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How many 75 ml bottles can be filled by 7 1/2 litres

First we transform all the units into litres: 

We know that milli means "one thousandth", i.e. 1/1000. Therefore 75 mL = 75L * 1/1000 = 75/1000 L = 0,075L

Also:

$$7 \frac{1}{2}L= 7L + \frac{1}{2}L =\frac{14}{2}L + \frac{1}{2}L = \frac{15}{2}L$$

So, how many times can you pour 75 mL in order to fill 7 1/2 L ?

Remember that when we divide one fraction by another, we multiply the first one by the multiplicative inverse of the last one, we get:

$$\frac{15}{2}L/\frac{75}{1000}L=\frac{15}{2}/\frac{75}{1000}= \frac{15}{2}*\frac{1000}{75}= 100$$

We see that the units cancels out and we get a dimentionless quantity, which is what is expected in this case.

Thank you so much!!!

2

one times one equals one.

plus one is two.

sin = y / r      cosec = r / y  

cos = x / r      sec = r / x

tan = y / x       cot  = x / y

Here, we know y = 3 and r = 7  so...using the Pythagorean Theorem, x  = √(7^2 - 3^2)  = √(49 - 9) = √40 = 2√10

Knowing these values....can you finish this one ???

Assuming there is no static friction (!) then the resistive force of the table on the upper mass is 0.6*5*9.8 Newtons.

The force on the lower mass is 5*9.8 Newtons.

The net force on the system is therefore 5*9.8 - 0.6*5*9.8 Newtons, or 0.4*5*9.8 Newtons.

The total mass of the system is 10kg, so, from Newton's second law of motion, the acceleration is net force/mass:

$${\mathtt{acceleration}} = {\frac{{\mathtt{0.4}}{\mathtt{\,\times\,}}{\mathtt{5}}{\mathtt{\,\times\,}}{\mathtt{9.8}}}{{\mathtt{10}}}} \Rightarrow {\mathtt{acceleration}} = {\mathtt{1.96}}$$

acceleration = 1.96 m/s²

.

I'm assuming you have

√2 x √3  ???

This just = √6

Here's a couple of sites that can help you with this topic....

Use this....

rads x  180/ pi  =  degrees....

For instance

2pi/3  x  180/ pi  = (2/3) x 180  = 120 degrees   (notice that the "pi's" will cancel...)

Since the cosine function only has a range from -1 to 1, inclusive......this is undefined...to put it in another way.......there is no angle whose cosine is equal to 23/9.....

I assume you might want to find the roots (zeros) of this??...if so, we have

2x^2 - 10x - 12 = 0     divide through by 2

x^2 - 5x - 6  = 0   factor

(x - 6) ( x + 1) = 0    and setting each factor to 0, we find that the roots are  x = -1  and  x = 6

3cos(x)/sin(x) can be simplified to 3cot(x)

TY

These diagonals bisect each other...so the total area is given by

(17)(28)(sin(67) + sin(113)) = about 876.32 sq units

I'm assuming that you're talking about just the positive integers......

This is just equal to the sum of the first 100 integers minus the sum of the first 9 integers, and is given by

(100)(101)/2 - (9)(10)/2 =

5050 - 45  =

5005

Since opposite angles of a paralleogram are supplemental, the other included angle between the sides is given by  (180 - 36)  = 144°

And the length of the longer diagonal - call it "D" -  will be opposite this angle......and we can find this using the Law of Cosines...so we have

D^2 = 30^2 + 20^2 - 2(20)(30)cos(144)

D^2 = 2270.82039325   take the square root of this

D = about 47.65 units

A. Since the diagonals of a parallelogram bisect each other, we can use the Law of Cosines to find one of the intersecting angles of the diagonals.....

So we have

!0^2 = 14^2 + 9^2 - 2(14)(9)cosΘ   simplify

100 = 196 + 81 - 252 cosΘ

cosΘ = (100-196 - 81) / (-252) = 0.7023809523809524

And, using the cosine inverse to find Θ, we have

cos^-1 ( 0.7023809523809524) = about 45.381658347154°  ...  and this is the smaller angle

B. The larger angle of the intersecting diagonals is given by (180 - 45.381658347154) = 134.618341652846°

And the total area of the parallelogram is given by  (14)(9)(sin 45.381658347154 + sin 134.618341652846) = 179.373911146416 sq cm

C. Using the Law of Cosines  we can find the length of the other side...call this "S"

S^2 = 14^2 + 9^2 - 2(14)(9)cos(134.618341652846)

S^2 = 454.000000000012  ....  take the square root of both sides

S = about 21.3 cm

I was thinking DavidQD, that old percent question died around about the Chrismas before last.

It went round in circles just like 10+9 does now.  Do you remember what the exact question was.   

Thank you very much :)

Okay - I said I would be back and here I am :)

If an impact sprinkler pumps 9.425 litres per minute over a circular lawn with perimeter of 37.7 metres, and the water is evenly distributed, how long do you need to water, to give the entire lawn 10mm of water?

First you need to know that  a container  1cm³  will hold 1ml of water.

So I will change all  the units to cm and ml.

the question has become

If an impact sprinkler pumps 9425 mL per minute over a circular lawn with perimeter of 3770 cm, and the water is evenly distributed, how long do you need to water, to give the entire lawn 1cm of water?

Now the perimeter is the circumference of the circle which is   $$C=2\pi r$$

so

$$\\3770=2\pi r\\
$divide both sides by 2\pi\\
r=\frac{3770}{2\pi}$$

I would prefer to keep r in this form until the end but you may prefer to get a figure for this.

$${\frac{{\mathtt{3\,770}}}{\left({\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{\pi}}\right)}} = {\mathtt{600.014\: \!135\: \!456\: \!445\: \!415\: \!8}}$$     

So the radius of the cirlce is very close to 600cm     (that is 6m which sounds reasonable for a sprinkler)

radius=600cm

Now, the area of the circle is    $$A=\pi r^2$$

$$\\A=\pi * 600^2 \quad cm^2\\
A=360000\pi\quad cm^2\\$$

$${\mathtt{360\,000}}{\mathtt{\,\times\,}}{\mathtt{\pi}} = {\mathtt{1\,130\,973.355\: \!292\: \!325\: \!565\: \!846\: \!6}}$$      

$$\\A=1130973\quad cm^2$$

Now think of this like a very shallow circular pond.  

The area of the base is 1130973 cm² and the water is 1 cm high.

So that is a volume of 1130973 x 1 cm³  = 1130973 cm³ 

And a pond this size will hold 1130973 mL

So NOW the question is redued to

An impact sprinkler pumps 9425 mL per minute.   How long will it take to pump out 1130973 mL?

So we have            

$$\frac{9425mL}{minute},\qquad 1130973\;ml,\qquad $and we want to know $\qquad ?\;minutes$$

I have my own pet method of dealing with rates and if you learn it then you can work out really complicated rates questions easily.

I want the answer to be minutes so I want to start with minutes on the top of the fraction.

if the sprinkler pumps 9425ml / minute THEN it is also true to say that in 1 minute it pumps 9425ml.

SO it is valid to turn the fraction upside down 

so I have      $$\frac{1 \;minute}{9425ml}$$     if I multiply this by  1130973 ml    the ml will cancel out leaving only minutes which is the answer that you want.

i.e.

$$\frac{1 \;minute}{9425ml}\times \frac{1130973\;ml}{1}=\frac{11310}{9425}\;minutes$$

$${\frac{{\mathtt{1\,130\,973}}}{{\mathtt{9\,425}}}} = {\mathtt{119.997\: \!135\: \!278\: \!514\: \!588\: \!9}}$$

So it will take 120 minutes which is 2 hours

Now let me know if you understand.  Maybe I can explain individual bits better.  

This one is a little trcky......

Call the longer diagonal D1 and the shorter one D2

And the area of the rhombus = (1/2) the product of the diagonals...so we have

250 = (1/2) (D1)(D2)    →  (D1)(D2) = 500   → D2 = 500/D1

And adjacent angles in  a rhombus are supplemental....so, the other angle is = (108- 37.25) = 142.75°

And the greater angle lies opposite its respective diagonal, D1...and the lesser angle lies opposite the shorter diagonal, D2  ...and the diagonals bisect both of these angles

So...using the Law of Sines, we have

sin(37.25/2) / (1/2)(D2)   =  sin(142.75/2)/(1/2)D1   and substituting for D2, we have....

sin(37.25/2) / (1/2)(500/D1)   =  sin(142.75/2)/(1/2)D1  simplify...

(D1)sin(18.625) / 500 = sin(71.375) / D1

(D1)^2  = 500sin(71.375)/sin(18.625) = 1483.5782595004849666   take the square root of both sides

D1 = about 38.5   and D2  = 500/D1 = 500/38.5 = about 12.99

And we can find the side length - S - using the Pythagorean Theorem....(1/2) of the length of each diagonal will be the "leg"  lengths, and the hypoteneuse will be the side length

So we have

S = √(19.25^2 + 6.495^2) = S = about 20.3...and that's the side length...!!!

Since the sides of a rhombus are equal, then one side must be = 10

And the diagonals bisect each other and are at right angles to each other...thus, we can use the Pythagorean Theorem to find the length (D) of the other diagonal.....the side length is the hypoteneuse and (1/2) the length of the known diagonal is a leg...so we have....

(1/2)D = √(10^2 - 6^2)  = √(100-36)  = √64 = 8

And we just need to double this to find the length of the other diagonal = 16

a. We can use the Law of Cosines to find the larger angle "Θ" between the sides...we have

160^2  = 100^2 + 100^2 - 2(100)(100)cosΘ  simplify

cosΘ = 5600/(-2*100^2) = -7/25  ... find Θ by using the cosine inverse..

cos^-1(-7/25)  = about 106.26°

And since adjacent angles in a rhombus are supplementary, the smaller angle between the sides = (180-106.26)= 73.74°

b. We can again use the Law of Cosines to find the length of the smaller diagonal.....call it "D"....so we have

D^2  = 100^2 + 100^2 - 2(100)(100)cos(73.74) = 14400.06859844   .. take the square root of this

D = about 120

c. The area is given by (1/2) the product of the diagonals  = (1/2)(160)(120) = 9600 sq units

We can use the Law of Cosines to find BC, we have

BC^2  = AC^2 + AB^2 - 2(AC)*AB)cos(BAC)

BC^2  = 100 + 49  - 2(10)(7)cos(36.07)

BC^2  = 35.83824141348     ...take the positive square root of both sides

BC = about 5.99

Correction : 37°25' sorry :D

Sorry I only just saw your reply.

I will go home soon and explain better

It will be easier when I am on my computer. :)