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Rina8888 [55]
3 years ago
9

Solve for y. 4y+5/3 = y-8/2 y = -4 y = -5 y = -8

Mathematics
1 answer:
horrorfan [7]3 years ago
5 0

Answer:

<h3>y = -34/5</h3>

Step-by-step explanation:

\dfrac{4y+5}{3}=\dfrac{y-8}{2}\qquad\text{multiply both sides by }\ 2\cdot3=6\\\\6\!\!\!\!\diagup^2\cdot\dfrac{4y+5}{\not3_1}=6\!\!\!\!\diagup^3\cdot\dfrac{y-8}{\not2_1}\\\\2(4y+5)=3(y-8)\qquad\text{use distributive property}\\\\(2)(4y)+(2)(5)=(3)(y)+(3)(-8)\\\\8y+10=3y-24\qquad\text{subtract 10 from both sides}\\\\8y=3y-34\qquad\text{subtract 3y from both sides}\\\\5y=-34\qquad\text{divide both sides by 5}\\\\\boxed{y=-\dfrac{34}{5}}

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Find the unknown side length. Write your answer in simplest radical form.
Eva8 [605]

Answer: 1. x= 20 units,   2. x= 5\sqrt{3} units    3. x= 4\sqrt{2} units.

Step-by-step explanation:

In a right triangle,

Hypotenuse^2= Perpendicular^2+Base^2  (Pythagoras theorem)

1. Hypotenuse = x , Base = 16, Perpendicular = 12

\Rightarrow\ x^2 =12^2+16^2\\\\ \Rightarrow\ x^2=144+256\\\\ \Rightarrow\ x^2=400\\\\ \Rightarrow\ x=\sqrt{400}=20

2. Hypotenuse = 14 , Base = 11, Perpendicular = x

\Rightarrow\ 14^2=x^2+11^2\\\\\Rightarrow\ 196=x^2+121\\\\\Rightarrow\ x^2=196-121\\\\\Rightarrow\ x^2 = 75\\\\\Rightarrow\ x=\sqrt{75}=\sqrt{25\times3}=5\sqrt{3}

3.  Hypotenuse = 9 , Base = x, Perpendicular =7

\Rightarrow\ 9^2=x^2+7^2\\\\\Rightarrow\ 81= x^2+49\\\\\Rightarrow\ x^2=81-49\\\\\Rightarrow\ x^2=32\\\\\Rightarrow\ x=\sqrt{32}=\sqrt{16\times2}\\\\\Rightarrow\ x=4\sqrt{2}

5 0
2 years ago
How can you get the square side area??
emmainna [20.7K]
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2 years ago
How to solve this problem 14-6= 10-6=?
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3 years ago
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gladu [14]

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5 0
2 years ago
Which biconditional is not a good definition?
Xelga [282]
<span>The <u>correct answers</u> are:

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For the first question, the first three answers are very specific and true:

A whole number is odd if it is not divisible by 2, and a number is not divisible by 2 if it is odd;
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For the second question, the Law of Detachment says if our conditional "if p, then q" is true and p is true, then q must also be true.

For this question, "I can go to the concert if I can afford to buy a ticket" is true as well as "I can go to the concert." This means "I can afford to buy a ticket" must be true as well.</span></span>
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3 years ago
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