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FromTheMoon [43]
3 years ago
5

What is x^2 + v = w when you have to make x the subjectim really stuck

Mathematics
2 answers:
ipn [44]3 years ago
6 0
X²+v=w (move v to w)
x²=w-v
x=√w-v (then square root w-v)
ArbitrLikvidat [17]3 years ago
5 0
Substract v to both sides of the equation: x^2+v-v = w-v
you get x^2=w-v
Now take the square root of both sides: root(x^2) = root(w-v)
you get x= sq root (w-v)   or        x= +/-  sq root(w-v)
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Determine the domain of the following graph
Lunna [17]

Answer:

-3 ≥ x ≥ 9

Step-by-step explanation:

The domain are the x-values of every point in your graph. Since it a continuous domain, it'll be represented with ≥. Your two extremities are -3 and 9. Therefore, your domain is -3 ≥ x ≥ 9.

8 0
3 years ago
Math question can somebody please help me with this math problem
Elodia [21]
You need to add up all the sides of the thing thank you
5 0
3 years ago
Read 2 more answers
Max wants to make a cute collar and tie for his dog. He needs to cut a 38 inches fabric into two pieces. The long longer piece m
CaHeK987 [17]
Do this.
38 divided by two
then use the equation
38=5+2x
8 0
3 years ago
dolores and four friends went to a buffet dinner.The total cost was at most $130 including the $20 dollar tip they left.How much
Levart [38]
So the actual meal was at most 130 dollars they paid minus the 20 dolars tip: 130-20=110 dollars.

it was 5 people: Dolores+5 friends, who presumably all paid the same (buffet).

then we have to divide the 110 among the 5 people: it's 110/5=22

so each of them paid at most 22 dollars:

x (the amount they paid)≤22

5 0
3 years ago
Solve the system: <br> 2/x - 3/y =-5; 4/x + 6\y =14
sertanlavr [38]

The solution is x = 2 \text{ and } y = \frac{1}{2}

<em><u>Solution:</u></em>

Let us assume,

a = \frac{1}{x}\\\\b = \frac{1}{y}

<em><u>Given system of equations are:</u></em>

\frac{2}{x} - \frac{3}{y} = -5

\frac{4}{x} + \frac{6}{y} = 14

<em><u>Rewrite the equation using "a" and "b"</u></em>

2a - 3b = -5 ------------ eqn 1

4a + 6b = 14 -------------- eqn 2

<em><u>Let us solve eqn 1 and eqn 2</u></em>

<em><u>Multiply eqn 1 by 2</u></em>

2(2a - 3b = -5)

4a - 6b = -10 ------------- eqn 3

<em><u>Add eqn 2 and eqn 3</u></em>

4a + 6b = 14

4a - 6b = -10

( - ) --------------------

8a = 4

a = \frac{4}{8}\\\\a = \frac{1}{2}

Substitute a = 1/2 in eqn 1

2(\frac{1}{2}) -3b = -5\\\\1 - 3b = -5\\\\3b = 6\\\\b = 2

Now let us go back to our assumed values

Substitute a = 1/2 in assumed values

a = \frac{1}{x}\\\\\frac{1}{2} = \frac{1}{x}\\\\x = 2

Substitute b = 2 in assumed value

b = \frac{1}{y}\\\\2 = \frac{1}{y}\\\\y = \frac{1}{2}

Thus the solution is x = 2 \text{ and } y = \frac{1}{2}

3 0
3 years ago
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