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3 years ago

Using the Rational Zeros Theorem and synthetic division, find all the real zeros

Mathematics

1 answer:

Dahasolnce [82]3 years ago

7 0

Answer:

mnvxzdftyujvbv

Step-by-step explanation:

step 1 whufsrew56u7uikhjhjgdre56tuykhjhgfd step 2 ngffdstruihkvbcxrtryuhjknbvcdfttuhk step 3 hgyuyihjcgtyukhjbvgtdfuykjgcg there u go

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Write the solution of the inequality in set builder notation 7-3d>28

IgorC [24]

D<-7
{d | d<-7}
Hope this helps!

6 0

3 years ago

Read 2 more answers

The Barnes family has a coupon for 10% off their dinner. If their bill comes to $82.50 and they wish to tip 15%, what will they

satela [25.4K]

Answer:

Total cost= $86.25

Step-by-step explanation:

Giving the following information:

Discount= 10%

Nill= $82.5

Tip= 15%

First, we need to calculate the discounted price:

Discount price= 82.5 / 1.1= $75

Now, the tip:

Tip= 75*0.15= $11.25

Finally, the total cost:

Total cost= 75 + 11.25

Total cost= $86.25

3 0

3 years ago

In Exercises 40-43, for what value(s) of k, if any, will the systems have (a) no solution, (b) a unique solution, and (c) infini

svet-max [94.6K]

Answer:

If k = −1 then the system has no solutions.

If k = 2 then the system has infinitely many solutions.

The system cannot have unique solution.

Step-by-step explanation:

We have the following system of equations

$x - 2y +3z = 2\\x + y + z = k\\2x - y + 4z = k^2$

The augmented matrix is

$\left[\begin{array}{cccc}1&-2&3&2\\1&1&1&k\\2&-1&4&k^2\end{array}\right]$

The reduction of this matrix to row-echelon form is outlined below.

$R_2\rightarrow R_2-R_1$

$\left[\begin{array}{cccc}1&-2&3&2\\0&3&-2&k-2\\2&-1&4&k^2\end{array}\right]$

$R_3\rightarrow R_3-2R_1$

$\left[\begin{array}{cccc}1&-2&3&2\\0&3&-2&k-2\\0&3&-2&k^2-4\end{array}\right]$

$R_3\rightarrow R_3-R_2$

$\left[\begin{array}{cccc}1&-2&3&2\\0&3&-2&k-2\\0&0&0&k^2-k-2\end{array}\right]$

The last row determines, if there are solutions or not. To be consistent, we must have k such that

$k^2-k-2=0$

$\left(k+1\right)\left(k-2\right)=0\\k=-1,\:k=2$

Case k = −1:

$\left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&-1-2\\0&0&0&(-1)^2-(-1)-2\end{array}\right] \rightarrow \left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&-3\\0&0&0&-2\end{array}\right]$

If k = −1 then the last equation becomes 0 = −2 which is impossible.Therefore, the system has no solutions.

Case k = 2:

$\left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&2-2\\0&0&0&(2)^2-(2)-2\end{array}\right] \rightarrow \left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&0\\0&0&0&0\end{array}\right]$

This gives the infinite many solution.

5 0

3 years ago

Calculus help, I am trying to do this questions.. someone teach me please 10 points

Nutka1998 [239]

Answer:

Step-by-step explanation:

5 0

2 years ago

three friends are in the same algebra class. Their scores on a recent test are three consecutive odd intergers whose sum is 279.

jasenka [17]

X= 1st integer
x+2= 2nd integer
x+4= 3rd integer

Add the integers together

x + (x + 2) + (x + 4)= 279
combine like terms
3x + 6= 279
subtract 6 from both sides
3x= 273
divide both sides by 3
x= 91 first integer

Substitute x=91 to find 2nd & 3rd integers

2nd Integer
=x+2
=91+2
=93

3rd Integer
=x+4
=91+4
=95

ANSWER: The three test scores are 91, 93 and 95.

Hope this helps! :)

7 0

3 years ago