The ends (tails) of the graph of a 6th degree polynomial will ________________

I need help is it A B C or D

lana [24]

I'm thinking b
or c is what i think hope i could help ^-^

3 0

3 years ago

Read 2 more answers

PLZZZZZZZZZ HELP!!!!!!!!!!!!! Ty!!!!!!!!!! ANYONE PLZ HELP!

marishachu [46]

Answer: its the first answer choice

ANSWER:x^3 -9x^2+25x

7 0

3 years ago

Jessica got a prepaid debit card with $20 on it. For her first purchase with the card, she bought some bulk ribbon at a craft st

faltersainse [42]

I don’t know but can you please help me out

5 0

2 years ago

For each pair of points below, think about the line that joins them. For which pairs is the line parallel to the y-axis? Circle

gogolik [260]

Answer:

Option B.

Step-by-step explanation:

If two lines are parallel then their slopes are always same.

Following this rule we can find the slope by the given pairs of coordinates of the options.

If the slope of the line is same as the slope of y axis then the line passing through these points will be parallel to the y axis.

Slope of y - axis = ∞

Option A). Slope = $\frac{y-y'}{x-x'}$

= $\frac{24-8.5}{3.22-3.2}$

= $\frac{15.5}{0.02}$

= 775

Therefore, line passing through points (3.2, 8.5) and (3.22, 24) is not parallel to y axis.

Option B). Slope of the line passing through $(\frac{40}{3},\frac{14}{3})$ and $(\frac{40}{3},7)$ will be

= $\frac{\frac{14}{3}-7}{\frac{40}{3}-\frac{40}{3}}$

= ∞

Therefore, line passing though these points is parallel to the y axis.

Option C). Slope of the line passing through $(2.9,5.4)$ and (7.2, 5.4)

= $\frac{5.4-5.4}{7.2-2.9}$

= 0

Therefore, slope of this line is not equal to the slope of y axis.

Option B. is the answer.

5 0

3 years ago

1)A System of equations is shown below . What is the solution to the system of equations? 5x+2y=-15 2x-2y=-6

meriva

Answer:

x= -3 and y= 0

Step-by-step explanation:

5x+2y=-15

2x-2y=-6

7x =-21

x= -3

Putting value of x in equation 1

5(-3) +2y=-15

-15+2y= -15

2y= 0

y= 0

This can be solved with the help of matrices

In matrix form the above equations can be written in the form

$\left[\begin{array}{ccc}5&2\\2&-2\/\end{array}\right]$ $\left[\begin{array}{ccc}x\\y\\\end{array}\right]$ = $\left[\begin{array}{ccc}-15\\-6\\\end{array}\right]$

Let

$\left[\begin{array}{ccc}5&2\\2&-2\/\end{array}\right]$ = A $\left[\begin{array}{ccc}x\\y\\\end{array}\right]$ = X and $\left[\begin{array}{ccc}-15\\-6\\\end{array}\right]$ = B

Then AX= B

or X= A⁻¹ B

where A⁻¹= adj A/ ║A║ where mod A≠ 0

adj A= $\left[\begin{array}{ccc}-2&-2\\-2&5\/\end{array}\right]$

║A║= ( 5*-2- 2*2)= -10-4= -14≠0

X= A⁻¹ B

$\left[\begin{array}{ccc}x\\y\\\end{array}\right]$ =- 1/14 $\left[\begin{array}{ccc}-2&-2\\-2&5\/\end{array}\right]$ $\left[\begin{array}{ccc}-15\\-6\\\end{array}\right]$

$\left[\begin{array}{ccc}x\\y\\\end{array}\right]$ =- 1/14 $\left[\begin{array}{ccc}-2*-15&+ -2*-6\\-2*-15&+ 5*-6\\\end{array}\right]$

$\left[\begin{array}{ccc}x\\y\\\end{array}\right]$ =- 1/14 $\left[\begin{array}{ccc} 30&+12\\30&+-30\\\end{array}\right]$

$\left[\begin{array}{ccc}x\\y\\\end{array}\right]$ =- 1/14 $\left[\begin{array}{ccc}42\\0\\\end{array}\right]$

$\left[\begin{array}{ccc}x\\y\\\end{array}\right]$ = $\left[\begin{array}{ccc}-42/14\\0/-14\\\end{array}\right]$

$\left[\begin{array}{ccc}x\\y\\\end{array}\right]$ = $\left[\begin{array}{ccc}-3\\0\\\end{array}\right]$

From here x= -3 and y= 0

Solution Set = [(-3,0)]

3 0

3 years ago

The ends (tails) of the graph of a 6th degree polynomial will _________________.