, g(x) has a greater y int.

3 0

3 years ago

What is the greatest common factor of 88 and a 121

Anton [14]

Answer: 11

Step-by-step explanation: The factors of 88 are 1,2,4,8,11,22,44,88; The factors of 121 are 1,11,121.

8 0

3 years ago

A tank contains 30 lb of salt dissolved in 500 gallons of water. A brine solution is pumped into the tank at a rate of 5 gal/min

Dmitry [639]

At any time $t$ (min), the volume of solution in the tank is

$500\,\mathrm{gal}+\left(5\dfrac{\rm gal}{\rm min}-5\dfrac{\rm gal}{\rm min}\right)t=500\,\mathrm{gal}$

If $A(t)$ is the amount of salt in the tank at any time $t$ , then the solution has a concentration of $\dfrac{A(t)}{500}\dfrac{\rm lb}{\rm gal}$ .

The net rate of change of the amount of salt in the solution, $A'(t)$ , is the difference between the amount flowing in and the amount getting pumped out:

$A'(t)=\left(5\dfrac{\rm gal}{\rm min}\right)\left(\left(2+\sin\dfrac t4\right)\dfrac{\rm lb}{\rm gal}\right)-\left(5\dfrac{\rm gal}{\rm min}\right)\left(\dfrac{A(t)}{50}\dfrac{\rm lb}{\rm gal}\right)$