Jackie

I purchased a sweater dress from Forever21. It comes down to about mid-thigh length. The sleeves are three quarter. The neck link is swooped. It is black with brown/gold leopard shaped spots on it. I want to wear it for a date night coming up. Unfortunately, I could not find a picture of it on the website for a visual.

I want to wear boots with it. Should I wear tall boots or shorter ankle booties? Also should I wear tights with it? Should the tights be black, tan or brown? I'm only 5 feet tall. So I'm not sure if wearing taller boots would be too much for me. Or I'm wondering if I'll be able to pull off the ankle boots. Another question: wedge boots or boots with a skinnier heel? Let me know what you think would look best with my lovely sweater dress. I'm trying to do something different than just wearing regular old pumps. Plus it's winter time which is why I am going for the boots with dress look.

Please provide a picture or website links if you think you have some boots in mind that might work for this outfit. Thank you for your help fashionistas!

Two questions: What is the ventilatory response to low oxygen? What is the ventilatory response to high carbon dioxide?

My notes and book are quite confusing. If anyone is able to explain these concepts in a sophisticated yet simple manner it would be much appreciated. Thank you!

I have two fine lines on my forehead. I have tried just about everything to get rid of them. I have tried Neutrogena's rapid repair treatments, various serums, and other such anti-aging night creams. I have even tried applying hyaluronic acid topically as well as in a supplement form. Nothing has worked thus far to get rid of these horrid fine lines. I don't want to hear the speal about how we all age and whatnot... I am only 23 and I have this problem! I am not sure how this has happened at such a young age. It seems outrageous. None of my friends who are my age have fine lines. Help!

So I am wondering what is a treatment that actually works? I am about to look into Juvederm... has anyone tried this? Or is there truly a cream or serum out there that actually works? Let me know what has helped you! I have been dealing with these lines for almost a year now & I am over them. Just about ready to try anything that will work. Thank you all!

So I just purchased Ulta's extreme wear gel eyeliner in black. I got an angled eyeliner brush to apply it. However, the gel seems kind of hard? I am wondering how to transfer the product to the brush. Does it need to be broken up or what? I don't want to press too hard with my brush because I do not want to break it. I thought gel was supposed to be soft? If anyone can help me with this issue or has Ulta's gel eyeliner... I would love tips! Thanks so much!

Does anyone know of a sports drink (a specific brand would be optimal) that contains only carbs and no electrolytes? I need it for a project I am working on. It must only be carbs- no electrolytes or protein. Let me know! Thanks!

A billboard that is 30 ft wide is placed 60 ft from a highway as shown in the figure. The billboard is easiest to read from the highway when the viewing angle α is larger than 7°.

**The figure shows a car driving down a road. The billboard is perpendicular to the road, so at this corner there is a 90° angle. The distance from the car to the billboard line is marked "x". The distance from the billboard to the road is 60 feet. The distance btw car and 30 ft billboard is marked angle alpha. I marked the other angle from alpha to the road as angle beta. And alpha + beta is angle gamma. I hope my description doesn't confuse you. I did my best to describe the figure from my book.

α = tan^-1(90/x) - tan^-1(60/x); beta = tan^-1(60/x); gamma = tan^-1(90/x) *(these are two other tangent equations I calculated to go with the figure I described earlier). Also x has to be greater than 25 feet to get an α larger than 7°.

QUESTION ASKS: For approx. how long does a motorist traveling at 35 mph have a viewing angle larger than 7°?

Hint given was: Answer is btw 3 and 4 seconds. How to I arrive at this answer?

Any insight as to how to complete this question would be greatly appreciated. Thanks!

A billboard that is 30 ft wide is placed 60 ft from a highway as shown in the figure. The billboard is easiest to read from the highway when the viewing angle α is larger than 7°.

a) Show that α = tan^-1 (90/x) - tan^-1 (60/x). ***The hint given with this was: Label more angles in the drawing and write two tangent equations involving them. I am not sure what my prof means by this...

b)Graph the function in part a) ***As long as I have an equation from part 1, I should be able to get this part from my calc.

c) For what approximate values of x is α greater than 7°? ***hint: The total distance traveled for this time interval is approx. 195 feet.

d) For approximately how long does a motorist traveling at 35 mph have a viewing angle larger than 7°? ***hint: Answer is between 3 and 4 seconds.

If you can answer all or even part of this question, that would be super helpful and very much appreciated! Math is not my strong point, so thank you in advance.

Here is the answer to the corresponding question: b) Find the length of the circular arc for a width of 36 and a height of 10.

(ANS): Pythagorean theorem a^2+b^2=r^2

a=r-10in.

b= 1/2*36in

(r-10)^2+18^2=r^2 -> 20r=100+18^2 -> r=21.2

The use the inverse sin function to find the angle of the right triangle at the focus.

sub angle= sin^-1(b/r) = 58.1degrees

The angle of the arc is 2*sub angle = 116.21 degrees.

This must be converted to radians if your calculator or table gave the angel in degrees.

360 degrees is 2*Pi radians. 116.21 degrees is 2.028 radians. a =2.028 radians

Now the formula s=ar can be used.

s=2.028 radians*21.2inches = 43 inches

My new question is how do I rework this problem to express the length of circular arc L in terms of w and h?

Thus far I have... a = r-h; b = 1/2*w --> (r-h)^2 + 1/2w^2 = r^2. But I am not sure where to go next... If anyone can help me, I would greatly appreciate it! Thank you!

The formula is: r = H/2 + W^2/8H. So using w=36 in. and h=10 in. The radius was 21.2 in.

So I tried to simplify this and then plug those numbers in to check my answer. I simplified it to (8H^2 + W^2)/ 24H. However, I was not getting the same number. Does anyone know what I am doing wrong? Thank you!

Is there any way to simplify this further?

r = H/2 + W^2/8H = (8H^2/16H) + (W^2/8H) = 8H^2 + W^2/24H...

Can this = 8H + W^2/24 = (H + W^2)/3 ?

The distance in feet a football travels when launched from the ground with an initial velocity of v0 ft/sec is given by: x = (v0^2 sinΘ cosΘ)/16, where Θ is the angle btw the trajectory and the ground.

a) the problem states: Use an identity to write the distance x as a function of 2Θ.

The hint given was: Use double angle identity sin(2x) = 2sin(x)cos(x).

Does this mean I plug "(v0^2 sinΘ cosΘ)/16" in for x in sin(2x) = 2sin(x)cos(x)? Help please?

Afteward, how how would I graph the fxn from part a) using an initial velocity of 50 ft/sec?

The distance in feet that a football travels when launched from the ground with an initial velocity of V0 ft/sec is given by: x = (v0 sinΘ cosΘ)/16, where Θ is the angle btw the trajectory and the ground.

a) Use an identity to write the distance x as a fxn of 2Θ. (the hint given was: use double angle identity sin(2x) = 2sin(x)cos(x).

b) How would I graph the fxn from part a) using an initial velocity of 50 ft/sec?

c) From your graph, determine the value of theta that maximizes x. Does this value of Θ max x for any velocity? Explain.

d) When Morten Anderson kicks a football 55 yd using the angle determined in part c), what is the initial velocity of the football in mph? (hint given was: justify answer 49.5 mph).

e) do you think the actual initial velocity for a 55-yd field goal is larger or smaller than that found in part d?

If you can answer all or even just parts of my question, that would be fabulous! Some help is better than none at all! So thank you in advance. Hopefully this will help me to better understand Trig. = )

Two questions: g) Find an expression in terms of V0 and Θ for the max height? & h) Show that y = (-16 sec^2 Θ/v0^2)* x^2 + x tanΘ.

The hint my prof gave me (which doesn't make sense to me) was g) subs half of the literal solution for t from part e) into the y(t) fxn and simplify. h) Solve the x(t) fxn for t literally and subs that into the y(t) fxn and simplify. Another hint I found somewhere else was: g)Plug in 270 for x in the x = ... equation, then solve for t

h) Replace sec^2 with 1/(cos^2) and tan with (sin/cos) and see if you get the original equations.

I also found out that the projectile is in the air for about 5.4 seconds. The projectile lands 270 feet from the gun. And that the max height is 117.2 feet. So I hope some of that info helps in solving. The solution from part e) was t = V0sinΘ/16. If anyone can help me, I will be forever thankful! Thank you in advance!

Okay so am I on the right track for part g)... is the answer either 270 = V0(5.4s)cosΘ = 50 = V0*cosΘ OR 270 = V0(2.7s)cosΘ = 100 = V0*cosΘ?

for h) if I sub (1/cos^2Θ) and (sinΘ/cosΘ) in for tanΘ where do I go from there?

Find an expression in terms of V0 and Θ for the max height?

Two questions: g) Find an expression in terms of V0 and Θ for the max height? & h) Show that y = (-16 sec^2 Θ/v0^2)* x^2 + x tanΘ.

The hint my prof gave me (which doesn't make sense to me) was g) subs half of the literal solution for t from part e) into the y(t) fxn and simplify. h) Solve the x(t) fxn for t literally and subs that into the y(t) fxn and simplify. Another hint I found somewhere else was: g)Plug in 270 for x in the x = ... equation, then solve for t

h) Replace sec^2 with 1/(cos^2) and tan with (sin/cos) and see if you get the original equations.

I also found out that the projectile is in the air for about 5.4 seconds. The projectile lands 270 feet from the gun. And that the max height is 117.2 feet. So I hope some of that info helps in solving. The solution from part e) was t = V0sinΘ/16. If anyone can help me, I will be forever thankful! Thank you in advance!

e) Find an expression in terms of v0 and theta for the time in the air.

f) Find an expression in terms of v0 and theta for the distance from the gun.

The coordinates for the projectile are x = v0tcosΘ and y = -16t^2 + v0t sinΘ. The projectile leaves the gun at 100 ft/s and theta = 60 degrees. I also figured out that the projectile is in the air for a total of 5.4 seconds and that the max height is 117.2 feet. The projectile also lands 270 ft from the gun. I hope some of that info is helpful.

My prof also gave this hint (however, his hint makes no sense to me): Literally solve for the elapsed time in the air. That is, solve for the y(t) fxn = to 0 for t. BUT... maybe this hint will help you. I also read that someone else said e) Ok y=0 so you have 0=-16*t^2 + v*t*sin(theta) (ignor t=0) so divide by t gives 0=-16*t + v*sin(theta) so you can get time in air from this. But I still can't quite get the answer. If you can help me I would be forever grateful!!

The coordinates of the projectile are x=v0tcosΘ and y=-16t^2 + v0tsinΘ. The problem also states that a projectile leaves the gun at 100 ft/sec and theta=60 degrees. The question: For how many seconds if the projectile in the air? I read a hint somewhere to plug 0 in for y and solve.

Would that look something like this: 0 = -16t^2 + 100t(sin60)? If so, how would I solve this? If not... does anyone know what the right answer is? Thank you so much for your help!

How do I simplify (2-√3)/(7-4√3)?

The volume of air in a swimmer's lungs (in cubic cm) is modeled by the equation:?

V=250 sin (76πt) +700 where t is the time in minutes. Find the minimum volume of air in the swimmer's lungs, and determine how many breaths the swimmer takes per minute.

The answers are: minimum volume is 450 cm cubed and 38 breaths per min. However... I do not know how to get to those answers! I would greatly appreciate your help! Thank you.

V=250 sin (76πt) +700 where t is the time in minutes. Find the minimum volume of air in the swimmer's lungs, and determine how many breaths the swimmer takes per minute.

The answers are: minimum volume is 450 cm cubed and 38 breaths per min. However... I do not know how to get to those answers! I would greatly appreciate your help! Thank you.

Okay the premise is that a polygon is inscribed in a circle with radius r. As n increases, the shape of the n-gon gets closer and closer to the circumference of the circle. The limit becomes closer to one.

Now the question is: What can you conclude is the formula for the area of a circle of radius r?

Step-by-step explaination would be greatly appreciated. Thank you in advance!