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fredd [130]
3 years ago
8

HELP!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Mathematics
2 answers:
inn [45]3 years ago
4 0
5/6a - 5/6b + 1/3a

1/3 x 2/2 = 2/6

5/6a + 2/6a = 7/6a

7/6a - 5/6a = 2/6a

2/6a = 1/3a
<span>
d. 1/3a is your answer 

hope this helps </span>
Kamila [148]3 years ago
3 0
A. 7/6a - 5/6b would be the answer
add all like terms = 5/6 + 2/6 (Change 1/3 to 2/6 so the denominators are the same) which is 7/6. 
You cannot add 5/6b and 7/6a because they are not like terms.
Your final answer would be 7/6a - 5/6b
Hope this helps!!
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Simplify the expression and combine like terms.
Alenkinab [10]

Answer:

5x + 16

Step-by-step explanation:

the first thing you do in this equation is to distribute the 2 into x + 6

you get

2x + 12 + 3x + 4 = ?

then you must put the 2x and the 3x together

5x + 12 + 4

and you can then add the 12 and 4

5x + 16

4 0
3 years ago
Type your answer and then click or tap Done.
masya89 [10]

Answer:

See below

Step-by-step explanation:

x(x-2y)-(y-x)2  

Final result :

-y2

Step by step solution :

Step  1  :

Equation at the end of step  1  :

x • (x - 2y) -  (y - x)2

Step  2  :

2.1     Evaluate :  (y-x)2   =    y2-2xy+x2  

Final result :

-y2

7 0
2 years ago
Help me out please guys
kolbaska11 [484]

Answer:

The answer is D,  g(x)=-2^x

Step-by-step explanation:

By graphing each one in a graphing tool like Desmos, you can determine which one it is.

If your looking for an actual explanation on how it is found, I can't really help you on that one.

4 0
3 years ago
Consider the first five steps of the derivation of the Quadratic Formula.
lara31 [8.8K]

Answer:

Full proof below

Step-by-step explanation:

\displaystyle ax^2+bx+c=0\\\\ax^2+bx=-c\\\\x^2+\biggr(\frac{b}{a}\biggr)x=-\frac{c}{a}\\\\x^2+\biggr(\frac{b}{a}\biggr)x+\biggr(\frac{b}{2a}\biggr)^2=-\frac{c}{a}+\biggr(\frac{b}{2a}\biggr)^2\\\\x^2+\biggr(\frac{b}{a}\biggr)x+\frac{b^2}{4a^2}=-\frac{c}{a}+\frac{b^2}{4a^2}\\ \\\biggr(x+\frac{b}{2a}\biggr)^2=-\frac{4ac}{4a^2}+\frac{b^2}{4a^2}\\ \\\biggr(x+\frac{b}{2a}\biggr)^2=\frac{b^2-4ac}{4a^2}\\ \\x+\frac{b}{2a}=\pm\frac{\sqrt{b^2-4ac}}{2a}\\ \\x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}

6 0
2 years ago
What is the horizontal asymptote for y(t) for the differential equation dy dt equals the product of 2 times y and the quantity 1
marta [7]
First, we need to solve the differential equation.
\frac{d}{dt}\left(y\right)=2y\left(1-\frac{y}{8}\right)
This a separable ODE. We can rewrite it like this:
-\frac{4}{y^2-8y}{dy}=dt
Now we integrate both sides.
\int \:-\frac{4}{y^2-8y}dy=\int \:dt
We get:
\frac{1}{2}\ln \left|\frac{y-4}{4}+1\right|-\frac{1}{2}\ln \left|\frac{y-4}{4}-1\right|=t+c_1
When we solve for y we get our solution:
y=\frac{8e^{c_1+2t}}{e^{c_1+2t}-1}
To find out if we have any horizontal asymptotes we must find the limits as x goes to infinity and minus infinity. 
It is easy to see that when x goes to minus infinity our function goes to zero.
When x goes to plus infinity we have the following:
$$\lim_{x\to\infty} f(x)$$=y=\frac{8e^{c_1+\infty}}{e^{c_1+\infty}-1} = 8
When you are calculating limits like this you always look at the fastest growing function in denominator and numerator and then act like they are constants. 
So our asymptote is at y=8.

3 0
3 years ago
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