All Answers 357331 Answers

Nope MRNT, 

A woodchuck would chuck as much wood as a woodchuck could chuck if a woodchuck could chuck wood

i remeber that one too

Scientists deduce that the universe is a 11D (our world is 3D) "cube".<-- Can't describe.....

Twinkle Twinkle Little Star XD

why r u answering to your own question

well my dog has had 2 sergeries one on his stomach because he had a twisted stomach. he almost ran away to hide and die but my brother found him. my family was so scared he was going to die but we got him to a very nice vet hospital .  he survived that twisted stomach only because we asked God  to heal him.to all of you who read this i hope you will understand that there is a God that can help and protect you and when you except Jesus as your Lord and savior you will understand how quick my Faith grew stronger in Jesus.thank you!

i am great (lol answering my brothers question when i could just talk to him)

Based on what has gone before the probablility of winning is 1 in 2

so 1/2

first 3 terms of arithmetic sequence

where the 21st term is -38 and the 50th term is -125

Let
$\begin{array}{ll} x=21 & t_x=t_{21}=-38 \\ y=50 & t_y=t_{50}=-125 \\ z=1 & t_z=t_1= ? \\ \end{array}$

$\begin{array}{|lrcll|} \hline \text{AP:} &\mathbf{ t_z } &\mathbf{=}& \mathbf{ t_x\cdot (\frac{z-y}{x-y}) +t_y\cdot (\frac{x-z}{x-y}) } \\ &t_{1} &=& (-38)\cdot (\frac{1-50}{21-50}) + (-125)\cdot (\frac{21-1}{21-50}) \\ &t_{1} &=& (-38)\cdot (\frac{-49}{-29}) + (-125)\cdot (\frac{20}{-29}) \\ &t_{1} &=& -38\cdot (\frac{49}{29}) + 125\cdot (\frac{20}{29}) \\ &t_{1} &=& \frac{-38\cdot 49+ 125\cdot 20}{29} \\ &t_{1} &=& \frac{638}{29} \\ & \mathbf{ t_{1} } & \mathbf{=} & \mathbf{22} \\ \hline \end{array}$

Let
$\begin{array}{ll} x=21 & t_x=t_{21}=-38 \\ y=50 & t_y=t_{50}=-125 \\ z=2 & t_z=t_2= ? \\ \end{array}$

$\begin{array}{|lrcll|} \hline \text{AP:} &\mathbf{ t_z } &\mathbf{=}& \mathbf{ t_x\cdot (\frac{z-y}{x-y}) +t_y\cdot (\frac{x-z}{x-y}) } \\ &t_{2} &=& (-38)\cdot (\frac{2-50}{21-50}) + (-125)\cdot (\frac{21-2}{21-50}) \\ &t_{2} &=& (-38)\cdot (\frac{-48}{-29}) + (-125)\cdot (\frac{19}{-29}) \\ &t_{2} &=& -38\cdot (\frac{48}{29}) + 125\cdot (\frac{19}{29}) \\ &t_{2} &=& \frac{-38\cdot 48+ 125\cdot 19}{29} \\ &t_{2} &=& \frac{551}{29} \\ & \mathbf{ t_{2} } & \mathbf{=} & \mathbf{19} \\ \hline \end{array}$

$\begin{array}{|lrcll|} \hline d&=&t_2-t_1 \\ d&=&19-22 \\ d&=&-3 \\\\ t_3 &=& t_2 +d \\ t_3 &=& 19-3 \\ \mathbf{ t_{3} } & \mathbf{=} & \mathbf{16} \\ \hline \end{array}$

The first 3 terms of arithmetic sequence: 22, 19, 16, ...

a5 = 7, a8 = 56. Find a11

$\begin{array}{|lrcll|} \hline \text{AP:} &\mathbf{ t_z } &\mathbf{=}& \mathbf{ t_x\cdot (\frac{z-y}{x-y}) +t_y\cdot (\frac{x-z}{x-y}) } \\ \text{GP:} &\mathbf{ t_z } &\mathbf{=}& \mathbf{ t_x^{\frac{z-y}{x-y}} \cdot t_y^{\frac{x-z}{x-y}} } \\ \hline \end{array}$

Let

$\begin{array}{ll} x= 5 & t_x=t_5=7 \\ y=8 & t_y=t_8=56 \\ z=11 & t_z=t_{11}=? \\ \end{array}$

$\begin{array}{|lrcll|} \hline \text{AP:} &\mathbf{ t_z } &\mathbf{=}& \mathbf{ t_x\cdot (\frac{z-y}{x-y}) +t_y\cdot (\frac{x-z}{x-y}) } \\ &t_{11} &=& 7\cdot (\frac{11-8}{5-8}) + 56\cdot (\frac{5-11}{5-8}) \\ &t_{11} &=& 7\cdot (\frac{3}{-3}) + 56\cdot (\frac{-6}{-3}) \\ &t_{11} &=& 7\cdot (-1) + 56\cdot 2 \\ &t_{11} &=& -7 + 112 \\ & \mathbf{ t_{11} } & \mathbf{=} & \mathbf{105} \\ \hline \end{array}$

$\begin{array}{|lrcll|} \hline \text{GP:} &\mathbf{ t_z } &\mathbf{=}& \mathbf{ t_x^{\frac{z-y}{x-y}} \cdot t_y^{\frac{x-z}{x-y}} } \\ &t_{11} &=& 7^{\frac{11-8}{5-8}} \cdot 56^{\frac{5-11}{5-8}} \\ &t_{11} &=& 7^{-1} \cdot 56^{2} \\ &t_{11} &=& \frac{3136}{7} \\ & \mathbf{ t_{11} } & \mathbf{=} & \mathbf{448} \\ \hline \end{array}$

sin(arcsin x + arccos x)

i)
$\begin{array}{lrcll} & \cos{(\varphi)} &=& x \\ \text{or}& \quad \varphi &=& \arccos{(x)}\\ \end{array}$

ii)
$\begin{array}{lrclcl} &\sin{(90^\circ-\varphi)} &=& \cos{(\varphi)} &=& x \\ \text{or}& \quad 90^\circ-\varphi && &=& \arcsin{(x)}\\ \end{array}$

iii)

$\begin{array}{rcll} && \sin\Big(\arcsin(x)+\arccos(x)\Big) \\ &=& \sin(90^\circ-\varphi+\varphi) \\ &=& \sin(90^\circ)\\ &=& 1\\ \end{array}$

This explains it pretty well :

https://en.wikipedia.org/wiki/Rule_of_72

If sin(A)=cos(A) was is the angle measurement?

$\small{ \begin{array}{rcl|rcl|rcl} \sin(A)&=&\cos(A) \\ \cos(A)-\sin(A) &=& 0 & \sin(45^{\circ}-A) &=& \sin(45^{\circ})\cdot\cos(A)- \cos(45^{\circ})\cdot\sin(A) & \sin(45^{\circ})=\cos(45^{\circ}) = \frac{\sqrt{2}}{2} \\ & & & \sin(45^{\circ}-A) &=& \frac{\sqrt{2}}{2}\cdot\cos(A)- \frac{\sqrt{2}}{2}\cdot\sin(A) \\ & & & \sin(45^{\circ}-A) &=& \frac{\sqrt{2}}{2}\Big(\cdot\cos(A)- \sin(A) \Big) \\ \frac{2}{\sqrt{2}}\cdot \sin(45^{\circ}-A)&=& 0 & \frac{2}{\sqrt{2}}\cdot \sin(45^{\circ}-A) &=& \cos(A)- \sin(A) \\ \sin(45^{\circ}-A)&=& 0 \\ 45^{\circ}-A &=& \arcsin(0) \\ 45^{\circ}-A &=& 0\pm n\cdot 360^{\circ} \qquad n \in \mathbb{N} \\ \mathbf{A} & \mathbf{=} & \mathbf{45^{\circ} \pm n\cdot 360^{\circ}} \\\\ \sin(45^{\circ}-A)=\sin(180^{\circ}-(45^{\circ}-A) ) &=& 0 \\ 180^{\circ}-(45^{\circ}-A) &=& \arcsin(0) \\ 180^{\circ}-45^{\circ}+A &=& \arcsin(0) \\ 135^{\circ} + A &=& 0\pm n\cdot 360^{\circ} \qquad n \in \mathbb{N} \\\\ A &=& -135^{\circ} \pm n\cdot 360^{\circ} \\ A &=& -135^{\circ} + 360^{\circ} \pm n\cdot 360^{\circ} \\ \mathbf{A} & \mathbf{=} & \mathbf{225^{\circ} \pm n\cdot 360^{\circ}} \\ \end{array} }$

Area of triangle  1  =

(1/2) (8)(14.2) sin (99°)   ≈  56.1 in^2

Area of triangle 2   =

 (1/2) (12) (14)sin(20°)  ≈  28.7  ft^2

Area of triangular lawn =

(1/2) ( 8) (12) sin (51°)  ≈ 37.3 m^2

Area of last triangle =

(1/2) (3.2) (4.7) sin ( 62°)  ≈  6.64  ft^2

What is the distance to the earth’s horizon

Tangent Secant Theorem

What is the distance to the earth’s horizon from point P?

Round to the nearest tenth. 

r = 3959 mi

$x^2 = 15.6 \cdot (15.6 + 2\cdot 3959) \\ x = 351.8\ \text{mi}$

A skydiver is at an altitude of 1.4 mi above the earth’s surface.

From the skydiver’s viewpoint, what is the distance to the horizon?

Round to the nearest tenth.

r = 3959 mi

$x^2 = 1.4\cdot (1.4 + 2\cdot 3959) \\ x = 105.3\ \text{mi}$

I think you have written this :

f (x)  = 27x^3        g(x)  =  1/3  + ∛x

These are not inverses

However.....I think you may mean this :

f(x)   =  27x^3          g(x)  = (1/3)∛x

To see if they are inverses....put f into g   and then  g into f.....if both results  = x...we have inverses

f into  g   =    (1/3)∛ (27x^3)   =  (1/3) (3 x)   =   x

g into f   =   27 [ (1/3)∛x ] ^3  =   27 (1/27) *x   =   x

Since  we get x in each instance, f and g are inverses....!!!!

We can only simplify this equation, since you haven't defined x.......

$15x - 0.3 + 100 \implies 15x + (100-0.3) \implies \boxed{15x + 99.7}$

15 x - 0.3+100 =

15x + 99.7