Question

Can anyone please do these input an attachment in i did some of the questions but ion know how to do the rest pleasee i need to turn it in asappp‍♀️

Answer 1

Question 3

`(a)\quad\boxed{\displaystyle y= -(2)/(3)x + b}`

`(b)\quad\boxed{y = (3)/(2)x + b}`

Question 4

`(a) \quad \boxed{40 x^8}`

`(b)\quad \boxed{3x^2-x-44}`


`(c) \quad\boxed{\quad 5x^2-6x+10xy-2y+1\quad}`

`(d) \quad \boxed{3x^3-3x^2 -3x +89}`

Explanation:

Question 3

Part a

The given line has an equation

`y = -(2)/(3)x - 5 \cdots\cdots[1]`

`\rm{The \: slope\: of \: this\: line =} \displaystyle -(2)/(3)`

(a) Equation of a parallel line
All lines parallel to the given line [1] will have the same slope

So equation of a line parallel to [1] would be

`\boxed{\displaystyle y= -(2)/(3)x + b}`
where b is the y-intercept.

We cannot determine the y-intercept unless we are given a point through which the parallel line passes, so leave it as is

Part b

Perpendicular lines will have a slope which is the negative of the reciprocal of the line given in [1]

`\sf{Reciprocal \;of \displaystyle -(2)/(3) = -(3)/(2)}\\\\\sf{Negative \; of\;the \;reciprocal\;= \displaystyle -\left(-(3)/(2)\right) = (3)/(2)`
So a perpendicular line will have the equation

`\boxed{y = (3)/(2)x + b}`

Again, b cannot be determined unless we are given a point through which the line passes

Question 4

Part a

`4x^3\left(10x^5\right) = 4\cdot 10 \cdot x^3(x^5) \\ \\= 40(x^(3 + 5))\\\\= \boxed{40 x^8}`

Part b

`\left(3x\:+\:11\right)\left(x\:-\:4\right)`

use the FOIL method (first-outer-inner-last) to multiply
First terms in both are 3x and x so for F we get
`\large \text{$ 3x \cdot x = 3x^2 $}`
Outer terms are 3x and - 4; multiply to get 3x × (-4) =
`-12x`

Inner terms are 11 and x; multiply to get
`11x`

last terms are 11 and -4; multiply to get
`-44`

So the result is

`3x^2-12x+11x-44`

`= \boxed{3x^2-x-44}`

Part c

`\left(x\:+2y\:-\:1\right)\left(5x\:-\:1\right)`

We can rewrite this as

`5x(x + 2y - 1) -1(x + 2y - 1)\\\\= 5x(x) +5x(2y) -5x(1) -1(x) - 1(2y) -1(-1)\\\\= 5x^2 + 10xy -5x -1x -2y + 1\\\\= 5x^2 + 10xy - 6x -2y + 1`

We usually write this in decreasing powers of x followed by the y terms and then the constants to get

`\boxed{\quad 5x^2-6x+10xy-2y+1\quad}`

Part d

`3x\left(x^2-5\right)-\left(3x^2-12x\:+\:11\right)+\:100`

Expand individual terms and then add everything together and simplify if necessary

`3x\left(x^2-5\right) = 3x(x^2) -3x(5) = 3x^3 -15x\\`

`-(3x^2 - 12x + 11) = -3x^2-(-12x) -(11) = -3x^2 + 12x - 11`

Adding all the terms and adding the constant 100 gives us:

`3x^3-15x -3x^2 + 12x - 11 + 100`

Gather like terms and simplify:

`3x^3 -3x^2 -15x + 12x -11 + 100\\\\`

`= \boxed{3x^3-3x^2 -3x +89}`