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N76 [4]
3 years ago
5

What is the sum of the last four terms of the series 7+9+11+...+21

Mathematics
1 answer:
Doss [256]3 years ago
7 0
<span>7 - 9 = 2 9 - 11 = 2 You should see that the series just increases by +2 therefore you can figure out what the last four terms are and then add them up one of the last terms given is 21</span>
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3 years ago
Solve x^3-7x^2+7x+15​
ruslelena [56]

Step-by-step explanation:

\underline{\textsf{Given:}}

Given:

\mathsf{Polynomial\;is\;x^3+7x^2+7x-15}Polynomialisx

3

+7x

2

+7x−15

\underline{\textsf{To find:}}

To find:

\mathsf{Factors\;of\;x^3+7x^2+7x-15}Factorsofx

3

+7x

2

+7x−15

\underline{\textsf{Solution:}}

Solution:

\textsf{Factor theorem:}Factor theorem:

\boxed{\mathsf{(x-a)\;is\;a\;factor\;P(x)\;\iff\;P(a)=0}}

(x−a)isafactorP(x)⟺P(a)=0

\mathsf{Let\;P(x)=x^3+7x^2+7x-15}LetP(x)=x

3

+7x

2

+7x−15

\mathsf{Sum\;of\;the\;coefficients=1+7+7-15=0}Sumofthecoefficients=1+7+7−15=0

\therefore\mathsf{(x-1)\;is\;a\;factor\;of\;P(x)}∴(x−1)isafactorofP(x)

\mathsf{When\;x=-3}Whenx=−3

\mathsf{P(-3)=(-3)^3+7(-3)^2+7(-3)-15}P(−3)=(−3)

3

+7(−3)

2

+7(−3)−15

\mathsf{P(-3)=-27+63-21-15}P(−3)=−27+63−21−15

\mathsf{P(-3)=63-63}P(−3)=63−63

\mathsf{P(-3)=0}P(−3)=0

\therefore\mathsf{(x+3)\;is\;a\;factor}∴(x+3)isafactor

\mathsf{When\;x=-5}Whenx=−5

\mathsf{P(-5)=(-5)^3+7(-5)^2+7(-5)-15}P(−5)=(−5)

3

+7(−5)

2

+7(−5)−15

\mathsf{P(-5)=-125+175-35-15}P(−5)=−125+175−35−15

\mathsf{P(-5)=175-175}P(−5)=175−175

\mathsf{P(-5)=0}P(−5)=0

\therefore\mathsf{(x+5)\;is\;a\;factor}∴(x+5)isafactor

\underline{\textsf{Answer:}}

Answer:

\mathsf{x^3+7x^2+7x-15=(x-1)(x+3)(x+5)}x

3

+7x

2

+7x−15=(x−1)(x+3)(x+5)

\underline{\textsf{Find more:}}

Find more:

6 0
3 years ago
Sin4x.sin5x+sin4x.sin3x-sin2x.sinx=0
andreev551 [17]

Recall the angle sum identity for cosine:

cos(<em>x</em> + <em>y</em>) = cos(<em>x</em>) cos(<em>y</em>) - sin(<em>x</em>) sin(<em>y</em>)

cos(<em>x</em> - <em>y</em>) = cos(<em>x</em>) cos(<em>y</em>) + sin(<em>x</em>) sin(<em>y</em>)

==>   sin(<em>x</em>) sin(<em>y</em>) = 1/2 (cos(<em>x</em> - <em>y</em>) - cos(<em>x</em> + <em>y</em>))

Then rewrite the equation as

sin(4<em>x</em>) sin(5<em>x</em>) + sin(4<em>x</em>) sin(3<em>x</em>) - sin(2<em>x</em>) sin(<em>x</em>) = 0

1/2 (cos(-<em>x</em>) - cos(9<em>x</em>)) + 1/2 (cos(<em>x</em>) - cos(7<em>x</em>)) - 1/2 (cos(<em>x</em>) - cos(3<em>x</em>)) = 0

1/2 (cos(9<em>x</em>) - cos(<em>x</em>)) + 1/2 (cos(7<em>x</em>) - cos(3<em>x</em>)) = 0

sin(5<em>x</em>) sin(-4<em>x</em>) + sin(5<em>x</em>) sin(-2<em>x</em>) = 0

-sin(5<em>x</em>) (sin(4<em>x</em>) + sin(2<em>x</em>)) = 0

sin(5<em>x</em>) (sin(4<em>x</em>) + sin(2<em>x</em>)) = 0

Recall the double angle identity for sine:

sin(2<em>x</em>) = 2 sin(<em>x</em>) cos(<em>x</em>)

Rewrite the equation again as

sin(5<em>x</em>) (2 sin(2<em>x</em>) cos(2<em>x</em>) + sin(2<em>x</em>)) = 0

sin(5<em>x</em>) sin(2<em>x</em>) (2 cos(2<em>x</em>) + 1) = 0

sin(5<em>x</em>) = 0   <u>or</u>   sin(2<em>x</em>) = 0   <u>or</u>   2 cos(2<em>x</em>) + 1 = 0

sin(5<em>x</em>) = 0   <u>or</u>   sin(2<em>x</em>) = 0   <u>or</u>   cos(2<em>x</em>) = -1/2

sin(5<em>x</em>) = 0   ==>   5<em>x</em> = arcsin(0) + 2<em>nπ</em>   <u>or</u>   5<em>x</em> = arcsin(0) + <em>π</em> + 2<em>nπ</em>

… … … … …   ==>   5<em>x</em> = 2<em>nπ</em>   <u>or</u>   5<em>x</em> = (2<em>n</em> + 1)<em>π</em>

… … … … …   ==>   <em>x</em> = 2<em>nπ</em>/5   <u>or</u>   <em>x</em> = (2<em>n</em> + 1)<em>π</em>/5

sin(2<em>x</em>) = 0   ==>   2<em>x</em> = arcsin(0) + 2<em>nπ</em>   <u>or</u>   2<em>x</em> = arcsin(0) + <em>π</em> + 2<em>nπ</em>

… … … … …   ==>   2<em>x</em> = 2<em>nπ</em>   <u>or</u>   2<em>x</em> = (2<em>n</em> + 1)<em>π</em>

… … … … …   ==>   <em>x</em> = <em>nπ</em>   <u>or</u>   <em>x</em> = (2<em>n</em> + 1)<em>π</em>/2

cos(2<em>x</em>) = -1/2   ==>   2<em>x</em> = arccos(-1/2) + 2<em>nπ</em>   <u>or</u>   2<em>x</em> = -arccos(-1/2) + 2<em>nπ</em>

… … … … … …    ==>   2<em>x</em> = 2<em>π</em>/3 + 2<em>nπ</em>   <u>or</u>   2<em>x</em> = -2<em>π</em>/3 + 2<em>nπ</em>

… … … … … …    ==>   <em>x</em> = <em>π</em>/3 + <em>nπ</em>   <u>or</u>   <em>x</em> = -<em>π</em>/3 + <em>nπ</em>

<em />

(where <em>n</em> is any integer)

5 0
3 years ago
What is the length of cloth needed to make 24 shirts if each shirt is made from .75 yards of material?
RideAnS [48]

Answer:

18 yards of material needed to make 24 shirts

Step-by-step explanation:

The unit rate is

        1 shirt

------------------------------ = 1 1/3 shirts per yard of material = 4/3 shirt/yard

 0.75 yards material

This rate could also be written as 3/4 yard per shirt

Multiply this latter rate by 24 shirts:

 3 yard

-------------- * 24 shirts = 18 yards of material needed to make 24 shirts

 4 shirt

4 0
2 years ago
Read 2 more answers
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