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3 years ago

What is the degree of the polynomial? Can someone please check my work?

Mathematics

1 answer:

worty [1.4K]3 years ago

7 0

Yes I agree with that because you combined the terms in the monomial correctly and the degree of the monomial was greater then the degree of any other mono/binomials in the polynomial

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Factor completely 3X^2 -25

Dominik [7]

In order to factor a polynomial, you have to find its roots

$x_1,\ x_2,\ldots,x_n$

So that you can factor it as

$p(x)=(x-x_1)(x-x_2)\cdots(x-x_n)$

So, in this case, we're looking for the solutions of

$3x^2-25=0 \iff 3x^2=25 \iff x^2 = \dfrac{25}{3} \iff x=\pm\sqrt{\dfrac{25}{3}}=\pm\dfrac{5}{\sqrt{3}}$

So, the polynomial can be written as

$3x^2-25 = \left(x-\dfrac{5}{\sqrt{3}}\right)\left(x+\dfrac{5}{\sqrt{3}}\right)$

5 0

3 years ago

8. A sequence can be generated by using 2n - 11. What are the first four terms in<br> the sequence?

Lapatulllka [165]

Answer:

$\bold{a_1=-9\,,\quad a_2=-7\,,\quad a_2=-5\,,\quad a_4=-3}$

Step-by-step explanation:

$a_n=2n-11\\\\a_1=2(1)-11=2-11=-9\\\\a_2=2(2)-11=4-11=-7\\\\a_3=2(3)-11=6-11=-5\\\\a_4=2(4)-11=8-11=-3$

7 0

3 years ago

Why does 2+2=ddddddddddddddddddddddddddddddddddd

Dahasolnce [82]

Answer:

Because 2+2 equals 4dd

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

Explain how to use the combine place values strategy to find 223 - 119

ololo11 [35]

the anwser is 104 because u subrtacted

8 0

3 years ago

For the following telescoping series, find a formula for the nth term of the sequence of partial sums {Sn}. Then evaluate limn→[

Ivenika [448]

Answer:

The following are the solution to the given points:

Step-by-step explanation:

Given value:

$1) \sum ^{\infty}_{k = 1} \frac{1}{k+1} - \frac{1}{k+2}\\\\2) \sum ^{\infty}_{k = 1} \frac{1}{(k+6)(k+7)}$

Solve point 1 that is $\sum ^{\infty}_{k = 1} \frac{1}{k+1} - \frac{1}{k+2}\\\\$ :

when,

$k= 1 \to s_1 = \frac{1}{1+1} - \frac{1}{1+2}\\\\$

$= \frac{1}{2} - \frac{1}{3}\\\\$

$k= 2 \to s_2 = \frac{1}{2+1} - \frac{1}{2+2}\\\\$

$= \frac{1}{3} - \frac{1}{4}\\\\$

$k= 3 \to s_3 = \frac{1}{3+1} - \frac{1}{3+2}\\\\$

$= \frac{1}{4} - \frac{1}{5}\\\\$

$k= n^ \to s_n = \frac{1}{n+1} - \frac{1}{n+2}\\\\$

Calculate the sum $(S=s_1+s_2+s_3+......+s_n)$

$S=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+.....\frac{1}{n+1}-\frac{1}{n+2}\\\\$

$=\frac{1}{2}-\frac{1}{5}+\frac{1}{n+1}-\frac{1}{n+2}\\\\$

When $s_n \ \ dt_{n \to 0}$

$=\frac{1}{2}-\frac{1}{5}+\frac{1}{0+1}-\frac{1}{0+2}\\\\=\frac{1}{2}-\frac{1}{5}+\frac{1}{1}-\frac{1}{2}\\\\= 1 -\frac{1}{5}\\\\= \frac{5-1}{5}\\\\= \frac{4}{5}\\\\$

$\boxed{\text{In point 1:} \sum ^{\infty}_{k = 1} \frac{1}{k+1} - \frac{1}{k+2} =\frac{4}{5}}$

In point 2: $\sum ^{\infty}_{k = 1} \frac{1}{(k+6)(k+7)}$

when,

$k= 1 \to s_1 = \frac{1}{(1+6)(1+7)}\\\\$

$= \frac{1}{7 \times 8}\\\\= \frac{1}{56}$

$k= 2 \to s_1 = \frac{1}{(2+6)(2+7)}\\\\$

$= \frac{1}{8 \times 9}\\\\= \frac{1}{72}$

$k= 3 \to s_1 = \frac{1}{(3+6)(3+7)}\\\\$

$= \frac{1}{9 \times 10} \\\\ = \frac{1}{90}\\\\$

$k= n^ \to s_n = \frac{1}{(n+6)(n+7)}\\\\$

calculate the sum: $S= s_1+s_2+s_3+s_n\\$

$S= \frac{1}{56}+\frac{1}{72}+\frac{1}{90}....+\frac{1}{(n+6)(n+7)}\\\\$

when $s_n \ \ dt_{n \to 0}$

$S= \frac{1}{56}+\frac{1}{72}+\frac{1}{90}....+\frac{1}{(0+6)(0+7)}\\\\= \frac{1}{56}+\frac{1}{72}+\frac{1}{90}....+\frac{1}{6 \times 7}\\\\= \frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{42}\\\\=\frac{45+35+28+60}{2520}\\\\=\frac{168}{2520}\\\\=0.066$

$\boxed{\text{In point 2:} \sum ^{\infty}_{k = 1} \frac{1}{(n+6)(n+7)} = 0.066}$

8 0

3 years ago