All categories

3 years ago

Can someone PLS solve this mystery story.

Mathematics

1 answer:

nata0808 [166]3 years ago

3 0

Answer:

ill start figuring it out now luv <3

Step-by-step explanation:

You might be interested in

This is an example of an arithmetic series: 6 + 8 + 10 + 12 + . . . + (4 + 2n) + . . .

nasty-shy [4]

True. A series is arithmetic if its terms come from an arithmetic sequence.

And a sequence is said to be arithmetic any of its two consecutive terms differ by a fixed difference, common to any consecutive pair.

As you can see, in this case, all terms differ by 2, so the sequence

$6,\ 8,\ 10,\ 12,\ \ldots,\ 4+2n$

is arithmetic, and thus the series

$6 + 8 + 10 + 12 + \ldots + (4+2n)$

is arithmetic as well.

5 0

3 years ago

Round 71.491 to the place of the underlined digit which is the 4

TiliK225 [7]

When rounding the 4 you look to the right of the number, which is the 4, and if the number is higher than 5 you would change the 4 to 5.
Therefore: 71.5

5 0

2 years ago

A Cessna 185 files an average of 208 km in one hour how many meters can the cessna fly in one hour?

Sonja [21]

This is quite basic actually,

208 kmh equals 208,000 meters per hour

This is because a km equates to 1000 meters.

8 0

3 years ago

Find an equation of the tangent plane to the given parametric surface at the specified point.

Neko [114]

Answer:

Equation of tangent plane to given parametric equation is:

$\frac{\sqrt{3}}{2}x-\frac{1}{2}y+z=\frac{\pi}{3}$

Step-by-step explanation:

Given equation

$r(u, v)=u cos (v)\hat{i}+u sin (v)\hat{j}+v\hat{k}$ ---(1)

Normal vector tangent to plane is:

$\hat{n} = \hat{r_{u}} \times \hat{r_{v}}\\r_{u}=\frac{\partial r}{\partial u}\\r_{v}=\frac{\partial r}{\partial v}$

$\frac{\partial r}{\partial u} =cos(v)\hat{i}+sin(v)\hat{j}\\\frac{\partial r}{\partial v}=-usin(v)\hat{i}+u cos(v)\hat{j}+\hat{k}$

Normal vector tangent to plane is given by:

$r_{u} \times r_{v} =det\left[\begin{array}{ccc}\hat{i}&\hat{j}&\hat{k}\\cos(v)&sin(v)&0\\-usin(v)&ucos(v)&1\end{array}\right]$

Expanding with first row

$\hat{n} = \hat{i} \begin{vmatrix} sin(v)&0\\ucos(v) &1\end{vmatrix}- \hat{j} \begin{vmatrix} cos(v)&0\\-usin(v) &1\end{vmatrix}+\hat{k} \begin{vmatrix} cos(v)&sin(v)\\-usin(v) &ucos(v)\end{vmatrix}\\\hat{n}=sin(v)\hat{i}-cos(v)\hat{j}+u(cos^{2}v+sin^{2}v)\hat{k}\\\hat{n}=sin(v)\hat{i}-cos(v)\hat{j}+u\hat{k}\\$

at u=5, v =π/3

$=\frac{\sqrt{3} }{2}\hat{i}-\frac{1}{2}\hat{j}+\hat{k}$ ---(2)

at u=5, v =π/3 (1) becomes,

$r(5, \frac{\pi}{3})=5 cos (\frac{\pi}{3})\hat{i}+5sin (\frac{\pi}{3})\hat{j}+\frac{\pi}{3}\hat{k}$

$r(5, \frac{\pi}{3})=5(\frac{1}{2})\hat{i}+5 (\frac{\sqrt{3}}{2})\hat{j}+\frac{\pi}{3}\hat{k}$

$r(5, \frac{\pi}{3})=\frac{5}{2}\hat{i}+(\frac{5\sqrt{3}}{2})\hat{j}+\frac{\pi}{3}\hat{k}$

From above eq coordinates of r₀ can be found as:

$r_{o}=(\frac{5}{2},\frac{5\sqrt{3}}{2},\frac{\pi}{3})$

From (2) coordinates of normal vector can be found as

$n=(\frac{\sqrt{3} }{2},-\frac{1}{2},1)$

Equation of tangent line can be found as:

$(\hat{r}-\hat{r_{o}}).\hat{n}=0\\((x-\frac{5}{2})\hat{i}+(y-\frac{5\sqrt{3}}{2})\hat{j}+(z-\frac{\pi}{3})\hat{k})(\frac{\sqrt{3} }{2}\hat{i}-\frac{1}{2}\hat{j}+\hat{k})=0\\\frac{\sqrt{3}}{2}x-\frac{5\sqrt{3}}{4}-\frac{1}{2}y+\frac{5\sqrt{3}}{4}+z-\frac{\pi}{3}=0\\\frac{\sqrt{3}}{2}x-\frac{1}{2}y+z=\frac{\pi}{3}$

5 0

2 years ago

Determine the corresponding general form of each of the following vertex form

SSSSS [86.1K]

Answer:

Step-by-step explanation:

(a + b)² =a² + 2ab + b²

(a -b)² = a² - 2ab + b²

1) y = (x -1)²

y= x² - 2*x*1 + 1

y = x² - 2x + 1

Ans: C

2)y = (x +4)² + 5

y = x² +2*x*4 + 4² + 5

= x² + 8x + 16 + 5

y = x² + 8x + 21

3) y = -(x + 9)²- 10

y = - [x² + 18x + 81] - 10

= -x² - 18x - 81 - 10

y =-x² - 18x - 91

4) y = 3(x + 2)² - 18

y =3 [x² + 4x + 4] - 18

y = 3x² + 12x + 12 - 18

y =3x² + 12x - 6

5) y = -2(x + 1)² - 16

= -2[x² + 2x + 1] -16

= -2x² - 4x - 2 - 16

y = -2x² - 4x - 18

6) y = 5(x + 5)²

=5[x²+ 10x + 25]

y = 5x² +50x + 125

7)y = (1/2)(x + 8)² - 8

y = (1/2) (x² + 16x + 64) - 8

$y = \dfrac{1}{2}*x^{2}+ \dfrac{1}{2}*16x + \dfrac{1}{2}*64 -8\\\\\\y =\dfrac{1}{2}x^{2}+8x +32 - 8\\\\\\y=\dfrac{1}{2}x^{2} +8x + 24$

8) y = (x + 3/2)² + 3/4

$y = x^{2} +2*x*\dfrac{3}{2}+\dfrac{9}{4}+\dfrac{3}{4}\\\\y=x^{2}+3x+\dfrac{12}{4}\\\\\\y=x^{2}+3x + 3$

9) y = 2[x² + 16x + 64] - 5x

y = 2x² + 32x + 64 - 5x

y =2x² + 27x + 6

4 0

2 years ago