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Since  x , y  ≥ 0.......we are only concerned with the lattice points intersected by the line in the first quadrant

So  we  have

y  = -2x + 18

The line has a y intercept  of  18.....so one lattice point  is  (0,18)

And letting  y  = 0, we can find the x intercept as

0 = -2x + 18

-18  = -2x

9  =  x ....so  a second lattice point is (9,0)

Now.....the slope  is  - 2.....which  means that y will change by -2  when x changes by 1

So.....the other lattice points intercepted by the line are

(1, 16)    (2, 14)   (3, 12)  (4, 10)  (5, 8)  (6, 6)  ( 7, 4)  (8,2)

So....10 lattice points

See the graph here :

https://www.desmos.com/calculator/climjgy2hb

7x^2  + 5x  = h

7x^2 + 5x - h  = 0

We will have one solution when the discriminant  =  0

So

5^2  - 4(7)(-h)   = 0

25 + 28h = 0

25 =  - 28h

-25/28 =  h

This graph shows that this is true :

https://www.desmos.com/calculator/jbpjhjdgt4

y^2  +  24y + 16 + k

Take  (1/2) of 24  =  12

Square this  = 144

So....this must be added to y^2 + 24y   to have a perfect square binomial

Which implies that

144 = 16 + k          subtract 16 from both sides

128  =  k

tan(46°)  = x / 12     multiply both sides by 12

12* tan (46°)  = x ≈  12.4

Let the height of the pyramid  =  h

And let the distance from the face of the base to the point directly beneath the apex = x

So....we have two equations

tan(36.4)  =  h / (350 + x)       multiply both sides by (350 + x)

(350 + x) tan (36.4)  = h      (1)

And the second one is

tan (51.9)  = h / x

x* tan(51.9)  = h     (2)

Equating (1)  and (2), we have

(350 + x) * tan(36.4)   =  x * tan (51.9)     simplify

350*tan(36.4)  + x * tan (36.4)  = x * tan (51.9)

350* tan(36.4)  =    x* tan(51.9) - x * tan (36.4)

350*tan (36.4)  = x [ tan(51.9) - tan(36.4) ]         

350 tan(36.4) / [ tan (51.9) - tan(36.4 ) ]   = x

x ≈ 479.558 ft

Using (2)  we can find the height as

479.558 * tan(51.9)  =  h ≈  611.6 ft

The distance, d, that the submarine descends in one minute can be expressed as

sin(5°)  =  d / 792      multiply  both sides by  792

792*sin(5°)  =  d

So...to find out how many minutes it will take the sub to reach a depth of 200 ft we can evaluate this

200 / [ 792*sin (5°) ]  ≈ 2.9 min ≈ 2 min 54 sec 

Your question is rather unusual in the sense that the Rabbit and the March Hare are running one after another, in the same direction, instead of running towards each other where they would meet at a certain time and distance.

Average speed = Distance covered / Time

320 km / 70 = 4 4/7 hours for the Rabbit to reach the Castle.

320 km / 20 = 16 hours for the March Hare to reach the Castle !!.

Since the March Hare started 2 hours after the Rabbit, then it means that they will meet at the Castle 16 + 2 = 18 hours after the Rabbit started running.

You can assign any time you want when the Rabbit started running. So, for example, if it started at 1:00 pm, then it would meet the March Hare 18 hours later. Or, 1 + 18 =19 hours -12 hours=7:00 am the following morning.

If 54+9f(x)+x^2(f(x))^3=0 and f(1)=−3, find f′(1)

\(If \;\;\;54+9f(x)+x^2(f(x))^3=0 \quad and \quad f(1)=−3, \quad find\;\;f′(1)\\ let\;\;y=f(x)\\ 54+9y+x^2y^3=0\\ \text{differentiate wrt x}\\ 9y'+2xy^3+3x^2y^2y'=0\\ y'(9+3x^2y^2)+2xy^3=0\\ sub\;\;in\;\;x=1\;\;and\;\;f(1)=-3\\ f'(1)(9+3*1^2*(-3)^2)+2*1*(-3)^3=0\\ f'(1)(-18)+-54=0\\ f'(1)(-18)=54\\ f'(1)=-3 \)

I have done this with many distractions so you better check my working.

Somebody please help 

Here's a similar question: http://www.chegg.com/homework-help/submarine-traveling-90-mph-descending-angle-depression-many-chapter-5.2-problem-58e-solution-9781424089161-exc

Two right angles: 90+90=180

Angle opposite to the 90-degree angles is 150 degrees.

So, the answer is 180-150=\(\boxed{30^\circ}\)

A box and whisker plot—also called a box plot—displays the five-number summary of a set of data.

The five-number summary is the minimum, first quartile, median, third quartile, and maximum.

In a box plot, we draw a box from the first quartile to the third quartile.

Hey Maplesnowy!

Since a full circle is 360 degrees and we are missing one angle, we can write an equation to solve for that angle.

Also, a right angle is 90 degrees.

90 + 90 + 150 + x = 360

330 + x = 360

x = 30

The angle you want to solve for is 30 degrees.

I hope this helped,

Gavin.

Using the Law of Sines

x / sin (32°)  = 11/ sin(47°)

Multiply both sides by  sin(32°)

x  =  11 * sin(32°) / sin(47°)

3.   Note that  triangle ABC is a 6-8-10 right triangle

And angle BCA  =  90

And angle PCA  =  45

So....angle BCP  = 45

Then.....PC  bisects  angle BCA

Let the intersection of PC and AB  =  D

So.....in triangle ABC, since angle BCA  is bisected, we have the following relationship

BC/CA   = BD/AD

6/8  = BD/AD

3/4  = BD/AD

So....since BA  = 10.....there  are 7 equal parts of BA

BD  =  (3/7)*10   =  30/7

AD = (4/7)* 10  = 40/7

Note....sin BAC  =  6/10  =  3/5

And  sin PCA = sin 45  =  1/√2

So.....using the Law of Sines

AD/ sin PCA  = CD/ sin BAC

(40/7)/ (1/√2)  =  CD / (3/5)

CD  =  (40/7)(3/5)*√2   =  (120/35)√2  =  (24/7)√2

And using the intersecting chord theorem

BD * AD  = CD * PD

(30/7) (40/7)  =  24√2 / 7  * PD

1200/7  =  24√2 * PD

PD  =  (1200) / ( 7 * 24 * √2) 

PD  =  (25/7)√2

So....CP  = 

PD + CD  =

(25/7)√2  + (24/7)√2  =

(49/7)√2 =

7√2