# Week 13 in Pre-Calculus 11

1. In this week we learned how to solve equations that have variables in their denominators. You cross multiply them. You multiply the nominator on the left with the denominator on the right, and you multiply the denominator on the left by the nominator on the right. You put the answers of these on two different sides of a equality sign. And then you solve the rest like any other equation.

# Week 12 in Pre-Calculus 11 In this week we learned about non-permissible values.You cant have 0 as your denominator. So if you have a trinomial with a variable in your denominator, that trinomial can’t value as zero. So your variable cant be something that makes the trinomial equal to zero. The numbers that your variable can’t be is called the non-permissible values.

# Week 9 in Pre-Calculus 11

In this week, I learned how to turn a trinomial into a parabola formula. If you have ax^2+bx+c, you put the c aside and then complete the square with ax^2+bx. after you completed the square, you calculate the sum of the number you added to complete the square with c. Then you’ll have your formula.

# Week 8 in Pre-Calculus

In this week I learned about parabolas and how to work with them.If we have ax^2, “a” shows the stretch or the compress. If it’s negative then it’s opened down, if positive it’s opening up. X^2 + p means the y intercept is at “p”. If we have (x – q)^2, the value of q shows the horizontal translation. If it’s negative then it means the parabola goes left, if positive it goes right.

# Week 7 in Pre-Calculus 11

In this week I learned how yo solve a quadratic equation with the quadratic formula. The formula is written on the picture attached to this post.

a= coefficient of x squared

b= coefficient of x

c = the integer

You just have to replace the a, b, and c with numbers and then you just have to calculate. # Week 6 in Pre-Calculus

In this week I learned how to solve a quadratic equation. A quadratic equation is a equation which the variable in it has a power of 2 or more. In order to solve it, we first have to make one side of the equation’s to zero. For ex. (x+5)(x-7)=0

Now we have a equation, which in it the, the multiplication of two brackets evens to zero. That means one the brackets worths zero. Either x+5=0 or x-7=0. we have to solve these to new equations.

x+5=0 ==> x=-5

x-7=0 ==> x=7

both of these answers are correct and it can be any of them.

# Week 5 in Pre-Calculus

In this week I learned how to work with slope and graph. I learned when we have Y=3x +4,

The 3x is actually 3/1x. The Nominator is the amount of the units you go up and the denominator is the amount of the units you go right. The 4 is the Y-intercept which means it’s the number you start with. So when you have Y=3/1x + 4,  the first point is 4 on the Y axis, (0,4). then you go 3 units up and 1 unit right. that’s the second point. you keep doing that until you have enough points to draw the line.