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babymother [125]
3 years ago
6

What is 3 in a+ b- in C written as a single logarithmn?

Mathematics
1 answer:
Leona [35]3 years ago
4 0

There are a lot of other step that will be in between but the answer will end up being.....

Answer: Log (a^2c/ b^3 d^4)

You might be interested in
|x - 1| > 4 is equivalent to which of the following?
Vsevolod [243]

Answer:

C. x > 5 or x < - 3

Step-by-step explanation:

  • |x - 1| > 4

Step 1:

x - 1 > 4

x > 4 + 1

x > 5

Step 2:

x - 1 < -4

x < -4 + 1

x < -3

HP: {x > 5 or x < -3} (c)

3 0
2 years ago
It says to solve each proportion which I forgot how to do. Please help.
skelet666 [1.2K]
You have to divide denominator by numerator to find the constant of proportionality.
7 0
3 years ago
Read 2 more answers
Can someone solve this?
ladessa [460]
  • First question:

Recall that \cos^2x+\sin^2x=1 and \sqrt{x^2}=|x| for all x. So

\sqrt{1-\cos^2x}=\sqrt{\sin^2x}=|\sin x|

\sqrt{1-\sin^2x}=\sqrt{\cos^2x}=|\cos x|

For 0, we expect both \cos x>0 and \sin x>0 (i.e. the sine and cosine of any angle that lies in the first quadrant must be positive). By definition of absolute value, |x|=x if x>0.

So we have

\dfrac{\sqrt{1-\cos^2x}}{\sin x}+\dfrac{\sqrt{1-\sin^2x}}{\cos x}=\dfrac{\sin x}{\sin x}+\dfrac{\cos x}{\cos x}=1+1=\boxed{2}

making H the answer.

  • Second question:

C is always true, because the inequality reduces to x > y.

6 0
3 years ago
For problem 5, use the ordered pairs to answer the following question. MUST SHOW YOUR WORK FOR FINDING SLOPE.
Firlakuza [10]

Answer:

graph them and then see if they intersect(but if you notice that the coordss of the second line are all smaller than the coords of the first line, meaning that they wont be touching at all)

Step-by-step explanation:

7 0
2 years ago
Find the expansion of cos x about the point x=0
ycow [4]

Answer:

Cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{1!} + ...

Step-by-step explanation:

We use Taylor series expansion to answer this question.

We have to find the expansion of cos x at x = 0

f(x) = cos x, f'(x) = -sin x, f''(x) = -cos x, f'''(x) = sin x, f''''(x) = cos x

Now we evaluate them at x = 0.

f(0) = 1, f'(0) = 0, f''(0) = -1, f'''(0) = 0, f''''(0) = 1

Now, by Taylor series expansion we have

f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)(x-a)^2}{2!} + \frac{f'''(a)(x-a)^3}{3!} + \frac{f''''(a)(x-a)^4}{4!} + ...

Putting a = 0 and all the values from above in the expansion, we get,

Cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{1!} + ...

8 0
3 years ago
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