What’s 1.2983 rounded to the nearest hundredth

Answers

Answer 1

1.30 :)
All you have to do is see if the number on your left is greater than 5 or not if it’s is round up 1 if not it stays the same !
Hope this helped !!!

Answer 2

Answer:

1.30 :)

All you have to do is see if the number on your left is greater than 5 or not if it’s is round up 1 if not it stays the same !

Hope this helped !!!

Step-by-step explanation:

Related Questions

Find the general solution of y''' + 9y'' + y' − 111y = 0 if it is known that y1 = e−6x cos(x) is one solution.

Answers

Answer:

[tex]\mathbf{y(x) = \dfrac{(c_1 + c_2) \ cos (x)}{e^{6x} }+ \dfrac{i(c_1 -c_2) \ sin (x)}{e^{6x}}+ c_3e^{3\ x}}[/tex]

Step-by-step explanation:

The objective of this question is to solve:

[tex]\dfrac{dy(x)}{dx}+ 9\dfrac{d^2y(dx)}{dx^2}+\dfrac{d^3y(x)}{dx^3}-111y(x) = 0 :[/tex]

Suppose the general solution is proportional to [tex]e^{\lambda x}[/tex] for [tex]\lambda[/tex] is constant; Then:

Let's replace [tex]y(x) = e^{\lambda\ x}[/tex] into the above equation:

i.e.

[tex]\dfrac{d^3}{dx^3}(e ^{\lambda x} )+ 9 \dfrac{d^2}{dx^2}(e ^{\lambda x} ) + \dfrac{d}{dx}(e ^{\lambda x} )- 111 \ e ^{\lambda x} = 0[/tex]

To Replace:

[tex]\dfrac{d^3}{dx^3}(e ^{\lambda x} )[/tex] with [tex]\lambda^3 e ^{\lambda x }[/tex]

[tex]\dfrac{d^2}{dx^2}(e ^{\lambda x} ) \ with \ \lambda^2 e^{\lambda\ x}[/tex]

[tex]\dfrac{d}{dx}(e ^{\lambda x} ) \ with \ \lambda e ^{\lambda \ x}[/tex]

Thus;

[tex]\lambda^3 e ^{\lambda x }[/tex] + [tex]9 \lambda^2 e^{\lambda\ x}[/tex] + [tex]\lambda e ^{\lambda \ x}[/tex] - 111 [tex]e ^{\lambda \ x}[/tex] = 0

[tex]e ^{\lambda \ x} (\lambda ^3 + 9 \lambda ^2 + \lambda - 111 )= 0[/tex]

∴

In as much as [tex]e^{ \lambda x}\neq 0[/tex] for any finite [tex]\lambda[/tex]; Then:

[tex]\lambda ^3 + 9 \lambda ^2 + 111 = 0[/tex]

By Factorization:

[tex](\lambda - 3) ( \lambda ^2 + 12 \lambda + 37) = 0[/tex]

[tex]\lambda = -6 + i \ or\ \lambda = -6 - i \ or \ \lambda = 3[/tex]

However;

The root [tex]\lambda = -6 \pm i[/tex] yield;

[tex]y_1 = (x) = c_1 e ^ {(-6+i)x}[/tex]

[tex]y_2 (x) = c_2e^{(-6-i)x}[/tex]

The root [tex]\lambda = 3[/tex] yield;

[tex]y_3(x) = c_3 e^{3x}[/tex]

∴

The general solution is:

[tex]y(x) = y_1(x) + y_2(x) + y_3(x) = \dfrac{c_1}{c^{(6-i)}x}+\dfrac{c_2}{c^{(6+i)}x}+ c_3e^{3x}[/tex]

Using Euler's Identity ;

[tex]e^{\alpha+i \beta} = e^\alpha \ cos (\beta ) + i \ e^\alpha \ sin ( \beta)[/tex]

[tex]y(x) = c_1 ( \dfrac{cos (x) }{e^{6x}}+ \dfrac{i \ sin x }{e^{6x} }) + c_2 ( \dfrac{cos (x)}{e^{6x}}- \dfrac{-i \ sin (x)}{e^{6x}})+c_3 e^{3x}[/tex]

[tex]\mathbf{y(x) = \dfrac{(c_1 + c_2) \ cos (x)}{e^{6x} }+ \dfrac{i(c_1 -c_2) \ sin (x)}{e^{6x}}+ c_3e^{3\ x}}[/tex]

What Is It HomeWork*.

Answers

Answer:

sumthing that we hate doing

Step-by-step explanation:ughhhhhhhhhh

The inequalities is 6 is greater than or equal to 6

find the values of the six trigonometric functions for angle theta in standard position if a point with the coordinates (1, -8) lies on its terminal side

Answers

Answer:

cosФ = [tex]\frac{1}{\sqrt{65}}[/tex] , sinФ = [tex]-\frac{8}{\sqrt{65}}[/tex] , tanФ = -8, secФ = [tex]\sqrt{65}[/tex] , cscФ = [tex]-\frac{\sqrt{65}}{8}[/tex] , cotФ = [tex]-\frac{1}{8}[/tex]

Step-by-step explanation:

If a point (x, y) lies on the terminal side of angle Ф in standard position, then the six trigonometry functions are:

cosФ = [tex]\frac{x}{r}[/tex]sinФ = [tex]\frac{y}{r}[/tex]tanФ = [tex]\frac{y}{x}[/tex]secФ = [tex]\frac{r}{x}[/tex]cscФ = [tex]\frac{r}{y}[/tex]cotФ = [tex]\frac{x}{y}[/tex]Where r = [tex]\sqrt{x^{2}+y^{2} }[/tex] (the length of the terminal side from the origin to point (x, y)You should find the quadrant of (x, y) to adjust the sign of each function

∵ Point (1, -8) lies on the terminal side of angle Ф in standard position

∵ x is positive and y is negative

→ That means the point lies on the 4th quadrant

∴ Angle Ф is on the 4th quadrant

∵ In the 4th quadrant cosФ and secФ only have positive values

∴ sinФ, secФ, tanФ, and cotФ have negative values

→ let us find r

∵ r = [tex]\sqrt{x^{2}+y^{2} }[/tex]

∵ x = 1 and y = -8

∴ r = [tex]\sqrt{x} \sqrt{(1)^{2}+(-8)^{2}}=\sqrt{1+64}=\sqrt{65}[/tex]

→ Use the rules above to find the six trigonometric functions of Ф

∵ cosФ = [tex]\frac{x}{r}[/tex]

∴ cosФ = [tex]\frac{1}{\sqrt{65}}[/tex]

∵ sinФ = [tex]\frac{y}{r}[/tex]

∴ sinФ = [tex]-\frac{8}{\sqrt{65}}[/tex]

∵ tanФ = [tex]\frac{y}{x}[/tex]

∴ tanФ = [tex]-\frac{8}{1}[/tex] = -8

∵ secФ = [tex]\frac{r}{x}[/tex]

∴ secФ = [tex]\frac{\sqrt{65}}{1}[/tex] = [tex]\sqrt{65}[/tex]

∵ cscФ = [tex]\frac{r}{y}[/tex]

∴ cscФ = [tex]-\frac{\sqrt{65}}{8}[/tex]

∵ cotФ = [tex]\frac{x}{y}[/tex]

∴ cotФ = [tex]-\frac{1}{8}[/tex]

Need help ASAP pls !!!!!

Answers

Answer:

growth and .09% increase I believe

Step-by-step explanation:

HELP PLSSSSS AAAAAAAAAAAAAAAAAAAAAA ILL GIVE BRAINLIEST B(

Answers

Answer:

a(n) = (3/5)a(n-1), where a(1) = 10 and n is 2 or greater.

Step-by-step explanation:

determine the common factor...The first term (10) by 3/5 results in the given second term (6); the third term is 3/5 of the second term (6), resulting in 18/5 (equivalent to 36/10 or 3.6/1, or just 3.6. Resulting in a(n) = (3/5)a(n-1).

the third term is 3.6, which, if mult. by (3/5), produces 2.16, as expected.

Thus, the recursive formula for this geometric sequence is a(n) = (3/5)a(n-1).

This is good only for a(1) = 10 and n = 2, 3, ....

Hope this helps you

Answer: