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son4ous [18]
3 years ago
15

Solve for x. Will mark brainliest

Mathematics
1 answer:
Serjik [45]3 years ago
4 0

Answer:

41) x = 8

Step-by-step explanation:

41)

Arc MN is 100 degrees because it is the double of 50 degrees.

12x+4=100

12x-4=100-4

12x=96

x/12=96/12

x=8

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What is the end behavior in the function y=2x^3-x
sasho [114]

Answer:

Step-by-step explanation:

When a question asks for the "end behavior" of a function, they just want to know what happens if you trace the direction the function heads in for super low and super high values of x. In other words, they want to know what the graph is looking like as x heads for both positive and negative infinity. This might be sort of hard to visualize, so if you have a graphing utility, use it to double check yourself, but even without a graph, we can answer this question. For any function involving x^3, we know that the "parent graph" looks like the attached image. This is the "basic" look of any x^3 function; however, certain things can change the end behavior. You'll notice that in the attached graph, as x gets really really small, the function goes to negative infinity. As x gets very very big, the function goes to positive infinity.

Now, taking a look at your function, 2x^3 - x, things might change a little. Some things that change the end behavior of a graph include a negative coefficient for x^3, such as -x^3 or -5x^3. This would flip the graph over the y-axis, which would make the end behavior "swap", basically. Your function doesn't have a negative coefficient in front of x^3, so we're okay on that front, and it turns out your function has the same end behavior as the parent function, since no kind of reflection is occurring. I attached the graph of your function as well so you can see it, but what this means is that as x approaches infinity, or as x gets very big, your function also goes to infinity, and as x approaches negative infinity, or as x gets very small, your function goes to negative infinity.

6 0
3 years ago
Over what interval is the function in this graph increasing?
Levart [38]
C because -3 is less than 2 and we’ll stay the same
6 0
3 years ago
This my last question my bad guys and I appreciate everybody that’s been helping me out!!!❤️
Elenna [48]

Answer:

quadrant 3

Step-by-step explanation:

Cuz x and y are in the negative if they’re both less than 0

4 0
3 years ago
Read 2 more answers
What is the compound interest of 10400 at 12.7% for 4 year
Naddik [55]

The Compound Interest of 10400 at 12.7% for 4 years is 6378.

The principal amount is given as 10400.

The rate of interest is given as 12.7%.

The time period to be calculated is given as 4 years.

The compound interest for the given above is to be calculated.

<h3>What is compound interest?</h3>

Compound interest is the interest that we earn both on the principal amount and the interest we earn.

The formula used to calculate compound interest is:                                  

   P [ (1 + \frac{R}{100} )^n - 1 ]

Where P = principal amount, R = rate of interest, and n = number of years.

We have,

P = 10400

R = 12.7%

n = 4 years

Compound interest:

P [ (1 + \frac{R}{100} )^n - 1 ]\\\\10400 [ (1 + \frac{12.7}{100} )^4 - 1 ]

Now,

10400 [ ( 1 + 0.12.7 )^2 - 1 ]

10400 [ 1.127^4 - 1 ]

10400 [ 1.61322 - 1 ]

10400 x 0.6132

6377.56

Rounding to the nearest whole number.

We have,

Compound Interest = 6378.

Thus the Compound Interest of 10400 at 12.7% for 4 years is 6378.

Learn more about Compound Interest here:

brainly.com/question/13155407

#SPJ1

8 0
2 years ago
Find sinϴ and cosϴ if tanϴ=1/4 and sinϴ&gt;0
eduard
If you're using the app, try seeing this answer through your browser:  brainly.com/question/2787701

_______________


\mathsf{tan\,\theta=\dfrac{1}{4}\qquad\qquad(sin\,\theta\ \textgreater \ 0)}\\\\\\&#10;\mathsf{\dfrac{sin\,\theta}{cos\,\theta}=\dfrac{1}{4}}\\\\\\&#10;\mathsf{4\,sin\,\theta=cos\,\theta\qquad\quad(i)}


Square both sides:

\mathsf{(4\,sin\,\theta)^2=(cos\,\theta)^2}\\\\&#10;\mathsf{4^2\,sin^2\,\theta=cos^2\,\theta}\\\\&#10;\mathsf{16\,sin^2\,\theta=cos^2\,\theta\qquad\qquad(but,~cos^2\,\theta=1-sin^2\,\theta)}\\\\&#10;\mathsf{16\,sin^2\,\theta=1-sin^2\,\theta}

\mathsf{16\,sin^2\,\theta+sin^2\,\theta=1}\\\\&#10;\mathsf{17\,sin^2\,\theta=1}\\\\&#10;\mathsf{sin^2\,\theta=\dfrac{1}{17}}\\\\\\&#10;\mathsf{sin\,\theta=\pm\,\sqrt{\dfrac{1}{17}}}\\\\\\&#10;\mathsf{sin\,\theta=\pm\,\dfrac{1}{\sqrt{17}}}


Since \mathsf{sin\,\theta} is positive, you can discard the negative sign. So,

\mathsf{sin\,\theta=\dfrac{1}{\sqrt{17}}\qquad\quad\checkmark}


Substitute this value back into \mathsf{(i)} to find \mathsf{cos\,\theta:}

\mathsf{4\cdot \dfrac{1}{\sqrt{17}}=cos\,\theta}\\\\\\&#10;\mathsf{cos\,\theta=\dfrac{4}{\sqrt{17}}\qquad\quad\checkmark}


I hope this helps. =)


Tags:   <em>trigonometric identity relation trig sine cosine tangent sin cos tan trigonometry precalculus</em>

7 0
3 years ago
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