Sergio Said:

Can somebody explain to me exactly how to solve this practice ACT math problem?

We Answered:

We have to use algebra here.
Let's call H the number of hours she works.
45 * H + 30 = 210
because they earn $45 per hour, with a guaranteed $30 for started. We make this equal to the $210 bill in this case.

When manipulating equations, we always do the same thing to both sides. First we subtract 30 from both sides:
45 * H + 30 - 30 = 210 - 30
45 * H = 180
This time divide both sides by 45:
45 * H / 45 = 180 / 45
H = 180 / 45
H = 4

Answer: B, 4 hours.

Tracy Said:

If I got 22 questions right on the ACT Math practice test, what is my raw and scale score?

We Answered:

the raw score is 22 or the # correct.
your scale score would probably pretty low.

the act site says a raw score of 50-52 has a scale score of about 22.

you should really try harder. there are 60 math questions!!!

Patrick Said:

sat/act math practice question!?

We Answered:

Ok, the rule of a triangle is:
No side can be longer than or equal to the length of the sum of two other sides. Therefore, if you are looking for the shortest possible side, then you'll take the congruent sides as small as possible. Let's call one of the congruent sides x:
Since the base cannot be longer than the sum of the other sides,
2x>13
x>6.5 You want an integer, so x=7
Shortest possible side is 7 units.

Brent Said:

ACT practice MATH HELP please answer this?

We Answered:

900 miles / 50 miles per hour = 18 hours : First we find out the time for the trip at 50 mph

But she wants to get there 3 hours earlier.

So she only wants to drive 18 - 3 = 15 hours.

So we need to find out how fast 900 miles / 15 hours is.

900 miles / 15 hours = 60 miles per hour.

So she drove 60 - 50 = 10 miles per hour faster.

Cody Said:

ACT MATH PORTION PRACTICE TEST HELP....NEED SOME HELP?

We Answered:

First, let's use "standard" typing notation for all of our math functions: "*" = multiply, "/" = "divide", and "^" = "to the power of".

As far as not giving you the answer, that's not always easy, as sometimes the explanation readily gives the answer. However, I won't just hand you the answer - we'll walk though the process of how to get the answer and hopefully that will help, even though the answer may be given at the end.

Here we go...

+++++
Problem 1:

Recall that "x%" can be written as "x/100". For example: 0.01 = 1% = 1/100

So you can substitute P/100 for P%. Now, we need only write the equation from the description, and then look at the choices to find the right one.

"M is P% of what number?" can be written as:

M = P% * x

Solving for x gives us:

x = M/(P%)

And then substituting the alternate way of presenting P%:

x = M/(P/100) ? x = 100*M/P


+++++
Problem 2:

I'm assuming it's suppose to be the quadratic equation:

x^2 + 8x + s = 0

So recall that given a quadratic equation in the form of:

a*x^2 + b*x + c = 0

That you can solve for x by using the quadratic formula:

x = [-b ±?(b^2 - 4*a*c)] / (2*a)

Now, in order for there to be only one solution, does that not imply that:

b^2 - 4*a*c = 0?

Yes. If b^2 - 4*a*c ? 0, then there would be TWO solutions.

So, in our quadratic equation, a = 1, b = 8 and c = s, giving us:

8^2 - 4*1*s = 0 ? 64 - 4s = 0

Now, simply solve for s.

{Here I won't actually give you s, but it's not a difficult equation to solve.}


+++++
Problem 3:

Because this could be "any" triangle, and not (necessarily) a "right" triangle, you need the "Law of Cosines" to solve this.

Recall "Law of Cosines":

For any given triangle with sides a, b, and c, and angles ?, ß, and ? (where ? is the angle opposite of a, ß is the angle opposite of b, and ? is the angle opposite of c), then:

a^2 = b^2 + c^2 - 2bc*cos(?),
b^2 = a^2 + c^2 - 2ac*cos(ß), and
c^2 = a^2 + b^2 - 2ab*cos(?)

The angle formed by the two shortest sides will be the angle opposite the longest side. So, assign whichever variable you like to be the longest side and solve for the corresponding angle.

Let's let a = longest side. Since you know the values for a, b, and c, simply re-write:

a^2 = b^2 + c^2 - 2bc*cos(?)

To solve for cos(?), then find ? [by taking the arccos of cos(?)]:

a^2 = b^2 + c^2 - 2bc*cos(?) ? 2bc*cos(?) = b^2 + c^2 - a^2 ?
cos(?) = (b^2 + c^2 - a^2) / (2bc)

and ? = arccos[cos(?)]

+++++
Problem 4:

These are matrices (plural of "matrix") and you're right, there is not really a good way of typing them correctly in this type of parser. But the math is simple enough.

Let A be the 2 x 2 matrix:

?...............?
? A11, A12 ?
? A21, A22 ?
?...............?

{where A11 = the value in the first row and first column of A, A12 is the value in the first row, second column of A, etc.}

and let B be the 2 x 2 matrix:

?...............?
? B11, B12 ?
? B21, B22 ?
?...............?

and x be a scalar, then:

x*A =

?.....................?
? x*A11, x*A12 ?
? x*A21, x*A22 ?
?.....................?

{i.e. each individual element is multiplied by the scalar "x".}

and A ± B =

?..............................?
? A11±B11, A12±B12 ?
? A21±B21, A22±B22 ?
?..............................?

{i.e. the corresponding indivual elements from each matrix are added or subtracted together.}

So then in the problem, 2A-B would be the matrix:

?.....................................…
? (2*A11) - B11, (2*A12) - B12 ?
? (2*A21) - B21, (2*A22) - B22 ?
?.....................................…

QED

*****
IF YOU DON'T WANT TO KNOW THE ACTUAL ANSWERS TO 2, 3, and 4, STOP READING NOW!!


I MEAN IT!!!


Heare are the answers:

Problem 2: s = 16

Problem 3: ? = 90°

Problem 4:

?........?
? -9, 8 ?
? 7, 13?
?........?