Help plz???////////////////////

lianna [129]

The slope is 1, so m = 1.
The given point is (x,y) = (1,5) which means x = 1 and y = 5 pair up together.

Plug the three pieces of info (m = 1, x = 1, y = 5) into the y = mx+b equation. Then solve for b

y = mx+b
y = 1*x + b ... replaced m with 1
5 = 1*1 + b .... replaced x with 1, replaced y with 5
5 = 1+b
1+b = 5
b+1 = 5
b+1-1 = 5-1
b = 4

Since m = 1 and b = 4, we know that y = mx+b turns into y = 1x+4 which simplifies to y = x+4

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Answer: y = x+4

8 0

3 years ago

Science help meee

saul85 [17]

Answer:

There are ways of slowing the rate at which we add greenhouse gases to the atmosphere that actually save money. These are familiar ideas like switching to more efficient cars, heating and air conditioning systems, light bulbs, and appliances; and putting better insulation in homes and buildings.

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

Select the postulate that is illustrated for the real numbers.

Marysya12 [62]

4. The Commutative postulate for addition. Hope this helped you :)

8 0

3 years ago

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Think you could help i don't understand

deff fn [24]

Answer: it is the alphabet

4 0

3 years ago

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I NEED IMMEDIATE HELP!!!!

DerKrebs [107]

<h2>Hello!</h2>

The answers are:

C) $(3+xz)(-3+xz)$

D) $(y^2-xy)(y^2+xy)$

F) $(64y^2+x^2)(-x^2+64y^2)$

To know which of the products results in a difference of square, we need to remember the difference of squares from:

The difference of squares form is:

$(a+b)(a-b)=a^{2}-b^{2}$

So, discarding each of the given options in order to find which products result in a difference of squares, we have:

$(x-y)(y-x)=xy-x^{2}-y^{2} +yx=-x^{2} -y^{2}$

So, the obtained expression is not a difference of squares.

$(6-y)(6-y)=36-6y-6y+y^{2}=y^{2}-12y+36$

So, the obtained expression is not a difference of squares.

$(3+xz)(-3+xz) =(xz+3)(xz-3)=(xz)^{2}-3xz+3xz-(3)^{2}\\\\(xz)^{2}-3xz+3xz-(3)^{2}=(xz)^{2}-(3)^{2}$

So, the obtained expression is a difference of squares since it matches with the form of the difference of squares.

$(y^2-xy)(y^2+xy)=(y^{2})^{2}+y^{2}*xy-y^{2}*xy-(xy)^{2} \\\\(y^{2})^{2}+y^{2}*xy-y^{2}*xy-(xy)^{2}=(y^{2})^{2}-(xy)^{2}$

So, the obtained expression is a difference of squares since it matches with the form of the difference of squares.

$(25x-7y)(-7y+25x)=-175xy+(25x)^{2}+49y^{2}-175xy\\\\-175xy+(25x)^{2}+49y^{2}-175xy=(25x)^{2}+49y^{2}-350xy$

So, the obtained expression is not a difference of squares

$(64y^2+x^2)(-x^2+64y^2)=(64y^2+x^2)(64y^2-x^2)\\\\(64y^2+x^2)(64y^2-x^2)=(64y^{2})^{2} -(x^{2}*64y^{2})+(x^{2}*64y^{2})-(x^{2})^{2}\\ \\(64y^{2})^{2} -(x^{2}*64y^{2})+(x^{2}*64y^{2})-(x^{2})^{2}=(64y^{2})^{2}-(x^{2})^{2}$

So, the obtained expression is a difference of squares since it matches with the form of the difference of squares.

Hence, the products that result in a difference of squares are:

C) $(3+xz)(-3+xz)$

D) $(y^2-xy)(y^2+xy)$

F) $(64y^2+x^2)(-x^2+64y^2)$

Have a nice day!

5 0

4 years ago