Derrick Said:

Can you help me with these math (11th grade) practice test problems?

We Answered:

1. draw a circle on a graph with the center of the circle at (0,0)

2.draw a line from the center of the circle, up 60 degrees from the x-axis (the horizontal one).

3.where it touches the circle, draw a line down.

4. now you will have formed a right angle triangle.

5. use sin and cos to solve for x and y
I drew a picture here it is

Brad Said:

What are the reference / coterminal angles? (11th grade math)?

We Answered:

1. 462 degrees.
Keep subtracting 360 until you get a number between 0 and 360.
462 - 360 = 102.
The reference angle will be 180 - 102 = 78

2. -315 degrees
Keep adding 360 until you get a number between 0 and 360.
-315+360 = 45. This is also your reference angle.

3. 240 degrees
The reference angle is 240-180 = 60.

4. 135 degrees
The reference angle is 180-135 = 45.

Kristin Said:

How much of Grade 11 Chemistry is mathematics and how would you compare it to Advanced Math 11?

We Answered:

Chemistry is really NOT math.
You need to understand logarithms, and fractions. But not hard at all mathwise. I mean, you are NOT going to be deriving Schroedinger's equation or anything.
You need to know how to use conversion factors, how to work logs, how to use scientific notation, which is just basically using decimal places.....EASY.
You will not do any algebra except for simple rearrangements of formulae for substitutions. Usually there is a chapter in the beginning of the book that covers any and all math you will need. So, if you have had algebra, you are fine.

Elizabeth Said:

11th grade math problem-very hard, Trigonometry, 2min problem for a mathmatition?

We Answered:

Cross multiply to get (x + yi)(7 + i) = 7(1 + i)

Multiply these out and use the fact that the real part both sides must be the same and the imaginary part both sides must be the same. Both x, y are fractions with denominator 25.

Lynn Said:

11th grade math problems for my nephew?

We Answered:

Use the formula:

D = RT (distance = rate * time)

Plane one going eastbound:
D = RT

other going westbound:
d = rt

Since they start at the same time, the times are the same so:

T = t

We're given the rates of both planes:

R = 560
r = 500

And they want to know when the distance between them is 2000 miles. since they're going in opposite directions, it's just the sum of their distances from the source location:

d + D = 2000

I'll solve this for D:

D = 2000 - d

We have everything we need now to solve this:

D = RT

substitute what we know:

2000 - d = 560t

and 2nd plane:

d = rt
d = 500t

we now have two unknowns and two equations. We really want the "t" known, so we'll first substitute the 'd' into the first equation and solve:

2000 - d = 560t
2000 - 500t = 560t
2000 = 1060t
t = 1.88679

This is in hours.

I'll multply this by 60 to get the number of minutes:

113.2

So it was 1 hour and 53 min. Since they took off at noon, the time they want for a solution is:

1:53 PM

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Let:
d = number of dimes
n = number of nickles
q = number of quarters

If she has 3 times as many dimes as nicles:
d = 3n

and twice as many quarters as nickes:
q = 2n

and she has $3.40 in all:
0.05n + 0.10d + 0.25q = 3.40

How many of each coin does she have. we have three unknowns and three equations. solve by substitution. plug in the d and q in terms of n into the final equation, solve for n, then work out d and q:

0.05n + 0.1(3n) + 0.25(2n) = 3.40
0.05n + 0.3n + 0.5n = 3.40
0.85n = 3.40
n = 4

d = 3n
d = 3(4)
d = 12

q = 2n
q = 2(4)
q = 8

She had 4 nickles, 12 dimes, and 8 quarters.

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Let:
a = number of adults
c = number of children

30 people paid $12.25 with the given fair prices per passenger:

a + c = 30
0.5a + 0.25c = 12.25

two unknowns, two equations.

a = 30 - c

0.5(30 - c) + 0.25c = 12.25
15 - 0.5c + 0.25c = 12.25
15 - 0.25c = 12.25
- 0.25c = -2.75
c = 11

a = 30 - c
a = 30 - 11
a = 19

There were 19 adults and 11 children.