1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Mila [183]
3 years ago
9

An equivalent expression for 12x+1 and 2(2x+5) please help​

Mathematics
1 answer:
zhannawk [14.2K]3 years ago
3 0
<h2>Answer:</h2>

8x+9

<h2>Step-by-step explanation:</h2>

12x+1 = 2(2x+5)

<h3><em><u>Handle the brackets first</u></em></h3>

12x+1 = 4x+10

<h3><em><u>Collect like terms</u></em></h3>

=8x+9

You might be interested in
Find two unit vectors orthogonal to a=⟨2,−2,−3⟩a=⟨2,−2,−3⟩ and b=⟨3,2,2⟩b=⟨3,2,2⟩ enter your answer so that the first non-zero c
Ronch [10]
We need the cross product of the two vectors.
a x b
=<2,-2,-3> x <3,2,2>
=
i    j   k
2 -2 -3
3  2  2
=<-4+6, -(4+9), 4+6>
=<2,-13,10>
The second vector is obtained by reversing the direction, namely <-2,13,-10>
Thus the two vectors are <2,-13,10> and <-2,13,-10>.
8 0
3 years ago
Simplify −√81 3 −2√48 3 +2√24 Work Needed
MissTica

Answer:

Step-by-step explanation:

6 0
2 years ago
Read 2 more answers
What is the value of x?
Ratling [72]

Answer:

4x+7 = 5x-20                  

(reason: vertical angles are congruent; property used: distributive)

7 = 1x-20

1x = 7+20

1x = 27

x = 27

4 0
3 years ago
 First CORRECT answer gets brainliest!
SpyIntel [72]
Https://www.illustrativemathematics.org/content-standards/tasks/703  check it out it might help
3 0
3 years ago
Use this list of Basic Taylor Series and the identity sin2θ= 1 2 (1−cos(2θ)) to find the Taylor Series for f(x) = sin2(3x) based
notsponge [240]

Answer:

The Taylor series for sin^2(3 x) = - \sum_{n=1}^{\infty} \frac{-9^{n}2^{2n-1}x^{2n}}{(2n)!}, the first three non-zero terms are 9x^{2} -27x^{4}+\frac{162}{5}x^{6} and the interval of convergence is ( -\infty, \infty )

Step-by-step explanation:

<u>These are the steps to find the Taylor series for the function</u> sin^2(3 x)

  1. Use the trigonometric identity:

sin^{2}(x)=\frac{1}{2}*(1-cos(2x))\\ sin^{2}(3x)=\frac{1}{2}*(1-cos(2(3x)))\\ sin^{2}(3x)=\frac{1}{2}*(1-cos(6x))

   2. The Taylor series of cos(x)

cos(y) = \sum_{n=0}^{\infty}\frac{-1^{n}y^{2n}}{(2n)!}

Substituting y=6x we have:

cos(6x) = \sum_{n=0}^{\infty}\frac{-1^{n}6^{2n}x^{2n}}{(2n)!}

   3. Find the Taylor series for sin^2(3x)

sin^{2}(3x)=\frac{1}{2}*(1-cos(6x)) (1)

cos(6x) = \sum_{n=0}^{\infty}\frac{-1^{n}6^{2n}x^{2n}}{(2n)!} (2)

Substituting (2) in (1) we have:

\frac{1}{2} (1-\sum_{n=0}^{\infty}\frac{-1^{n}6^{2n}x^{2n}}{(2n)!})\\ \frac{1}{2}-\frac{1}{2} \sum_{n=0}^{\infty}\frac{-1^{n}6^{2n}x^{2n}}{(2n)!}

Bring the factor \frac{1}{2} inside the sum

\frac{6^{2n}}{2}=9^{n}2^{2n-1} \\ (-1^{n})(9^{n})=(-9^{n} )

\frac{1}{2}-\sum_{n=0}^{\infty}\frac{-9^{n}2^{2n-1}x^{2n}}{(2n)!}

Extract the term for n=0 from the sum:

\frac{1}{2}-\sum_{n=0}^{0}\frac{-9^{0}2^{2*0-1}x^{2*0}}{(2*0)!}-\sum_{n=1}^{\infty}\frac{-9^{n}2^{2n-1}x^{2n}}{(2n)!}\\ \frac{1}{2} -\frac{1}{2} -\sum_{n=1}^{\infty}\frac{-9^{n}2^{2n-1}x^{2n}}{(2n)!}\\ 0-\sum_{n=1}^{\infty}\frac{-9^{n}2^{2n-1}x^{2n}}{(2n)!}\\ sin^{2}(3x)=-\sum_{n=1}^{\infty}\frac{-9^{n}2^{2n-1}x^{2n}}{(2n)!}

<u>To find the first three non-zero terms you need to replace n=3 into the sum</u>

sin^{2}(3x)=\sum_{n=1}^{\infty}\frac{-9^{n}2^{2n-1}x^{2n}}{(2n)!}\\ \sum_{n=1}^{3}\frac{-9^{3}2^{2*3-1}x^{2*3}}{(2*3)!} = 9x^{2} -27x^{4}+\frac{162}{5}x^{6}

<u>To find the interval on which the series converges you need to use the Ratio Test that says</u>

For the power series centered at x=a

P(x)=C_{0}+C_{1}(x-a)+C_{2}(x-a)^{2}+...+ C_{n}(x-a)^{n}+...,

suppose that \lim_{n \to \infty} |\frac{C_{n}}{C_{n+1}}| = R.. Then

  • If R=\infty, the the series converges for all x
  • If 0 then the series converges for all |x-a|
  • If R=0, the the series converges only for x=a

So we need to evaluate this limit:

\lim_{n \to \infty} |\frac{\frac{-9^{n}2^{2n-1}x^{2n}}{(2n)!}}{\frac{-9^{n+1}2^{2*(n+1)-1}x^{2*(n+1)}}{(2*(2n+1))!}} |

Simplifying we have:

\lim_{n \to \infty} |-\frac{(n+1)(2n+1)}{18x^{2} } |

Next we need to evaluate the limit

\lim_{n \to \infty} |-\frac{(n+1)(2n+1)}{18x^{2} } |\\ \frac{1}{18x^{2} } \lim_{n \to \infty} |-(n+1)(2n+1)}|}

-(n+1)(2n+1) is negative when n -> ∞. Therefore |-(n+1)(2n+1)}|=2n^{2}+3n+1

You can use this infinity property \lim_{x \to \infty} (ax^{n}+...+bx+c) = \infty when a>0 and n is even. So

\lim_{n \to \infty} |-\frac{(n+1)(2n+1)}{18x^{2} } | \\ \frac{1}{18x^{2}} \lim_{n \to \infty} 2n^{2}+3n+1=\infty

Because this limit is ∞ the radius of converge is ∞ and the interval of converge is ( -\infty, \infty ).

6 0
3 years ago
Other questions:
  • A segway travels 12.5 miles every 1/2 hour. What is the unit rate of miles per hour?
    8·1 answer
  • Twice the sum of a number and 5 equals 9 as a equation
    7·1 answer
  • a sail that is in the form of a right triangle is four times as high as it is wide.The sail is made from 8 square meters of mate
    6·1 answer
  • Subtract:
    9·2 answers
  • A beetle was running along a number line…
    12·1 answer
  • Use additive inverses and their properties to find the equivalent expressions.
    12·1 answer
  • Alexa ran 2/3 of a mile. Caleb ran 5/8 of a mile. Who ran farther?
    13·1 answer
  • PLEASE ANSWER THE PIC BELOW (answer a, b and c)
    5·1 answer
  • In a bag of marbles, there are 8 blue marbles and 5 green marbles.
    13·2 answers
  • Deion is purchasing cat food. The brand he plans to buy comes in two seizes.
    15·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!