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Solnce55 [7]
3 years ago
5

4(g+2 )+8g=56 show work

Mathematics
2 answers:
mihalych1998 [28]3 years ago
7 0
4(g+2)+8g=56
4g+8+8g=56
12g+8=56
12g-8=56-8
12g= 48
12g/12=48/12
g=4

Steps
1. Distribute by multiply 4 times g and 4 time 2
2.do the opposite of either add or subtraction in this case is opposite of add which is subtract
3.divide both side by the coefficient where it has the letter for example in this case is 12g so you divide 12g/12=12/48
Which gives you the answer 4
Bingel [31]3 years ago
5 0
4g+8+8g=56
12g+8=56
12g+-8=-8
12g=48
DIVIDE 12 TO BOTH 
g=4
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In the following proportion, what does w equal? 3/4 = w/28<br><br> A. 7<br> B. 8<br> C. 18<br> D. 21
DIA [1.3K]
W=21
To get from 4 to 28 you multiply by 7, so 3x7=21.
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Use the diagram and information provided to find the missing ares and angles in<br> degrees).
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Step-by-step explanation:

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2 years ago
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Option 1

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8 0
3 years ago
Make 'h' as the subject. Remember to rationalize the denominator while solving.
Sliva [168]

Hello there BeGuiLE !

To solve your question,  we must first bring 'h' to one side of the equation.  Let's bring all the terms with the variable 'h' to the left side of the equation. So,
\large{\sf\sqrt{ 3  }  h =  1.6+h}
→ \large{\sf\sqrt{3}h-h=1.6 }

Now, let's combine the 2 terms containing the variable 'h'.
\large{\sf\sqrt{3}h-h=1.6 }
→ \large{\sf\left(\sqrt{3}-1\right)h=1.6 }

Now, we need to leave the variable 'h' alone in the left side of the equation & bring (√3 + 1) to the other side of the equation. This is done in order to make 'h' as the subject of the equation. So, we'll get it as,
\large{\sf\left(\sqrt{3}-1\right)h=1.6 }
→ \large{\sf\:h=\frac{1.6}{\sqrt{3}-1} }

Now, let's rationalize it. Rationalizing is a process where we move the root from the denominator of a fraction to the numerator for easier calculation. So,
\large{\sf\:h=\frac{1.6}{\sqrt{3}-1} }
→ \large{\sf\:h=\frac{1.6(\sqrt{3}+1)}{(\sqrt{3}-1 )(\sqrt{3}+1}) }
Now, use the algebraic identity, (a + b)(a - b) = a² - b²
→ \large{\sf\:h=\frac{1.6(\sqrt{3}+1)}{3-1} }
→ \large{\sf\:h=\frac{1.6(\sqrt{3}+1)}{2} }
→ \boxed{\large{\sf\:h=0.8(\sqrt{3}+1)}}

Now, we've made 'h' as the subject of the equation. If you want to solve it more, then,
\large{\sf\:h=0.8(\sqrt{3}+1)}
Take √3 as 1.73 (approx. value up to 2 decimal places)
→ \large{\sf\:h=0.8(1.73+1)}
→ \large{\sf\:h=0.8(2.73)}
→ \boxed{\boxed{\huge{\bf{\:h=2.184 \: (approx.)}}}}

I hope this will help you.

Please refer to the attached image if the explanation shows some error.

_______

Check out more links which will help you understand the topic better :

■ brainly.com/question/21406377

■ brainly.com/question/696184

_______
\mathfrak{Lucazz}


4 0
2 years ago
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Can you answer these show work
ArbitrLikvidat [17]

Hello!

1) 108

2) 48

3) 10

I hope it helps!

4 0
3 years ago
Read 2 more answers
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