Answer:
287 students were eligible for the discount :)
Step-by-step explanation:
Answer:
In order to have ran 33 miles, Bobby would have to attend <em>32 track practices.</em>
Step-by-step explanation:
Solving this problem entails of uncovering the amount of track practices Bobby must attend in order to have ran 33 miles. Start by reading the problem carefully to break down the information provided.
You can see that Bobby has already ran one mile on his own. This is important to remember for later. The problem also states that he expects to run one mile at every track practice.
Setting up an equation will help us solve. Here is how we could set up the equation:
(<em>amount of miles already ran</em> = 1) + (<em>number of track practices</em> = x) = (<em>total miles to run</em> = 33)
1 + x = 33
The equation is now in place. You can solve this, or isolate <em>'x',</em> by using the subtraction property of equality. This means we will subtract one from both sides of the equation, thus isolating the variable.
1 + x = 33
1 - 1 + x = 33 - 1
x = 32
The variable is the only term left on the left side of the equation. This means Bobby must attend track practice <em>32 times</em> in order to have ran 33 miles.
So there are two ways.
First you could solve for y and plug it into the calculator. one equation for y1 and another for y2. Then go 2nd trace and press 5 and then press enter 3 times to find the intersection.
the second way is to solve by hand. I would suggest solve the first equation for y and plug the equation solved for y into the other equation. so y = -3x + 9 and 3x-5y=15. Then you could do 3x - 5(-3x+9) = 15. Finally solve for x and plug in x into both equations to see if you get the same y value.
The answer should be (10/3, -1)
Answer:
Yes
Step-by-step explanation:
The HL theorem basically says that two triangles are congruent if their corresponding hypotenuses and one leg are equal.
Here, the hypotenuse of both triangles are each marked with two dashes meaning they're equal. Similarly, QE and ET are both equal.
Therefore we can see that they can be proved congruent by the HL Theorem.
The most reasonable answer is b
