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3 years ago

HURRY PLZ! also please find tan as well!!! will mark brainliest!!

Mathematics

1 answer:

vazorg [7]3 years ago

3 0

I honestly don’t know man but don’t worry, if you get a bad score it won’t matter in 30 years :)

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What is the equation of this graphed line? Enter your answer in slope-intercept form in the box. -6,-3 6,-7

-BARSIC- [3]

The answer is at the bottom hope this helps

7 0

3 years ago

Read 2 more answers

Differentiating a Logarithmic Function in Exercise, find the derivative of the function. See Examples 1, 2, 3, and 4.

Leni [432]

Answer: $\dfrac{2x^2-1}{x(x^2-1)}$

Step-by-step explanation:

The given function : $y=\ln(x(x^2 - 1)^{\frac{1}{2}})$

$\Rightarrow\ y=\ln x+\ln (x^2-1)^{\frac{1}{2}}$ [ $\because \ln(ab)=\ln a +\ln b$ ]

$\Rightarrow y=\ln x+\dfrac{1}{2}\ln (x^2-1)}$ [ $\because \ln(a)^n=n\ln a$ ]

Now , Differentiate both sides with respect to x , we will get

$\dfrac{dy}{dx}=\dfrac{1}{x}+\dfrac{1}{2}(\dfrac{1}{x^2-1})\dfrac{d}{dx}(x^2-1)$ (By Chain rule)

[Note : $\dfrac{d}{dx}(\ln x)=\dfrac{1}{x}$ ]

$\dfrac{1}{x}+\dfrac{1}{2}(\dfrac{1}{x^2-1})(2x-0)$

[ $\because \dfrac{d}{dx}(x^n)=nx^{n-1}$ ]

$=\dfrac{1}{x}+\dfrac{1}{2}(\dfrac{1}{x^2-1})(2x) = \dfrac{1}{x}+\dfrac{x}{x^2-1}\\\\\\=\dfrac{(x^2-1)+(x^2)}{x(x^2-1)}\\\\\\=\dfrac{2x^2-1}{x(x^2-1)}$

Hence, the derivative of the given function is $\dfrac{2x^2-1}{x(x^2-1)}$ .

8 0

3 years ago

For the Parabolay = (x – 7)2 – 3. the equation for the Line Of Symmetry is

devlian [24]

Answer:

x = 7

Step-by-step explanation:

y = (x – 7)^2 – 3

This equation is in vertex form

y = a(x-h)^2 +k

where (h,k) is the vertex

For a vertical parabola, the line of symmetry is x=h

x = 7

5 0

3 years ago

Evaluate the determinant for the following matrix: [- 5 1 1 4] A. –21 B. 19 C. –17 D. –24

mariarad [96]

Answer:

-21

Step-by-step explanation:

I assume this is a 2 by 2 matrix.

$\begin{bmatrix} -5 & 1\\1 & 4 \end{bmatrix}$

determinant = -5 * 4 - 1 * 1 = -20 - 1 = -21

6 0

3 years ago

Read 2 more answers

If every student in a large Statistics class selects peanut M&M’s at random until they get a red candy, on average how many

kotykmax [81]

Question:

According to Masterfoods, the company that manufactures M&M's, 12% of peanut M&M's are brown, 15% are yellow, 12% are red, 23% are blue, 23% are orange and 15% are green. You randomly select peanut M&M's from an extra-large bag looking for a red candy.

If every student in a large Statistics class selects peanut M&M’s at random until they get a red candy, on average how many M&M’s will the students need to select? (Round your answer to two decimal places.).

Answer:

If every student in the large Statistics class selects peanut M&M’s at random until they get a red candy the expected value of the number of M&M’s the students need to select is 8.33 M&M's.

Step-by-step explanation:

To solve the question, we note that the statistical data presents a geometric mean. That is the probability of success is of the form.

The amount of repeated Bernoulli trials required before n eventual success outcome or

The probability of having a given number of failures before the first success is recorded.

In geometric distribution, the probability of having an eventual successful outcome depends on the the completion of a certain number of attempts with each having the same probability of success.

If the probability of each of the preceding trials is p and the kth trial is the first successful trial, then the probability of having k is given by

Pr(X=k) = $(1-p)^{k-1}p$

The number of expected independent trials to arrive at the first success for a variable Xis 1/p where p is the expected success of each trial hence p is the probability for the red and the expected value of the number of trials is 1/p or where p = 12 % which is 0.12

1/p = 1/0.12 or 25/3 or 8.33.

4 0

3 years ago