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morpeh [17]
3 years ago
14

Ezekiel Elliot rushed for 68, 112, -24, 32, -14, and 127 yards in his first six football games. What was his total number of rus

hing yards in the first six games?

Mathematics
1 answer:
kondor19780726 [428]3 years ago
6 0

The solution is in the attachment

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Write an equation for the nth term of the arithmetic sequence.
tresset_1 [31]

Answer:

<h3>              \bold{a_n=6n-3}\\\\\bold{a_{50}=297} </h3>

Step-by-step explanation:

3, 9, 15, 21,  ...

9-3 = 6;   15-9 = 6;   21-15 = 6   ⇒  d = 6

first term:  a = 3

difference:  d = 6

so the formula:

                         a_n=a+d(n-1)\\\\a_n=69+6(n-1)\\\\a_n =3+6n-6\\\\\underline{a_n=6n-3}                        

and:

   a_{50}=6\cdot50-3=300-3=297

5 0
3 years ago
Solve these linear equations by Cramer's Rules Xj=det Bj / det A:
timurjin [86]

Answer:

(a)x_1=-2,x_2=1

(b)x_1=\frac{3}{4} ,x_2=-\frac{1}{2} ,x_3=\frac{1}{4}

Step-by-step explanation:

(a) For using Cramer's rule you need to find matrix A and the matrix B_j for each variable. The matrix A is formed with the coefficients of the variables in the system. The first step is to accommodate the equations, one under the other, to get A more easily.

2x_1+5x_2=1\\x_1+4x_2=2

\therefore A=\left[\begin{array}{cc}2&5\\1&4\end{array}\right]

To get B_1, replace in the matrix A the 1st column with the results of the equations:

B_1=\left[\begin{array}{cc}1&5\\2&4\end{array}\right]

To get B_2, replace in the matrix A the 2nd column with the results of the equations:

B_2=\left[\begin{array}{cc}2&1\\1&2\end{array}\right]

Apply the rule to solve x_1:

x_1=\frac{det\left(\begin{array}{cc}1&5\\2&4\end{array}\right)}{det\left(\begin{array}{cc}2&5\\1&4\end{array}\right)} =\frac{(1)(4)-(2)(5)}{(2)(4)-(1)(5)} =\frac{4-10}{8-5}=\frac{-6}{3}=-2\\x_1=-2

In the case of B2,  the determinant is going to be zero. Instead of using the rule, substitute the values ​​of the variable x_1 in one of the equations and solve for x_2:

2x_1+5x_2=1\\2(-2)+5x_2=1\\-4+5x_2=1\\5x_2=1+4\\ 5x_2=5\\x_2=1

(b) In this system, follow the same steps,ust remember B_3 is formed by replacing the 3rd column of A with the results of the equations:

2x_1+x_2 =1\\x_1+2x_2+x_3=0\\x_2+2x_3=0

\therefore A=\left[\begin{array}{ccc}2&1&0\\1&2&1\\0&1&2\end{array}\right]

B_1=\left[\begin{array}{ccc}1&1&0\\0&2&1\\0&1&2\end{array}\right]

B_2=\left[\begin{array}{ccc}2&1&0\\1&0&1\\0&0&2\end{array}\right]

B_3=\left[\begin{array}{ccc}2&1&1\\1&2&0\\0&1&0\end{array}\right]

x_1=\frac{det\left(\begin{array}{ccc}1&1&0\\0&2&1\\0&1&2\end{array}\right)}{det\left(\begin{array}{ccc}2&1&0\\1&2&1\\0&1&2\end{array}\right)} =\frac{1(2)(2)+(0)(1)(0)+(0)(1)(1)-(1)(1)(1)-(0)(1)(2)-(0)(2)(0)}{(2)(2)(2)+(1)(1)(0)+(0)(1)(1)-(2)(1)(1)-(1)(1)(2)-(0)(2)(0)}\\ x_1=\frac{4+0+0-1-0-0}{8+0+0-2-2-0} =\frac{3}{4} \\x_1=\frac{3}{4}

x_2=\frac{det\left(\begin{array}{ccc}2&1&0\\1&0&1\\0&0&2\end{array}\right)}{det\left(\begin{array}{ccc}2&1&0\\1&2&1\\0&1&2\end{array}\right)} =\frac{(2)(0)(2)+(1)(0)(0)+(0)(1)(1)-(2)(0)(1)-(1)(1)(2)-(0)(0)(0)}{4} \\x_2=\frac{0+0+0-0-2-0}{4}=\frac{-2}{4}=-\frac{1}{2}\\x_2=-\frac{1}{2}

x_3=\frac{det\left(\begin{array}{ccc}2&1&1\\1&2&0\\0&1&0\end{array}\right)}{det\left(\begin{array}{ccc}2&1&0\\1&2&1\\0&1&2\end{array}\right)}=\frac{(2)(2)(0)+(1)(1)(1)+(0)(1)(0)-(2)(1)(0)-(1)(1)(0)-(0)(2)(1)}{4} \\x_3=\frac{0+1+0-0-0-0}{4}=\frac{1}{4}\\x_3=\frac{1}{4}

6 0
3 years ago
Please help! This was due yesterday :(
Julli [10]

Answer mon

 

Step-by-step explanation:

6 0
3 years ago
What is the general form of the equation for the given circle centered at [0, 0)?
sergejj [24]

Answer:

x^2+y^2=r^2 is quation of circle whose centre is (0,0)

3 0
3 years ago
Pythagorean theorem If l = 12 cm and m = 35 cm, what is the length of n?
poizon [28]
If n is the hypotenuse, then it is 37. L can't be the hypotenuse and if m is the hypotenuse then n is 32.878
4 0
3 years ago
Read 2 more answers
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