Enable contrast version

# Tutor profile: Deb D.

Inactive
Deb D.
Homeschool Mom and Tutor
Tutor Satisfaction Guarantee

## Questions

### Subject:Trigonometry

TutorMe
Question:

A 12-foot ladder resting against a wall makes a 75 degree angle with the ground. How high up the wall does the ladder reach?

Inactive
Deb D.

If you drew a picture of what is described in the question, you would see that the ladder, the grou nd and the wall form a triangle. You can assume that the wall and the ground meet at a 90° angle, thus making the ladder the hypotenuse of the triangle since the ladder is opposite the largest angle. Since we are given two variables, we can use the equation for the sin of the angle to solve for our unknown (how high up the wall the ladder reaches). We know that the sin of an angle is equal to the side opposite the angle divided by the hypotenuse. So in this case we have: sin75 = ht up the wall/12 ft Multiplying both sides by 12, we get: 12(sin75) = ht up the wall 12(0.9659) = ht up the wall 11.59 ft = the height up the wall

### Subject:Chemistry

TutorMe
Question:

Balance the following chemical reaction: FeBr3 + H2SO4 → Fe2(SO4)3 + HBr

Inactive
Deb D.

In balancing chemical reactions, it is often easiest to start with the more complicated compound first. In this case, we will start with Fe2(SO4)3. Since we have two Fe's on the right side of the equation, we will add a 2 in front of the FeBr3 and we get: 2FeBr3 + H2SO4 → Fe2(SO4)3 + HBr Next, in the same compound, we have 3 sulfate ions on the right of the equation, so we multiply the H2SO4 on the left by 3. 2FeBr3 + 3H2SO4 → Fe2(SO4)3 + HBr Now, we see that we have 6 Br's on the left and 6 H's on the left. So, we can easily finish balancing the equation by multiplying the HBr on the right by 6. 2FeBr3 + H2SO4 → Fe2(SO4)3 + 6HBr Which leads to all of the elements on the left equal to all of them on the right.

### Subject:Algebra

TutorMe
Question:

An exam worth 145 points contains 50 questions. Some are worth 2 points and some are worth 5 points. How many questions of each type are on the exam?

Inactive
Deb D.

If a = the number of questions worth 2 points, and b = the number of questions worth 5 points, Then we can write two equations with the two unknowns and solve for both of them. First, the number of 2 point questions plus the number of 5 point questions is equal to the total number of questions, or: a + b = 50 Second, the total number of points (145) is equal to the sum of the number of points of a and the number of points of b, or: 145 = 2a + 5b When we subtract the first equation from the second, we want to elimate one of the variables so that we can solve for the second variable. In this instance, we multiply the first equation by -2 and add it to the second equation: 145 = 2a + 5b 145 = 2a + 5b -2( 50 = a + b) -100 = -2a - 2b ______________ ____________ 45 = 3b Then, divide both sides by 3 to find b. b = 15 questions worth 5 points We then solve for a by substituting b into either equation. Using the first equation will be the easiest. a + 15 = 50 -15 -15 __________ a = 35 questions worth 2 points We can check the answer by substituting both a and b into the equation we didn't use to solve for the second unknown to make sure it is correct: 145 = 35(2) + 15(5) 145 = 70 + 75 145 = 145

## Contact tutor

Send a message explaining your
needs and Deb will reply soon.
Contact Deb