HELP ME WITH 2 MATH ASSIGNMENTS MY FINAL ARE GONNA HURT IF I DONT DO IT! THEY ARE IN A WEEK!!

[pale lodge]

Check this video it will help you: https://youtube.com/watch?v=FWbcKade3rw&pp=ygUhZ3JhcGhpbmcgYSBzeXN0ZW0gb2YgaW5lcXVhbGl0aWVz

[storm arch]

DOESNT HELP

[storm arch]

it doesnt help i dont understand this

[timber snow]

Welp then just retake the class next year if possible

[storm arch]

nooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo

[timber snow]

Maybe you can get consults with your math teacher

If they're willing

[storm arch]

handins are due monday

[timber snow]

Does the italic I stand for the set of integers

In qn 5

@storm arch

[storm arch]

Sorry just doing somthing and i have no clue want a better photo

[timber snow]

Firstly let's look at qn 4

[storm arch]

ok

[timber snow]

So we're trying to graph the inequality for -2x+1>5

Firstly let's draw the line y=-2x+1 on the axes

[storm arch]

yes

so

we need are x and y

y= -2(0)+1

so x is 0 and 1 is y?

???

[timber snow]

Man

Do you not know how to sketch a line?

[storm arch]

YES I SUCK AT GRAPHING

[timber snow]

Take each square to be one unit

[storm arch]

yeah

[timber snow]

Alright so

The y intercept of this line is 1

[storm arch]

yes

[timber snow]

And since the gradient of this line is -2,for every one step we move to the right

We move 2 units down

[timber snow]

So these two points

[storm arch]

yes

[timber snow]

Alright so draw the line

Actually nvm mb

I think I mightve misread the question 😭

Since -2x+1>5

Then -2x>4
So x<-2

So all you really have to do is draw a semi solid line x=-2

Then shade the region to the left

Mb😭

@storm arch

[storm arch]

oh

[loud epoch]

I agree, the inequation has no y in it, so the solution will be in x only.

@Kit, Q4 is asking you for:

$\big{(x,,y):-2x+1>5,\ x\in\mathbb{R},\ y\in\mathbb{R}\big}$

Do you understand what all that notation means?

[frigid wind]

Jay

[timber snow]

Yea my bad

[loud epoch]

It means that you need to find all points with

coordinates $(x,,y)$, where both $x$ and $y$ are

real numbers, subject to the restriction that

$-2x+1>5$.

In other words, whatever the $x$ coordinate is,

when it is multiplied by $-2$ and then 1 is added,

the result must be bigger than 5.

So, trying $x = 4$, for example, gives

$-2x+1=-2\times 4 + 1 = -7$ and since $-7 > 5$

is NOT true, you can't have any point with $x= 4$.

Trying, $x = -4$, as another example, gives

$-2x+1=-2\times -4 + 1 = 8 + 1 = 9$ and since

$9 > 5$ IS true, you can have points with $x= -4$.

Now, which points with $x=-4$? Well, we have no

restrictions on $y$ apart from it being real, so any

point $(-4,, y)$ will be a member of the set we seek.

This should get you a long way to finding all the

solutions, noting that you need to find all possible

$x$-values by solving the inequation $-2x + 1 > 5$.

Does this make any sense?

[frigid wind]

Jay

[loud epoch]

It's all good, everyone makes mistakes, and you did reflect on what you had said and catch it 🙂

[loud epoch]

@Kit, Q2, in the second image, covers material you also

need for Q5.

In Q2, you are given the sketch already showing a

boundary, which is a straight line. If we assume that

that line is the boundary of the region we seek to find,

we need only test a point that is NOT on the line.

For example, test the origin, $(0,,0)$, in which case:

$\text{LHS}\ = 4y + 8x = 4(0) + 8(0) = 0 + 0 = 0$

$\text{RHS}\ = 2$

So, the statement $\text{LHS}\ \le\ \text{RHS}$ is the same as $0 \le 2$,

and so is TRUE. Thus, $(0,,0)$ is in the desired region,

and we would shade on that side of the boundary.

Unfortunately, looking closely, the line drawn is strange.

For the region $4y +8x \le 2$, the boundary will be the

equality case, with equation $4y + 8x = 2$, or $2y + 4x = 1$.

This line has intercepts at $\left(0,, \frac{1}{2}\right)$ and at $\left(\frac{1}{4},,0\right)$,

and has a gradient of $m = -2$.

The line shown in the diagram has intercepts at $(0,,2)$ and

$(1,,0)$, and a gradient of $m=-2$... and so is parallel to

the actual boundary, but above it.

FYI, the line shown on the diagram is the boundary line

of the region $4y + 2x \le 8$.

[frigid wind]

Jay

[storm arch]

NO