All Answers 358501 Answers

A family vacation trip took 12 hours. Of the total distance, 4/5 was in a car and 1/5 was in a boat, and the car traveled 5 times as fast as the boat. How much more time, in hours, was spent in the boat than in the car?

Call the total distance, D  ...call the boat's rate, R   and the car's rate 5R

And ........ D / Rate    = Time

So we have that

(4/5)D / [5R]  +  (1/5)D  / R  =  12

(4/25)D + (1/5)D = 12R

(9/25)D = 12R

D =  (100/3)R

So....the time in the boat was    (1/5)(100/3)R / R  =  100/15  =  20/ 3  hrs

And the time in the car was  (4/5)(100/3)R /[ 5R]  =   400 / 75  = 16/3  hrs

So....the time spent in the boat was   (20/3) - (16/3)  = 

4/3 hours more than in the car =

1 hr 20 min more

4/5D - Distance travelled by car
1/5D -Distance travelled by boat
S= Speed of boat
5S=Speed of car
Time=Distance/Speed
(4/5D)/(5S) + (1/5D) /S =12, solve for S
S =3D/100 Boat speed
15D/100 car speed
[(1/5D)/(3D/100)] =Boat time =20/3 =6 2/3 hours
[(4/5D)/(15D/100)] =Car time =16/3 =5 1/3 hours
20/3 - 16/3=4/3 =1 1/3 extra hours spent  on boat. 
Note: Regardless of the distance traveled or the speeds of the car and boat, there will always be a difference of 1 1/3 hours extra spent on the boat, given the conditions stipulated in the question.

Possibilities with two spins

(1,1)    (2,1)   (3,1)   (4,1)

(1,2)    (2,2)   (3,2)   (4,2)

(1,3)    (2/3)   (3,3)   (4,3)

(1,4)    (2,4)   (3,4)   (4,4)

Probability  of a "4" on two spins is  3/16

Probability of a "5"  on two spins is  4/16  = 1/4

There is a 1/4 probability  of landing on any number with one spin

So....the probability of landing on "1"  three times in a row is  (1/4) (1/4) (1/4)  = 1/64

The slope is    -1/2

And the point  (2,0) is on the graph

So......the equation of the line is

y = (-1/2)(x - 2)

y = (-1/2)x + 1     multiply through by 2

2y  =  -1x + 2 

1x + 2y   =   2    ⇒  A = 1, B = 2 and C  = 2  

5x - 12y = 60

y intercept  is   (0, - 5)

x intercept is  (12, 0)

The distance between these two points is

sqrt ( 5^2 + 12^2)  = sqrt (169)  =  13  units

3x - 4y= 7

8x + Ay = B

The slope of the first line is 3/4

So....the slope of the second line will be  -4/3 

So  we can write the second line as

Ay  =  -8x + B  ⇒  y  =  (-8/A)x + B/A

So......this implies that   

(-8/A)  = -4/3

A /-8  =  -3/4

A = 24/ 4   = 6

So we have that

2 =  (-4/3) (5) + B/6

2 = -20/3 + B/6

12 = -40 + B

52  = B

So  A + B =    6 + 52  =  58

Here's a graph : https://www.desmos.com/calculator/pgimsvymds

Thanks to hectictar for spotting my earlier mistake ....she always keeps me straight  !!!

They aren't really numbered, you could say dot 1 is just 1 and dot 2 is just 2

Hi ProMagma,

I have had to make some assumptions here. They are stated in my last post.

I have drawn this to scale.

\(cos\theta=\frac{2.5}{6}=\frac{5}{12}\\ \theta \approx 65.376^\circ\\ \therefore \angle AOB\approx 130.751^\circ\)

\(Area\;\;of\;\; \triangle AOB\approx 0.5*6*6*sin130.751^\circ\approx 13.636 u^2\\ Area\;\;of\;\; sector \;\;AOB\approx \frac{130.751}{360}\pi *6^2 \approx 41.077u^2\\ Area\;\;of\;\; minor \;\; segment \;\;AB\approx 41.077-13.636 = 27.441u^2 \)

This is the same as the Red region inside the intersection of the circles because both are half of the total intersection.

Now the area of the union of the 2 circles is   \(2\pi r^2 - 2*27.441 \approx 226.195 - 54.882\approx 171.313\)

Half of this area (because the rectangle is split in half) is \( \approx 85.857\)

The area of the whole rectangle is  \(20*15=300 inch^2\)

The are of half the rectangle is  \(150inch^2\)

So

\(Original\;red\;shaded\:area\;\;\approx  150 - 85.857 + 27.441 \approx 91.584 \;inch^2\)

I think that is right but my logic, and arithemetic does need to be double checked.

ARe the dots supposed to be numbered 1 to 10?

I mean, how can they equal 4 when they have no numbers attached to them ?

1. Open the compasses to a distance greater than half the distance between F and G.

2. Put the compass point on F and draw a semicircle that crosses line FG.

3. Repeat 2. with the compass point on G.

4. The two semicircles will intersect at two points, so use the straight edge to draw a line between them.  This line will be perpendicular to line FG.

. 

Note that the first question is simpler if you first rearrange the equation:

\(\sqrt{k+9}-\sqrt k=\sqrt3\)

\(\sqrt{k+9}=\sqrt k + \sqrt3\)

Now square both sides:

\(k+9=k+3+2\sqrt{3k}\rightarrow 3=\sqrt{3k}\)

Square both sides:

\(9=3k\text{ so: }k=3\)

.

I think there is still something missing because you have not told us how big the overlap is.

I think to get a number answer we would need that, or something that would allow us to find it..

I suppose I can assume there is a 1.5inch border around the circles which would mean that the circles are tight inside a rectangle  12 inches high and 17inches long 

You link doesn't work.

Both Circles are the same size, the box is made of perfect 90° angles,

Here is an update, left something out

2400 x 85% =2,040 Kg -the weight of cement,sand and stone.
1/[1+2+5] x 2,040 =255 kg is cement.
2/[1+2+5] x 2,040 =510 kg is sand
5/[1+2+5] x 2,040 =1,275 kg is stone

Is exactly what I got....

The file doesn't work

For the first one, it would have been easier computationally speaking if you rewrote \(\left(\sqrt[3]{27}\right)^4=3^4=3*3*3*3=9*9=81\)

Here's probably the easist method of all!
 

Both parts took me approximately the same amount of time (30 minutes).

27^(4/3) =

(27^4)^1/3 =

(531,441)^1/3 =81 - It is 27 to the 4th power and then take the cube root.

4^3/2 =

(4^3)^1/2 =

(64)^1/2 = 8 - it is the cube of 4 and then take the square root.

Solve for a:
w = 3 (2 a + b) - 4

w = 3 (2 a + b) - 4 is equivalent to 3 (2 a + b) - 4 = w:
3 (2 a + b) - 4 = w

Expand out terms of the left hand side:
-4 + 6 a + 3 b = w

Subtract 3 b - 4 from both sides:
6 a = 4 - 3 b + w

Divide both sides by 6:
a = 2/3 - b/2 + w/6

2400 x 85% =2,040 Kg -the weight of cement,sand and stone.
1/[1+2+5] x 2,040 =255 kg is cement.
2/[1+2+5] x 2,040 =510 kg is sand
5/[1+2+5] x 2,040 =1,275 kg is stone

Rearrange  w  =  3(2a + b ) - 4  to make  a  the subject.

w  =  3(2a + b) - 4

                                    Add  4  to both sides of the equation.

w + 4  =  3(2a + b)

                                    Divide both sides of the equation by  3 .

(w + 4)/3  =  2a + b

                                    Subtract  b  from both sides.

(w + 4)/3 - b  =  2a

                                           Divide both sides by  2 .

[ (w + 4)/3 - b ] / 2  =  a

a  =  [ (w + 4)/3 - b ] / 2