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Semenov [28]
3 years ago
6

How do i solve this equation

Mathematics
2 answers:
olga2289 [7]3 years ago
8 0
Your starting equation is: x/6 + 4 = 15 The first step would be to subtract 4 from both sides, to move all of the constants to the right side of the equation. x/6 = 11 The final step is to multiply both sides by 6, to get rid of the fraction on the left side of the equation. x = 66 Hope this helps!
Sedaia [141]3 years ago
6 0
\dfrac{x}{6}+4=15 

\dfrac{x}{6}=15-4    <span>Subtract \ 4 \ from \ both \ sides
</span>
\dfrac{x}{6}=11 

x=11*6&#10;&#10; &#10;     Multiply \ both \ sides \ by \ 6 

x=66          Solution
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(08.02)a pair of equations is shown below. 3x − y = 9 y = −2x + 11 if the two equations are graphed, at what point do the lines
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3 years ago
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3 years ago
Whats the answer to 2y^0-(3y)^0
mrs_skeptik [129]

Answer:

1

Step-by-step explanation:

To answer this question, we need to that any number (except zero) raised to power 0 is equal to 1. With that in mind, let's solve this:

2y^0 - (3y)^0

= 2(1) - (1)

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<em>N</em><em>o</em><em>t</em><em>e</em><em>:</em><em> </em><em>2</em><em>y</em><em>^</em><em>0</em><em> </em><em>=</em><em> </em><em>2</em><em>(</em><em>y</em><em>^</em><em>0</em><em>)</em><em> </em><em>=</em><em> </em><em>2</em><em>(</em><em>1</em><em>)</em><em> </em><em>=</em><em> </em><em>2</em><em> </em><em>w</em><em>h</em><em>e</em><em>r</em><em>e</em><em>a</em><em>s</em>

<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>(</em><em>2</em><em>y</em><em>)</em><em>^</em><em>0</em><em> </em><em>=</em><em> </em><em>1</em>

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3 years ago
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