geno3141

1/9 + 2/3  =

The denominators are 9 and 3.

Make both of the denominators 9 by multiplying the second fraction by 3/3.

1/9 + (2/3)(3/3)

=  1/9 + 6/9

=  (1 + 66)/9

=  7/9

4(t - 5)  =  8

Use the distributive law:

4(t) - 4(5)  =  8

4t - 20  =  8

Add 20 to both sides:

4t  =  28

Divide both sides by 4:

4t/4  =  28/4

t  =  7

The mode is the number used most often. Each of the two numbers is used just once; this data set has no mode.

The Richter scale is an exponential scale with a base of 10.

Thus, a magnitude of 9.5 means 10^9.5 and a magnitude of 7.9 means 10^7.9.

Comparing these:  10^9.5 / 10^7.9  =  10^1.6  which is about 39,8 or 40.   (Subtract exponents when you divide.)

So the Valdivia earthquake was about 40 times stronger than the Madrid earthquake.

(To find how many times larger one number is than another, you should divide.)

Translating won't change the size of the figure.

The length of the line segment from (-4,2) to (-1,2) is 3.  (These points are on a horizontal line and the distance from -4 to -1 is 3.)

So the length of the image is also 3.

Let  A = e^10x  and let  B = e^-10x.

Ignoring the outside exponent for a while,  the problem becomes [ (A - B) / 2 ] - [ (A + B) / 2 ]

   =  [A/2 - B/2] - [A/2 + B/2]  =  A/2 - B/2 - A/2 - B/2  =  - 2B/2  =  - B     which is     -e^-10x

Bringing the exponent back into the problem:  (-e^-10x)¹⁰  =  e^-100x

because a negative number to an even exponent is positive, and you need to multiply the exponents.

There are 100 choices for the first cube and 24 choices for this cube:  (100 x 24).

There are 99 choices for the second cube and 24 choices for this cube:  (99 x 24).

There are 98 choices for the third cube and 24 choices for this cube:  (98 x 24).

...

For the last cube:  there is 1 choice for the one-hundredth  cube and 24 choices for this coube:  (1 x 24).

Putting these together:  (100 x 24) x (99 x 24) x (98 x 24) x ... x (2 x 24) x (1 x 24)

                               =     (100 x 99 x 98 x ... x 2 x 1) x (24 x 24 x 24 x ... x 24 x 24)

                               =      100! x 24¹⁰⁰.

a)  To show that v₀ sqrt(2) / 32 x [sin(2A) - cos(2A) - 1]  equals  v₀ sqrt(2) / 16 x cos(A) x [sin(A) - cos(A)]

      --->              v₀ sqrt(2) / 16 x (1/2)[sin(2A) - cos(2A) - 1]  equals v₀ sqrt(2) / 16 x cos(A) x [sin(A) - cos(A)]

      --->              (1/2)[sin(2A) - cos(2A) - 1]  equals    cos(A) x [sin(A) - cos(A)]          

      --->  But:  sin(2A)  =  2sin(A)cos(A)   and  cos(2A)  =  2cos²(A) - 1

      --->  Left side:          

                       (1/2)[2sin(A)cos(A) - (2cos²(A) - 1) - 1]

                                =  (1/2)[2sin(A)cos(A) - 2cos²(A) + 1 - 1]

                                =  (1/2)[2sin(A)cos(A) - 2cos²(A)]

                                =  sin(A)cos(A) - cos²(A)

                                =  cos(A)[sin(A) - cos(A)    <--- which equals the right side.

b)  sin(2A) + cos(2A )  =  0

     --->   sin(2A)  =  -cos(2A)             (now divide both sides by cos(2A):

     --->   sin(2A) / cos(2A)  =  -1

     --->   tan(2A)  =  -1

     --->  2A  =  135°

     --->     A  =  67.5°

c)  Place these values into the equation, and you will find the maximum distance is 0.586 feet (approx).

To have two boys on the team you must choose one of the 8 boys (out of 19 total persons) followed by choosing one of the 7 remaining boys (out of 18 remaining persons):

The probability of getting two boys:  8/19 x 7/18  =  56/342.

To have two girls on the team you must choose one of the 11 girls (out of 19 total persons) followed by choosing one of the 10 remaining girls (out of 18 remaining persons);

The probability of getting two girls:  11/19 x 10/18  =  110/342,

The probability of getting either two boys or two girls is found by adding these independent probabilities together:

     56/342 + 110/342  =  166/342  =  83/171.

You would need a z-score of approximately 0.842 to separate the top 20% from the bottom 89%.

0.842  =  (z - 85)/5     --->     z > 89