For this case we propose a system of equations. We have to:
x: Let the variable that represents the number of dimes
y: Let the variable that represents the number of quaters
We know that:
One dime equals 10 cents, $0.10
A quater equals 0.25 cents, $0.25
According to the statement we have:
We multiply the first equation by -0.10:
We have the following equivalent system:
We add the equations:
Approximately 3 quater coins
And two dimes
Answer:
3 quater
2 dimes
Answer: Assuming the function is :
The x-intercept is .
The y-intercept is .
The horizontal asymptote is .
The vertical asymptote is .
Step-by-step explanation:
I'm going to assume the function is: and not
.
So we are looking at .
The x-intercept is when y is 0 (when g(x) is 0).
Replace g(x) with 0.
A fraction is only 0 when it's numerator is 0. You are really just solving:
Subtract 1 on both sides:
Divide both sides by 2:
The x-intercept is .
The y-intercept is when x is 0.
Replace x with 0.
.
The y-intercept is .
The vertical asymptote is when the denominator is 0 without making the top 0 also.
So the deliminator is 0 when x-3=0.
Solve x-3=0.
Add 3 on both sides:
x=3
Plugging 3 into the top gives 2(3)+1=6+1=7.
So we have a vertical asymptote at x=3.
Now let's look at the horizontal asymptote.
I could tell you if the degrees match that the horizontal asymptote is just the leading coefficient of the top over the leading coefficient of the bottom which means are horizontal asymptote is . After simplifying you could just say the horizontal asymptote is
.
Or!
I could do some division to make it more clear. The way I'm going to do this certain division is rewriting the top in terms of (x-3).
So you can think it like this what value will y never be here.
7/(x-3) will never be 0 because 7 will never be 0.
So y will never be 2+0=2.
The horizontal asymptote is y=2.
(Disclaimer: There are some functions that will cross over their horizontal asymptote early on.)
Answer: The answer is (B) ∠SYD.
Step-by-step explanation: As mentioned in the question, two parallel lines PQ and RS are drawn in the attached figure. The transversal CD cut the lines PQ and RS at the points X and Y respectively.
We are given four angles, out of which one should be chosen which is congruent to ∠CXP.
The angles lying on opposite sides of the transversal and outside the two parallel lines are called alternate exterior angles.
For example, in the figure attached, ∠CXP, ∠SYD and ∠CXQ, ∠RYD are pairs of alternate exterior angles.
Now, the theorem of alternate exterior angles states that if the two lines are parallel having a transversal, then alternate exterior angles are congruent to each other.
Thus, we have
∠CXP ≅ ∠SYD.
So, option (B) is correct.
