Answer:
7.6 seconds
Step-by-step explanation:
To solve this problem we can use the following equation:
S = So + 113*t + at^2/2
Where S is the final position, So is the inicial position, Vo is the inicial speed, a is the acceleration and t is the time.
In this problem, we have that S = 0, So = 65, Vo = 113 and a = -32 ft/s2 (The acceleration of gravity)
So we have that:
0 = 65 + 113t - 16t2
16t2 - 113t - 65 = 0
Using Bhaskara's formula, we have:
Delta = 113^2 + 4*65*16 = 16929
sqrt(Delta) = 130.11
t = (113 + 130.11) / (2*16) = 7.597 seconds
Rounding to nearest tenth, we have t = 7.6 seconds
Probability that you will draw exactly 1 king
P(draw both jacks) = 1/3*1/3 = 1/9P(no jacks) = 2/3*2/3 = 4/9P(1 jack) = 1 - 1/9 - 4/9 = 4/9
Each cube has a volume of (1/4*1/4*1/4), or (1/64). Next, you take 2304 and divide by one sixteenth. 2304/(1/16)=36864
Answer: 138
Step-by-step explanation:
by the corresponding angles theorem.
First, we have to make sure that the number of columns in the first matrix is equal to the number of rows in the second matrix.
![\left[\begin{array}{cc}1&-3&2&0\\\end{array}\right] * \left[\begin{array}{ccc}2&3&4\\1&2&3\end{array}\right]](https://tex.z-dn.net/?f=%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D1%26-3%262%260%5C%5C%5Cend%7Barray%7D%5Cright%5D%20%2A%20%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D2%263%264%5C%5C1%262%263%5Cend%7Barray%7D%5Cright%5D%20)
Since this is true, we can continue to solve the problem.
To multiply two matrices, multiply each row element in the first matrix by each column element in the second matrix. For example:
1*2 = 2
-3*1=-3
Then we add them to get our new matrix element.
-3+2=
-1Then we move to the next column of the second matrix.
1*3=3
-3*2=-6
-6+3=
-3Then the final column of the second matrix.
1*4=4
-3*3=-9
-9+4=-5
Our matrix so far:
![\left[\begin{array}{ccc}-1&-3&-5\\x&x&x\end{array}\right]](https://tex.z-dn.net/?f=%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-1%26-3%26-5%5C%5Cx%26x%26x%5Cend%7Barray%7D%5Cright%5D%20)
We do the same for the bottom row of the first matrix.
<em>First Column</em>
2*2=4
0*1=0
4+0=
4<em>Second Column
</em>2*3=6
0*2=0
6+0=
6
<em>Third Column</em>
2*4=8
0*3=0
8+0=
8Our final matrix is:
![\left[\begin{array}{ccc}-1&-3&-5\\4&6&8\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-1%26-3%26-5%5C%5C4%266%268%5Cend%7Barray%7D%5Cright%5D)
:)