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3 years ago

Find an explicit rule for the nth term of the sequence. 7, 11, 15, 19,

Mathematics

1 answer:

ioda3 years ago

8 0

9x4=36

so thats the answer

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Will mark brainliest!! Please help asap!!

Anna007 [38]

Answer:

the answer is A

Step-by-step explanation:

8 0

3 years ago

HELP! Find the value of sin 0 if tan 0 = 4; 180 < 0< 270

BabaBlast [244]

Hi there! Use the following identities below to help with your problem.

$\large \boxed{sin \theta = tan \theta cos \theta} \\ \large \boxed{tan^{2} \theta + 1 = {sec}^{2} \theta}$

What we know is our tangent value. We are going to use the tan²θ+1 = sec²θ to find the value of cosθ. Substitute tanθ = 4 in the second identity.

$\large{ {4}^{2} + 1 = {sec}^{2} \theta } \\ \large{16 + 1 = {sec}^{2} \theta } \\ \large{ {sec}^{2} \theta = 17}$

As we know, sec²θ = 1/cos²θ.

$\large \boxed{sec \theta = \frac{1}{cos \theta} } \\ \large \boxed{ {sec}^{2} \theta = \frac{1}{ {cos}^{2} \theta} }$

And thus,

$\large{ {cos}^{2} \theta = \frac{1}{17}} \\ \large{cos \theta = \frac{ \sqrt{1} }{ \sqrt{17} } } \\ \large{cos \theta = \frac{1}{ \sqrt{17} } \longrightarrow \frac{ \sqrt{17} }{17} }$

Since the given domain is 180° < θ < 360°. Thus, the cosθ < 0.

$\large{cos \theta = \cancel\frac{ \sqrt{17} }{17} \longrightarrow cos \theta = - \frac{ \sqrt{17} }{17}}$

Then use the Identity of sinθ = tanθcosθ to find the sinθ.

$\large{sin \theta = 4 \times ( - \frac{ \sqrt{17} }{17}) } \\ \large{sin \theta = - \frac{4 \sqrt{17} }{17} }$

Answer

sinθ = -4sqrt(17)/17 or A choice.

4 0

3 years ago

List a time when it might be useful to know the unit rate.

Andrej [43]

Answer:

a good time is when u are in math class and a teacher assigns u a homework of unit rate.

Step-by-step explanation:

3 0

3 years ago

15 is 10 % of what number

Len [333]

15*10=150

15 is 10% of 150

4 0

3 years ago

1. (a) Solve the differential equation (x + 1)Dy/dx= xy, = given that y = 2 when x = 0. (b) Find the area between the two curves

erastova [34]

(a) The differential equation is separable, so we separate the variables and integrate:

$(x+1)\dfrac{dy}{dx} = xy \implies \dfrac{dy}y = \dfrac x{x+1} \, dx = \left(1-\dfrac1{x+1}\right) \, dx$

$\displaystyle \frac{dy}y = \int \left(1-\frac1{x+1}\right) \, dx$

$\ln|y| = x - \ln|x+1| + C$

When x = 0, we have y = 2, so we solve for the constant C :

$\ln|2| = 0 - \ln|0 + 1| + C \implies C = \ln(2)$

Then the particular solution to the DE is

$\ln|y| = x - \ln|x+1| + \ln(2)$

We can go on to solve explicitly for y in terms of x :

$e^{\ln|y|} = e^{x - \ln|x+1| + \ln(2)} \implies \boxed{y = \dfrac{2e^x}{x+1}}$

(b) The curves y = x² and y = 2x - x² intersect for

$x^2 = 2x - x^2 \implies 2x^2 - 2x = 2x (x - 1) = 0 \implies x = 0 \text{ or } x = 1$

and the bounded region is the set

$\left\{(x,y) ~:~ 0 \le x \le 1 \text{ and } x^2 \le y \le 2x - x^2\right\}$

The area of this region is

$\displaystyle \int_0^1 ((2x-x^2)-x^2) \, dx = 2 \int_0^1 (x-x^2) \, dx = 2 \left(\frac{x^2}2 - \frac{x^3}3\right)\bigg|_0^1 = 2\left(\frac12 - \frac13\right) = \boxed{\frac13}$

7 0

2 years ago