All Answers 358272 Answers

5+5=10

thats easy

1111111111111111111111111111111111111112

I wouldn't mind some coffee right now!

Derivative:

diff( 2*y+6*x, x )

Integral:

int( 2*y+6*x, x )

Roots:

solve( 2*y+6*x = 0, x )

Haha, thanks FAFA! I'm great, you?

I can't have coffee... 

I'm great, thank you. I feel your pain!!! I need cofee too!!!!!!!

Yeah I just noticed that. But thx!

Your Welcome

I'm good but I could go for some coffee 

As follows:

\(-2\frac{1}{2} \rightarrow -(\frac{4}{2}+\frac{1}{2}) \rightarrow-\frac{5}{2}\\\\ -1\frac{3}{5} \rightarrow -(\frac{5}{5}+\frac{3}{5})\rightarrow-\frac{8}{5}\)

The product of two negative numbers is positive so:

\(-\frac{5}{2}\times-\frac{8}{5}\rightarrow\frac{5}{2}\times\frac{8}{5}\rightarrow \frac{8}{2}\rightarrow4\)

So:  \(-2\frac{1}{2}\times -1\frac{3}{5}=4\)

Hello asinus!

Thank you very much.

Hello heureka!

It is "top",
how you represent the trigonometric relationships.
I myself have not considered the arc cos 0.4 is ambiguous,
and your explanation with the unit circle is very obvious.

In my answer is wrong A = 0.916515138992,
correct is sin A = 0.916515138992

Greetings asinus :- )

 !

Cylinders X and Y are right cylinders with equal volumes.

Cylinder X has a height of 5 inches and a radius of 6 inches.

Cylinder Y has a height of 9 inches.

What is the radius of Cylinder Y, in inches

Radius of Cylinder X \(= r_x\)
Height of Cylinder X \(= h_x\)

Radius of Cylinder Y \(= r_y\)
Height of Cylinder Y \(= h_y\)

\(\begin{array}{rclcll} V &=& \pi \cdot r_y^2 \cdot h_y &=& \pi \cdot r_x^2 \cdot h_x \\ && \not{\pi} \cdot r_y^2 \cdot h_y &=& \not{\pi} \cdot r_x^2 \cdot h_x \\ && r_y^2 \cdot h_y &=& r_x^2 \cdot h_x \qquad & | \qquad : h_y\\ && r_y^2 &=& r_x^2 \cdot \frac{ h_x } { h_y } \qquad & | \qquad \sqrt{}\\ && r_y &=& r_x \cdot \sqrt{ \frac{ h_x } { h_y } } \qquad & | \qquad h_x = 5\ in. \qquad h_y = 9\ in. \qquad r_x = 6\ in.\\ && r_y &=& 6\ in. \cdot \sqrt{ \frac{ 5\ in. } { 9\ in. } } \\ && r_y &=& 6 \cdot \sqrt{ \frac{ 5 } { 9 } } \ in.\\ && r_y &=& 6 \cdot \frac{ \sqrt{ 5 } } { 3 } \ in.\\ && r_y &=& 2 \cdot \sqrt{ 5 } \ in.\\ && \mathbf{r_y} &\mathbf{=}& \mathbf{4.472135955 \ in.} \end{array}\)

Input        Output

  0                  3

  1                  5

  2                  7

Please help, I have been working on this problem for a while now. I can't seem to figure it out!

\(\begin{array}{|c|c|} \hline \text{Input} & \text{Input } + 1 = \text{ Index } \\ \hline 0 & 1 \\ \hline 1 & 2 \\ \hline 2 & 3 \\ \hline \end{array}\)

 Arithmetic sequence:

\(\begin{array}{rcl} a_n &=& a_1 + (n-1)\cdot d \qquad n = \text{input }+1 \qquad a_1 = 3 \qquad d= 2 \\ a_n &=& 3 + ( \text{input }+1 - 1 ) \cdot 2 \\ a_n &=& 3 + \text{input } \cdot 2 \\ \text{output } &=& 3 + \text{input } \cdot 2 \\ \end{array}\)

Eric jogged

3 1/4 miles on Monday,

5 5/8 miles on Tuesday, and

8 miles on Wednesday.

Suppose he continues the pattern for the remainder of the week.

How far will Eric jog on Friday?

\(\begin{array}{lrrll} \text{Monday} & 3 & \frac{2}{8} & \text{ miles} \\ &&&& +2\frac{3}{8}\\ \text{Tuesday} & 5 & \frac{5}{8} & \text{ miles} \\ &&&& +2\frac{3}{8}\\ \text{Wednesday} & 7 & \frac{8}{8} & \text{ miles} \\ &&&& +2\frac{3}{8}\\ \text{Tursday} & 9 & \frac{11}{8} & \text{ miles} \\ &&&& +2\frac{3}{8}\\ \text{Friday} & 11 & \frac{14}{8} & \text{ miles} \\ \end{array}\)

Eric will jog on Friday \(11\frac74 \ \text{miles}\)

You need to supply the ordered pairs!

.

pi*r²*h gives  pi*(5.3/2)²*9 m³   Don't forget to square the radius!

.

Find sin A if cos A =7/25

I. Quadrant:

\(\sin{(A)} = + \sqrt{1-\cos^2{(A)}}\\ \sin{(A)} = + \sqrt{1-(\frac{7}{25})^2}\\ \sin{(A)} = + \sqrt{1-\frac{7^2}{25^2}}\\ \sin{(A)} = + \sqrt{ \frac{25^2-7^2}{25^2}}\\ \sin{(A)} = + \frac{ \sqrt{ 25^2-7^2 } } {25}\\ \sin{(A)} = + \frac{24} {25}\\ \sin{(A)} = 0.96\)

IV. Quadrant:

\(\sin{(A)} = - \sqrt{1-\cos^2{(A)}}\\ \sin{(A)} = -\sqrt{1-(\frac{7}{25})^2}\\ \sin{(A)} = - \sqrt{1-\frac{7^2}{25^2}}\\ \sin{(A)} = - \sqrt{ \frac{25^2-7^2}{25^2}}\\ \sin{(A)} = - \frac{ \sqrt{ 25^2-7^2 } } {25}\\ \sin{(A)} = - \frac{24} {25}\\ \sin{(A)} = -0.96\)

Like this:

.

Find sin A if cos A = 0.4

I. Quadrant:

\(\sin{(A)} = + \sqrt{1-\cos^2{(A)}}\\ \sin{(A)} = + \sqrt{1-0.4^2}\\ \sin{(A)} = 0.91651513899 \)

IV. Quadrant:

\(\sin{(A)} = - \sqrt{1-\cos^2{(A)}}\\ \sin{(A)} = - \sqrt{1-0.4^2}\\ \sin{(A)} = -0.91651513899\)

Find sin A if cos A = 0.4

A = arc cos 0.4

sin A = sin (arc cos 0.4) = 0.9165

A = 0.916515138992

!