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VikaD [51]
3 years ago
7

Solve the equation a/a^2-16+2/a-4=2/a+4

Mathematics
1 answer:
nirvana33 [79]3 years ago
4 0
A=1/24
hope this helps 
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I don’t really understand what to do on these questions and could you tell me the ratio to solve the rest?
SVEN [57.7K]

Step-by-step explanation:

Assuming 10 cups of lemon-lime soda and 5 cups of orange soda make 15 cups of punch, we can write proportions for each problem.

x / 130 cups punch = 10 cups lemon-lime / 15 cups punch

x = 86.67 cups lemon-lime soda

y / 130 cups punch = 5 cups orange soda / 15 cups punch

y = 43.33 cups orange soda

x / 65 cups punch = 10 cups lemon-lime / 15 cups punch

x = 43.33 cups lemon-lime soda

y / 65 cups punch = 5 cups orange soda / 15 cups punch

y = 21.67 cups orange soda

x / 195 cups punch = 10 cups lemon-lime / 15 cups punch

x = 130 cups lemon-lime soda

y / 195 cups punch = 5 cups orange soda / 15 cups punch

y = 65 cups orange soda

8 0
3 years ago
Suppose y varies directly as x write a direct variation equation that relates x and y
Oksana_A [137]

Answer:

y=2/3x

Step-by-step explanation:

A direct variation equation is y=kx. if k=2/3, then if x was 9,y would be 6.

6 0
3 years ago
Read 2 more answers
Answer pleasee tyyy.
sveticcg [70]

HOPE THIS IS THE ANSWER FOR YOUR QUESTION

5 0
3 years ago
For the following pairs of equations explain why the equations are NOT equivalent.
Serga [27]
If you were to make both x's equivalent, you'd have to multiply the first equation by 2 to get rid of the coefficient, 1/2. However, you'd have to multiply 2 onto -8 as well, therefore the equation would turn into x - 16 = 18. That is not equivalent to x - 8 = 18.
7 0
3 years ago
Determine the equations of the vertical and horizontal asymptotes, if any, for g(x)=x-2/x^2+4x+3
KatRina [158]

Answer:

<h2>B. x = -1, x = -3, y = 0</h2>

Step-by-step explanation:

g(x)=\dfrac{x-2}{x^2+4x+3}\\\\vertical\ asymptote:\\\\x^2+4x+3=0\\x^2+x+3x+3=0\\x(x+1)+3(x+1)=0\\(x+1)(x+3)=0\iff x+1=0\ \vee\ x+3=0\\\\\boxed{x=-1\ \vee\ x=-3}\\\\horizontal\ asymptote:\\\\\lim\limits_{x\to\pm\infty}\dfrac{x-2}{x^2+4x+3}=\lim\limits_{x\to\pm\infty}\dfrac{x^2\left(\frac{1}{x}-\frac{2}{x^2}\right)}{x^2\left(1+\frac{4}{x}+\frac{3}{x^2}\right)}=\lim\limits_{x\to\pm\infty}\dfrac{\frac{1}{x}-\frac{2}{x^2}}{1+\frac{4}{x}+\frac{3}{x^2}}=\dfrac{0}{1}=0\\\\\boxed{y=0}

4 0
3 years ago
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