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3 years ago

Which step brings you closest to getting the variable alone on one side of the equation

Mathematics

2 answers:

Oliga [24]3 years ago

5 0

Answer:

Adding 5 from both sides then dividing by 7

Step-by-step explanation:

The best way to isolate x is to add 5 to both sides first:

21 = 7x

Then to isolate x by itself divide both sides by 7

x = 3

That would be the value of x and your final answer.

Anna007 [38]3 years ago

3 0

16 = 7x-5
+5 +5
21 divide by 7 = 7x divide by 7
3 = x

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I need help with #3 plz

SVEN [57.7K]

I think it's to late, but...

Just turn the second equation into y equals mx plus b form

Then graph both equations

Pretty easy so far

After graphing them, you just have to find where they intersect, and that is your answer.

If they're parrallel, then there is no solution.

If it's the same line, then it's many solutions.

HOPE THIS HELPED

3 0

3 years ago

Y=4x-2, y=x+3 the solution to the system of equation

ella [17]

Y=4x-2=x+3
3x=5
x=5/3
y=5/3+3=14/3

6 0

3 years ago

one winter day the temperature ranged from a high of 20 f to a low of -25 f. by how many degrees did the temperature change?

Mamont248 [21]

Answer:

45 degrees

Step-by-step explanation:

you would do 20-(-25) then that turns to 20+25 and you get 45

6 0

3 years ago

What is the conjecture sentence for the pattern 1, ,4, 9 ,16?

nexus9112 [7]

Answer:

Square numbers

Step-by-step explanation:

if 1 is squared, you get 1.

if 2 is squared, you get 4.

if 3 is squared, you get 9.

if 4 is squared, you get 16.

And thus square numbers.

8 0

3 years ago

Suppose a and b are both non zero real numbers. Find real numbers c and d such that 1/a+ib= c+id

Thepotemich [5.8K]

$\begin{gathered} c=\frac{a}{a^2+b^2} \\ d=\frac{-b}{a^2+b^2} \end{gathered}$

Explanation

$\frac{1}{a+bi}=c+di$

Step 1

multiplicate by the conjugate

$\begin{gathered} \frac{1}{a+bi}\cdot\frac{a-bi}{a+bi}=\frac{a-bi}{(a+bi)(a-bi)}=\frac{a-bi}{a^2-(bi)^2} \\ \frac{1}{a+bi}\cdot\frac{a-bi}{a+bi}=\frac{a-bi}{(a+bi)(a-bi)}=\frac{a-bi}{a^2-(-b^2)}=\frac{a-bi}{a^2+b^2} \end{gathered}$

notice that

$\begin{gathered} \frac{1}{a+bi}=\frac{a-bi}{a^2+b^2}=\frac{a}{a^2+b^2}-\frac{b}{a^2+b^2}i \\ \frac{a}{a^2+b^2}-\frac{b}{a^2+b^2}i=c+di \\ so \\ \end{gathered}$ $\begin{gathered} c=\frac{a}{a^2+b^2} \\ d=\frac{-b}{a^2+b^2} \end{gathered}$

I hope this helsp you

6 0

1 year ago