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2 years ago

SOLVE 6^3^-X=6^2 FOR X A.X=-6B.X=1C.X=-1D.X=9

Mathematics

1 answer:

Murrr4er [49]2 years ago

6 0

<h3>Answer: Choice B) x = 1</h3>

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Explanation:

The problem should read 6^(3-x) = 6^2. The 3-x must be in parenthesis to indicate that all of "3-x" is in the exponent and not just the 3.

Because both bases are 6, and because both sides are equal, this means that the exponents must be equal as well. Therefore 3-x = 2. Let's solve for x

3-x = 2

3-x+x = 2+x .... adding x to both sides

3 = 2+x

2+x = 3 ..... flipping both sides

x+2 = 3

x+2-2 = 3-2 .... subtracting 2 from both sides

You can use logarithms to solve this problem, but the fact that the bases are both the same allows us to use this faster shortcut.

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The disctance between -8 and 5 on the number line can be found by which expression??

Ilia_Sergeevich [38]

It's A. First of all, one of the only 2 things in math that will NEVER be negative is distances/lengths and times. You have a distance here, so because one of your 2 numbers is negative, you have to take the absolute value of the difference. -8-5 = -11, but the absolute value of that is 11. Put the numbers on a number line and count the units between -8 and +5 and there are 11 units between those 2 numbers.

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2 years ago

Which function is the same as y = 3 cosine (2 (x startfraction pi over 2 endfraction)) minus 2? y = 3 sine (2 (x startfraction p

kirza4 [7]

The function which is same as the function y = 3cos(2(x +π/2)) -2 is: Option A: y= 3sin(2(x + π/4)) - 2

<h3>How to convert sine of an angle to some angle of cosine?</h3>

We can use the fact that:

$\sin(\theta) = \cos(\pi/2 - \theta)\\\sin(\theta + \pi/2) = -\cos(\theta)\\\cos(\theta + \pi/2) = \sin(\theta)$

to convert the sine to cosine.

<h3>Which trigonometric functions are positive in which quadrant?</h3>

In first quadrant (0 < θ < π/2), all six trigonometric functions are positive.
In second quadrant(π/2 < θ < π), only sin and cosec are positive.
In the third quadrant (π < θ < 3π/2), only tangent and cotangent are positive.
In fourth (3π/2 < θ < 2π = 0), only cos and sec are positive.

(this all positive negative refers to the fact that if you use given angle as input to these functions, then what sign will these functions will evaluate based on in which quadrant does the given angle lies.)

Here, the given function is:

$y= 3\cos(2(x + \pi/2)) - 2$

The options are:

$y= 3\sin(2(x + \pi/4)) - 2$
$y= -3\sin(2(x + \pi/4)) - 2$
$y= 3\cos(2(x + \pi/4)) - 2$
$y= -3\cos(2(x + \pi/2)) - 2$

Checking all the options one by one:

Option 1: $y= 3\sin(2(x + \pi/4)) - 2$

$y= 3\sin(2(x + \pi/4)) - 2\\y= 3\sin (2x + \pi/2) -2\\y = -3\cos(2x) -2\\y = 3\cos(2x + \pi) -2\\y = 3\cos(2(x+ \pi/2)) -2$

(the last second step was the use of the fact that cos flips its sign after pi radian increment in its input)
Thus, this option is same as the given function.

Option 2: $y= -3\sin(2(x + \pi/4)) - 2$

This option if would be true, then from option 1 and this option, we'd get:
$-3\sin(2(x + \pi/4)) - 2= -3\sin(2(x + \pi/4)) - 2\\2(3\sin(2(x + \pi/4))) = 0\\\sin(2(x + \pi/4) = 0$

which isn't true for all values of x.

Thus, this option is not same as the given function.

Option 3: $y= 3\cos(2(x + \pi/4)) - 2$

The given function is $y= 3\cos(2(x + \pi/2)) - 2 = 3\cos(2x + \pi) -2 = -3\cos(2x) -2$

This option's function simplifies as:

$y= 3\cos(2(x + \pi/4)) - 2 = 3\cos(2x + \pi/2) -2 = -3\sin(2x) - 2$

Thus, this option isn't true since $\sin(2x) \neq \cos(2x)$ always (they are equal for some values of x but not for all).

Option 4: $y= -3\cos(2(x + \pi/2)) - 2$

The given function simplifies to: $y= 3\cos(2(x + \pi/2)) - 2 = 3\cos(2x + \pi) -2 = -3\cos(2x) -2$

The given option simplifies to:

$y= -3\cos(2(x + \pi/2)) - 2 = -3\cos(2x + \pi ) -2\\y = 3\cos(2x) -2$

Thus, this function is not same as the given function.

Thus, the function which is same as the function y = 3cos(2(x +π/2)) -2 is: Option A: y= 3sin(2(x + π/4)) - 2

Learn more about sine to cosine conversion here:

brainly.com/question/1421592

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1 year ago

Read 2 more answers

The store paid $2. 70 for a book and sold it for $6. 20. What is the profit as a percent?.

bogdanovich [222]

Answer:

profit = 129.63% (to the nearest hundredth) as a percent of the price the store paid for the book

Step-by-step explanation:

Use the percentage change formula:

percent change = [ (difference between the initial value and the final value) ÷ initial value] x 100

= [ (6.20 - 2.70) ÷ 2.70 ] x 100

= [ 3.5 ÷ 2.70 ] x 100

= 37/27 x 100

= 129.6296296...

= 129.63% (to the nearest hundredth)

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1 year ago

What is the sum of the first 50 terms of the series 2 + 17 + 32 + 47+ ...?

Alex

Answer:

$\huge\boxed{18,475}$

Step-by-step explanation:

The terms of this sum make the arithmetic sequence.

The fomula of a sum of <em>n</em> terms of an arithmetic sequence:

$S_n=\dfrac{[2a_1+(n-1)d]\cdot n}{2}$

We have

$a_1=2\\d=17-2=32-12=47-32=15\\n=50$

Substitute:

$S_{50}=\dfrac{[2\cdot2+(50-1)\cdot15]\cdot50}{2}=(4+49\cdot15)\cdot25=(4+735)\cdot25\\\\=739\cdot25=18,475$

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2 years ago

What decimal is 1/10 of 0.08

Luden [163]

The answer is .008

Ok so lets explain this with an easier number like 1.

1 times 1/10 is 1/10 or in decimals its .1

So the decimal just got moved further back. Same thing happens with any number.

Brainliest if this helps you out?

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2 years ago

SOLVE 6^3^-X=6^2 FOR X A.X=-6B.X=1C.X=-1D.X=9​

SOLVE 6^3^-X=6^2 FOR X A.X=-6B.X=1C.X=-1D.X=9