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2 years ago

Please help with this precalc!!! i desperately need help

Mathematics

1 answer:

Paladinen [302]2 years ago

6 0

Step-by-step explanation:

4 is the A or Ampltiude

To find the period set it equal to 2 pi/B

$\frac{2\pi}{b} = \frac{6}{1}$

$\frac{b}{2\pi} = \frac{1}{6}$

$\frac{2\pi}{6}$

$\frac{\pi}{3}$

There is no phase shift

The midline or d or vertical shift is 1.

Our equation for 8 is

$4 \sin( \frac{\pi}{6} x) + 1$

9.Use sin function. The a value is 2 or -2 your choice.

Set up the period equal to the equation

$\frac{2\pi}{b} = \frac{\pi}{3}$

$\frac{b}{2\pi} = \frac{3}{\pi}$

$b = \frac{3}{\pi} \times 2\pi$

Which equal 6.

B=6

There is no C and the D and Vertical Shift is 1.

Our equation is

$2 \sin(6x) + 1$

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A hydro company is laying down power lines in the ground. They have two sections of cable so far that are 325.0 m and 430.0 m lo

Hoochie [10]

The length of extra cable that is required to connect the two pieces of existing cable is equal to 182 meters.

<h3>How to determine the length of extra cable?</h3>

In order to determine the length of extra cable that is required to connect the two pieces of existing cable, we would apply the law of cosine as follows:

B² = A² + C² - 2(A)(C)cosB

Substituting the given parameters into the formula, we have;

B² = 325.0² + 430.0² - 2(325.0)(430.0)cos23

B² = 105,625 + 184,900 - 279,500(0.9205)

B² = 290,525 - 257,279.75

B² = 33,245.25

B = √33,245.25

B = 182.33 ≈ 182 meters.

Read more on cosine law here: brainly.com/question/11000638

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3 0

2 years ago

Of all the soft drink consumers in a particular sales region, 30% prefer Brand A and 70% prefer Brand

victus00 [196]

Answer:

Option D -0.67

Step-by-step explanation:

Given : Preference of Brand A =30% ⇒ $P(A)=\frac{30}{100}=0.3$

Preference of Brand B =70% ⇒ $P(B)=\frac{70}{100}=0.7$

Prefer Brand A and are Female = 20% ⇒ $P(A/F)=\frac{20}{100}=0.2$

Prefer Brand B and are Female = 40% ⇒ $P(B/F)=\frac{40}{100}=0.4$

To find : Selected consumer is female, given that the person prefers Brand A $P(F/A)$

Solution : Using Bayes' theorem, which state that

$P(A/B)=\frac{P(B/A)P(A)}{P(B)}$

where, P(A) and P(B) are probabilities of observing A and B.

P(B/A)= is a conditional probability where event B occur and A is true

P(A/B)= also a conditional probability where event A occur and B is true.

Now, applying Bayes' theorem,

$P(F/A)=\frac{P(A/F)P(A)}{P(B)P(B/F)+P(A)P(A/F)}$

$P(F/A)=\frac{(0.2)(0.3)}{(0.7)(0.4)+(0.3)(0.2)}$

$P(F/A)=\frac{0.6}{0.28+0.6}$

$P(F/A)=\frac{0.6}{0.88}$

$P(F/A)=0.68$

Therefore, Option D is correct probability that a randomly selected consumer is female, given that the person prefers Brand A -0.67

3 0

3 years ago

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shutvik [7]

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6 0

1 year ago

Which of the following are not polynomials

tankabanditka [31]

Answer:

B and E

Step-by-step explanation:

A polynomial cant have a variable as a denominator nor a negative/fractional exponent

5 0

3 years ago

CALC- limits<br> please show your method

gladu [14]

A. Factor the numerator as a difference of squares:

$\displaystyle\lim_{x\to9}\frac{x-9}{\sqrt x-3}=\lim_{x\to9}\frac{(\sqrt x-3)(\sqrt x+3)}{\sqrt x-3}=\lim_{x\to9}(\sqrt x+3)=6$

c. As $x\to\infty$ , the contribution of the terms of degree less than 2 becomes negligible, which means we can write

$\displaystyle\lim_{x\to\infty}\frac{4x^2-4x-8}{x^2-9}=\lim_{x\to\infty}\frac{4x^2}{x^2}=\lim_{x\to\infty}4=4$

e. Let's first rewrite the root terms with rational exponents:

$\displaystyle\lim_{x\to1}\frac{\sqrt[3]x-x}{\sqrt x-x}=\lim_{x\to1}\frac{x^{1/3}-x}{x^{1/2}-x}$

Next we rationalize the numerator and denominator. We do so by recalling

$(a-b)(a+b)=a^2-b^2$
$(a-b)(a^2+ab+b^2)=a^3-b^3$

In particular,

$(x^{1/3}-x)(x^{2/3}+x^{4/3}+x^2)=x-x^3$
$(x^{1/2}-x)(x^{1/2}+x)=x-x^2$

so we have

$\displaystyle\lim_{x\to1}\frac{x^{1/3}-x}{x^{1/2}-x}\cdot\frac{x^{2/3}+x^{4/3}+x^2}{x^{2/3}+x^{4/3}+x^2}\cdot\frac{x^{1/2}+x}{x^{1/2}+x}=\lim_{x\to1}\frac{x-x^3}{x-x^2}\cdot\frac{x^{1/2}+x}{x^{2/3}+x^{4/3}+x^2}$

For $x\neq0$ and $x\neq1$ , we can simplify the first term:

$\dfrac{x-x^3}{x-x^2}=\dfrac{x(1-x^2)}{x(1-x)}=\dfrac{x(1-x)(1+x)}{x(1-x)}=1+x$

So our limit becomes

$\displaystyle\lim_{x\to1}\frac{(1+x)(x^{1/2}+x)}{x^{2/3}+x^{4/3}+x^2}=\frac{(1+1)(1+1)}{1+1+1}=\frac43$

3 0

3 years ago