All Algebra 1 Resources
Example Questions
Example Question #81 : New Sat Math Calculator
A blue train leaves San Francisco at 8AM going 80 miles per hour. At the same time, a green train leaves Los Angeles, 380 miles away, going 60 miles per hour. Assuming that they are headed towards each other, when will they meet, and about how far away will they be from San Francisco?
Around 2:45AM, about 200.15 miles away from San Francisco
Around 10:43AM, about 217.12 miles away from San Francisco
Around 3AM the next day, about 1,520 miles away from San Francisco
The two trains will never meet.
Around 10:43AM, about 217.12 miles away from San Francisco
This system can be solved a variety of ways, including graphing. To solve algebraically, write an equation for each of the different trains. We will use y to represent the distance from San Francisco, and x to represent the time since 8AM.
The blue train travels 80 miles per hour, so it adds 80 to the distance from San Francisco every hour. Algebraically, this can be written as .
The green train starts 380 miles away from San Francisco and subtracts distance every hour. This equation should be .
To figure out where these trains' paths will intersect, we can set both right sides equal to each other, since the left side of each is .
add to both sides
divide both sides by 140
Since we wrote the equation meaning time for , this means that the trains will cross paths after 2.714 hours have gone by. To figure out what time it will be then, figure out how many minutes are in 0.714 hours by multiplying . So the trains intersect after 2 hours and about 43 minutes, so at 10:43AM.
To figure out how far from San Francisco they are, figure out how many miles the blue train could have gone in 2.714 hours. In other words, plug 2.714 back into the equation , giving you an answer of .
Example Question #201 : Systems Of Equations
We have three dogs: Joule, Newton, and Toby. Joule is three years older than twice Newton's age. Newton is Toby's age younger than eleven years. Toby is one year younger than Joules age. Find the age of each dog.
Joule: 9 years
Newton: 3 years
Toby: 8 year
Joule: 12 years
Newton: 1 year
Toby: 5 year
Joule: 8 years
Newton: 4 years
Toby: 8 year
Joule: 5 years
Newton: Not born yet
Toby: 1 year
none of these
Joule: 9 years
Newton: 3 years
Toby: 8 year
First, translate the problem into three equations. The statement, "Joule is three years older than twice Newton's age" is mathematically translated as
where represents Joule's age and is Newton's age.
The statement, "Newton is Toby's age younger than eleven years" is translated as
where is Toby's age.
The third statement, "Toby is one year younger than Joule" is
.
So these are our three equations. To figure out the age of these dogs, first I will plug the third equation into the second equation. We get
Plug this equation into the first equation to get
Solve for . Add to both sides
Divide both sides by 3
So Joules is 9 years old. Plug this value into the third equation to find Toby's age
Toby is 8 years old. Use this value to find Newton's age using the second equation
Now, we have the age of the following dogs:
Joule: 9 years
Newton: 3 years
Toby: 8 years
Example Question #1 : Integer Operations
What is ?
Divide the absolute values to get the magnitude of the answer.
Remember the following rules:
Positive divided by negative is negative.
Positive divided by positive is positive.
Negative divided by negative is positive.
Since we are dealing with two negatives, our answer will be positive.
Example Question #1 : Real Numbers
This is a classic order of operations question, and if you are not careful, you can end up with the wrong answer!
Remember, the order of operations says that you have to go in the following order of operations: Parentheses, Exponents, Multiplication, Division, Addition, Subtraction (also known as PEMDAS). In this equation, you will start with the parentheses. In the parentheses, we have
.
But within the parentheses, you still need to follow PEMDAS. First, we will solve the exponent, and the square of 2 is 4. Then, we'll divide 16 by 4, which gives us 4, so we can rewrite our original equation as
.
We can now divide into , which gives us
.
The last step is to add and subtract the numbers above, paying careful attention to negative signs. In the end, we end up with because added to equals , and minus equals .
Example Question #1 : Integer Operations
If a number is divisible by 9 and 12, it must also be divisible by...
I. 36
II. 48
III. 144
I and II only
III and II only
I only
III only
II only
I only
For a number A to be divisible by another number B, A must share all of the prime factors of B. For example, 100 is divisible by 10 because the prime factors of 10 (5 and 2) are found in the prime factors of 100 (2, 2, 5, and 5).
In this problem, we have a number A that is divisible by 9 and 12. First, find the prime factors of 9 and 12. The factors of 9 are (3, 3). The factors of 12 are (2,2,3). So, to be divisible by 9 and 12, the number A must have the factors (2,2,3,3). Multiply those together, and we get 2*2*3*3= 36.
So, A must be divisible by 36. It could be divisible by the other answer choices, but since the question asked which choice must be right, we can only choose 36.
Example Question #3 : Real Numbers
Rewrite 100 as a number in base six.
One way to do this:
Divide 100 by 6. The remainder will be the last digit.
Now divide the quotient by 6. This remainder will be the second-to-last digit.
The quotient is less than 6, so it is the first digit. The base-six equivalent of 100 is .
Example Question #1 : Integer Operations
Divide and express the quotient in scientific notation:
All of these choices are equivalent to the correct answer. But only is in scientific notation.
Example Question #2461 : Algebra 1
Divide:
Write the factors of the two numbers.
Example Question #1 : Integer Operations
If a number is divisible by , which of the following is always true?
I. Divisible by
II. Divisible by
III. Divisible by
I and II
I only
II only
None true
All true
I and II
If a number is divisble by some number , then it is also divisible by all of its factors. Hence, if a number is divisible by , it is also divisible by , and . Thus, I and II are both true.
Example Question #2 : Real Numbers
Divide .
Rewrite the expression in fraction form.
Write out the common factors of both the numerator and denominator.
We can then see that the fraction can be reduced to without having to conduct long division.
The answer is .