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3 years ago

How to solve a multiplication problem with a variable?

2 answers:

7 0

In order to move a number that is multiplied or divided by the variable, you must do the inverse to both sides. That means, if the number is multiplied, you must divide it by both sides. If it's divided, you must multiply it by both side. steps so that you can apply them to more complex problems.

ki77a [65]3 years ago

3 0

Example Question: 6x + 4 = 22

The goal in this is to get the variable alone and these are the steps how to. What you have to do is subtract 4 from 4 so that you can get 6x alone. But when you subtract 4 from one side, you have to subtract 4 from the other side too which leaves us with 6x = 18. Now, we have to do the opposite of multiplication which is division. So we are going to do 6x divided by 6 and the two sixes cancel each other out. But remember, if you do something to one side you have to do it to the other as well. So we have to do 18/6 which leaves us with the answer.

x = 3

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Find the range of the function y=1/2x+3 when the domain is (-2,0,2,4)

Triss [41]

Answer:

{2,3,4,5}

Step-by-step explanation:

Domain is the set of x-values for which the function is defined.

Range is the set of y-values for which the function is defined (set of values corresponding to the set of x-values).

The function is given as $y=\frac{1}{2}x+3$

The domain is (-2,0,2,4)

That means, we need to plug-in the 4 numbers given into x of the function and find the 4 corresponding y values (the range).

So, lets do this:

putting x = -2,

$y=\frac{1}{2}(-2)+3=2$

putting x = 0

y = 3

putting x = 2

$y=\frac{1}{2}(2)+3=4$

putting x= 4

$y=\frac{1}{2}(4)+3=5$

Thus, the range is {2,3,4,5}

5 0

3 years ago

Which type of graph would be best for showing how the population of a country changes each month?

Serjik [45]

Answer:

A. Line graph

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

Please solve this with the steps.

bija089 [108]

Subtract 9/4 from both sides
3/4x=6 3/4
Multiply them by 4/3
X=9

5 0

4 years ago

Read 2 more answers

Find the complex zeros of the polynomial function, write f in factored form. f(x) =x^3-9x^2+31x-39

bezimeni [28]

x = 3

zeros are: 3 + 2i, 3 - 2i

6 0

3 years ago

Express the following relations in the set builder notation. Then, determine whether it is reflexive, symmetric, transitive. Ple

pshichka [43]

Answer:

a)Reflexive, not symmetric, transitive

b)Reflexive, not symmetric, transitive

c)Not reflexive, symmetric, not transitive

d)Reflexive, not symmetric, transitive

Step-by-step explanation:

$R=\left \{ (a,b)\epsilon \mathbb{R} \times \mathbb{R} \mid a \leq b\right \}$

The relation R is reflexive for

$a\leq a$ for every real number a

it is not symmetric because 0 is less than 1, but 1 is not less than 0

it is transitive

$a\leq$ and $b\leq c\Rightarrow a\leq c$

So if aRb and bRc, then aRc

$R=\left \{ (m,n)\epsilon \mathbb{Z} \times \mathbb{Z} \mid \exists k\in \mathbb{Z} \ni m=kn \right \}$

R is reflexive because m=1.m for every integer m

R is not symmetric: 2 is a factor of 4, but 4 is not a factor of 2

R is transitive: if mRn and nRp if m=kn and n=qp, so m=(kq)p and kq is an integer , so mRp

$R=\left \{ (m,n)\epsilon \mathbb{Z} \times \mathbb{Z} \mid m\neq n\right \}$

R is obviously not reflexive because all numbers equals themselves

R is symmetric: if a different to b, then b different to a

R is not transitive: 1R2 and 2R1 (because 1 different to 2), but 1 = 1

$R=\left \{ A,B\mid A\subseteq B \right \}$

R is reflexive for every set A is a subset of itself

R is not symmetric {1,2} is a subset of {1,2,3} but {1,2,3} is not a subset of {1,2}

R is transitive: if A is subset of B and B is subset of C, then A is subset of C

8 0

4 years ago