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2 years ago

Solve for x. y=7x−7z−9

2 answers:

6 0

You have to isolate the variable so you need to add 7z and 9 to both sides. You then get y+ 7z+ 9= 7x. Then divide 7x from both sides. The equation becomes x= y/7+ z + 9/7.
I hope this helps love! :)

AleksandrR [38]2 years ago

5 0

Answer:

The value of x is $x=\frac{y+7z+9}{7}$

Step-by-step explanation:

Consider the provided equation.

$y=7x-7z-9$

We need to solve the equation for x.

Add 7z to both the sides.

$y+7z=7x-7z+7z-9$

$y+7z=7x-9$

Add 9 to both the sides.

$y+7z+9=7x-9+9$

$y+7z+9=7x$

Divide both sides by 7.

$x=\frac{y+7z+9}{7}$

Hence the value of x is $x=\frac{y+7z+9}{7}$ .

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Give an example of each of the following or explain why you think such a set could not exist.

Oksanka [162]

Answer:

a.No

b.No

c.No

Step-by-step explanation:

a.No,Such set does not exist .A set of natural numbers is N

Every point of this set is an isolated point but no accumulation point

Accumulation point:It is defined as that point a of set Swhich every neighborhood contains infinitely many distinct point of set

$(a-\epsilon,e+\epsilon)\cap S-{a}\neq\phi$

Isolated point : it is defined as that point a of set S which neighborhood does not contain any other point of set except itself

$(a-\epsilon0,a+\epsilon)\cap S-{a}=\phi$

Interior point of set :Let $a\in S$ .Then a is called interior point of set when its neighborhood is a subset of set S.

$a\in(a-\epsilon,e+\epsilon)\subset S$

When a set is uncountable then interior point exist it is necessary for interior points existance .

Boundary points :Let $a\in S$ .If every non empty neighborhood of a intersect S and complement of S.

Every member of a set is a boundary point

b.No, such set does not exist .A non empty set with isolated point then the set have no interior points .By definition of interior point and isolated point .For example.set of natural numbers

c.No, Such set does not exist ,for example set of natural every point is an isolated point and boundary point.By definition of boundary point and isolated point

6 0

2 years ago

I want a way to solve this question

noname [10]

Answer:

Step-by-step explanation:

$cos~L=\frac{m^2+n^2-l^2}{2mn} \\=\frac{5^2+8^2-7^2}{2*5*8} \\=\frac{25+64-49}{80} \\=\frac{40}{80} \\=\frac{1}{2} \\=cos 60\\$