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3 years ago

Y^8+6y^4+8/y^8+5y^4+6

1 answer:

7 0

Equation: <span>Y^8+6y^4+8/y^8+5y^4+6

y^8 + (6y^4 + 5y^4) + 8/y^8 + 6

y^8 + 11y^4 + 8/y^8 + 3

Final Answer:

y</span>⁸ + 11y⁷ + 8/y⁸ + 6

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The tabel shows the amount of miles of Cheryl travels 4 each day.About how many miles did she travel during the 4 days?

OLga [1]

4 times 4=16 So in four days, Cheyrl traveled 16 miles.

8 0

2 years ago

You have a $3175 student loan. Suppose you pay the loan off in two and a half years (30 months). Estimate your monthly payment b

finlep [7]

Answer:

$106

Step-by-step explanation:

The formula given for Monthly payment of a loan =

P × [ r (1 + r)/(1 + r)^n - 1

Where

r = interest rate

n = number of monthly payments

P = Present value of the loan

From the question,

r = interest rate, we were told to ignore hence, r = 0

P = $3,175

n = 30

Hence,

Amount to be paid monthly = P/n

= $3175/30

= $105.83

Approximately to the nearest dollars

= $106

6 0

3 years ago

4. Name the first five terms of the arithmetic sequence. a1 = -35, d=4

eduard

Answer:

Step-by-step explanation:

We first have to write the equation for the sequence, then finding the first five terms will be easy. Follow the formatting:

$a_n=a_1+d(n-1)$ and we are given enough info to fill in:

$a_n=-35+4(n-1)$ and

$a_n=-35+4n-4$ and

$a_n=-39+4n$ or in linear format:

$a_n=4n-39$ where n is the position of the number in the sequence. We already know the first term is -35.

The second term:

$a_2=4(2)-39$ so

$a_2=8-39$ and

$a_2=-31$

The third term:

$a_3=4(3)-39$ and

$a_3=12-39$ so

$a_3=-27$ and we could go on like this forever, but the nice thing about this is when we know the difference all we have to do is add it to each number to get to the next number.

That means that the fourth term will be -27 + 4 which is -23.

The fifth term then will be -23 + 4 which is -19. You can check yourself by filling in a 5 for n in the equation and solving:

$a_5=4(5)-39$ and

$a_5=20-39$ so

$a_5=-19$

7 0

2 years ago

The system of equations may have a unique solution, an infinite number of solutions, or no solution. Use matrices to find the ge

Leno4ka [110]

Answer:

Infinite number of solutions.

Step-by-step explanation:

We are given system of equations

$5x+4y+5z=-1$

$x+y+2z=1$

$2x+y-z=-3$

Firs we find determinant of system of equations

Let a matrix A= $\left[\begin{array}{ccc}5&4&5\\1&1&2\\2&1&-1\end{array}\right]$ and B= $\left[\begin{array}{ccc}-1\\1\\-3\end{array}\right]$

$\mid A\mid=\begin{vmatrix}5&4&5\\1&1&2\\2&1&-1\end{vmatrix}$

$\mid A\mid=5(-1-2)-4(-1-4)+5(1-2)=-15+20-5=0$

Determinant of given system of equation is zero therefore, the general solution of system of equation is many solution or no solution.

We are finding rank of matrix

Apply $R_1\rightarrow R_1-4R_2$ and $R_3\rightarrow R_3-2R_2$

$\left[\begin{array}{ccc}1&0&1\\1&1&2\\0&-1&-3\end{array}\right]$ : $\left[\begin{array}{ccc}-5\\1\\-5\end{array}\right]$

Apply $R_2\rightarrow R_2-R_1$

$\left[\begin{array}{ccc}1&0&1\\0&1&1\\0&-1&-3\end{array}\right]$ : $\left[\begin{array}{ccc}-5\\6\\-5\end{array}\right]$

Apply $R_3\rightarrow R_3+R_2$

$\left[\begin{array}{ccc}1&0&1\\0&1&1\\0&0&-2\end{array}\right]$ : $\left[\begin{array}{ccc}-5\\6\\1\end{array}\right]$

Apply $R_3\rightarrow- \frac{1}{2}$ and $R_2\rightarrow R_2-R_3$

$\left[\begin{array}{ccc}1&0&1\\0&1&0\\0&0&1\end{array}\right]$ : $\left[\begin{array}{ccc}-5\\\frac{13}{2}\\-\frac{1}{2}\end{array}\right]$

Apply $R_1\rightarrow R_1-R_3$

$\left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right]$ : $\left[\begin{array}{ccc}-\frac{9}{2}\\\frac{13}{2}\\-\frac{1}{2}\end{array}\right]$

Rank of matrix A and B are equal.Therefore, matrix A has infinite number of solutions.

Therefore, rank of matrix is equal to rank of B.

4 0

3 years ago

You start driving north for 21 miles, turn right, and drive east for another 20 miles.

Wewaii [24]

Answer:

Step-by-step explanation:

add 21 + 20

hope this helps :)

8 0

3 years ago