My invertebrate project was produced on sway, about the Giant Pacific Octopus.

https://sway.office.com/gwlD2nSS6l8iZQum?ref=Link

My brainstorm sheet was handed in person.

Skip to content
# Daniel's Blog

## Email

## Invertabrate Project

## Pre Calc-Week 15

## pre Calc Week 14

## Pre Calc-Week 13

## Pre Calc-Week 12

## Pre Calc- weeek 11

## Pre Calc-Week 10

## Pre Calc- Week 9

## Pre Calc- Week 8

## Week 7-Pre Calc

Welcome to the blog

132-dbandi@sd43.bc.ca

My Riverside Rapid Digital Portfolio

My invertebrate project was produced on sway, about the Giant Pacific Octopus.

https://sway.office.com/gwlD2nSS6l8iZQum?ref=Link

My brainstorm sheet was handed in person.

This week we worked on Sine Law

Insert Image

p/sin P=q/sin = r/sin R

Substitute the given measures: q=8.1, P=75Degrees, R=39Degrees

No 2 Ratios have only one unknown quantity, so use the sum of the angles in a triangle to Calculate Q.

Q=180Degrees – (75Degrees + 39Degrees)

Q=66degrees

Use the first 2 ratios in the Sine Law, and substitute for Q

p/Sin 75Degrees = 8.1/sin 66Degrees Solve for p. Multiply each side by Sin 75Degrees

p=8.1sin75Degrees/sin 66degrees

p=8.56

This week we had begun the Trigonometry unit, We began we Angles in Standard Position in Quadrant 1

(Determining the Trigonometric Ratios of an Angle Given a Terminal point)

*Diagram*

Too determine the distance r from the origin to P

Use the Pythagorean Theorem in the right Triangle

r(Squared) = 4(Squared) + 7 (Squared)

r(Squared)=65

r=*Root* 65

Too determine the primary trigonometric ratios of Fata

Sine Fata = Opp/Hyp

Sine Fata =7/*Root*65

Cose Fata=adj/hyp

Cose Fata 4/*Root*65

Tan Fata *Opp/Adj

Tan Fata= 7/4 or 1.75

Determine the measurement to the nearest degree

Use any trigonometric ratio from part b

Tan fata = 1.75

Fata=tan-1(1.75)

Fata=60.255

Fata is approximately 60 Degrees

This week in Pre calc we learned how to Solve Rational Equations with Binomial Denominators

3x+1/x(squared)-1=-x/x+1

We would start by factoring the denominator

3x+1/(x-1)(x+1)=-x/x+1

The non permissible values are (X=1) and (X=-1)

A common denominator is :(x-1)(x+1)

3x+1/(x-1)(x+1)=-x/x+1 You would then need to multiply each side by (x-1)(x+1)

(x-1)(x+1)(3x+1)(x-1)(x+1)=(x-1)(x+1)(-x/x+1)

Remove the (x-1) and (x+1) factors which match with each other basically remove the number of (x+1)(x-1) pairs.

3x+1=-x(squared)+x

make right side 0 by bringing right side to left side.

x(squared)+2x+1=o

(x+1)(x+1)=0

x=-1

x=-1 is a non permissible value of the variable, so the equation has no solution

We started instead on Chapter 6 instead of Chapter 5

Determining Non-Permissible Values

x(Squared)+2/ x(squared)-x-6

Equate the denominator to 0, then solve the equation.

x(squared)-x-6=0 Factor

(x-3)(x+2)=0 or x+2=0

x=3 x=-2

The none-permissible values of x(squared) +2/x(squared)-x-6 are x=3 and x=-2

x/x(squared) +1

Since the square of a number is always positive, x(squared) > 0.

so, x(squared) +1 >0

since the denominator cannot equal 0, there are no non-permissible values.

Using interval Notation

x<1

X is less than 1.

On the number line (Photo I), for less than use and open circle

In inverted notation

The critical value 1 is not included in the solution.

So use curved brackets

All values less than 1 are included in the solution, so use curved brackets.

All values less than 1 are included in the solution, so use the symbol -(Infinite) with a curved bracket.

X < -3 or X > 1

X is less than -3, or X is greater than 1.

On the number line for less than -3 use an open circle. For greater than 1 use an open circle.

For greater than 1 use an open circle.

This Week we learned about, Determining Characteristics from the Factored Form of the equation

y=0.5x(Squared)-1.5x-5 Too begin remove 0.5 as a common factor then multiply each term by 2.

y=0.5(xSquared-3x-10)

y=0.5(x-5)(x+2)

Them compare it to Factored fourm y=a(x-x1)(x-x2)

We first find the x-intercepts which are 5 and -2

Now we would find the axis of symmetry which would be between 5,-2

5-2/2=1.5

now we would substitute x=1.5

y=0.5(1.5)Squared -1.5(1.5)-5

y=-6.125 So the coordinates of the vertex are (1.5,-6.125)

We learned about Analyzing Quadratic Functions of the Form y= a(x-p) Squared + q

We need to begin by identifying the coordinates of the Vertex of the graph of each function (Graph)

Substitue the Values in the location P/Q

Choose a different point on the graph and substitute its coordinates in the equation, we chose (5,4) in this situation, we would substitute x=5,y=4

4 = a(5-3) squared + 2

subtract what is inside the bracket

4=a(2)Squared+ 2

4=4a + 2

take the 2 over and subtract

2=4a

a=1/2

which makes the equation y=1/2(x-3)squared + 2

In week 8 of Math 11 we learned about Properties of a Quadratic Function, too begin a quadratic function is any function that can be written in the form y= ax(squared) + bx + c, and c E R and a (a =/ 0). This is called general form of the equation of a quadratic function. We can now break down the graph, every Quadratic equation has a curve called a Parabola, the parts of the Parabola include the Vertex which is the highest or lowest point, the Vertex may be a Minimum point or a Maximum point. There also is the Axis of symmetry which is best shown through the image.

This week in Math 11 we learned how interpret a Discriminant, First we need to know what a discriminant is (Photo 1) in (Photo 2) it shows a Interpret. The Formula B(Squared)-4ac

If the equation ends up being a positive it means that the Equation will have two real answers, if it equals 0 it becomes exactly one real root, if its a negative it has no real roots.

Photo 1

Photo 2