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3 years ago

Instructions: Find the missing length indicated.

Mathematics

1 answer:

ikadub [295]3 years ago

4 0

Answer:

X = 25

Step-by-step explanation:

Set up as a ratio:

Simplify ratio, multiply by 5, divide by 12

$\frac{60}{144} = \frac{x}{60} \\\frac{5}{12}=\frac{x}{60} \\ 300 = 12x\\12x=300\\x=25$

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PLEASE HELP MEEE

antoniya [11.8K]

Answer:

(6a-4)°=(2a+68)°

6a-2a=68+4

4a=72

a=72/4

a=18

6 0

2 years ago

Read 2 more answers

Karen batting average was 0.444 She was at bat 45 times.how many hits did she get?

Simora [160]

<span>Karen got an average of 0.444 hits out of 45 bats.
We need to find the number of hits did Karen got.
=> 45 bats all in all
=> 0.444 average hits
x = how many hit did she get
Now, let’s multiply the total bats to Karen’s average hit:
=> 45 x 0.444 …. (repeating decimals)
=> 20
Therefore, Karen hit 20 times out of 45 bats
Let’s check our answer:
=> 20 / 45
=> 0.44444…. (repeating decimals)

</span>

4 0

2 years ago

Read 2 more answers

Which expressions show the expanded form for 200.06? Choose all answers that are correct. A. 2 × 100 + 0 × 1 + 0 × 1/10 + 6 × 1/

saveliy_v [14]

Answer:

Therefore,

Correct options are

A) 2 × 100 + 0 × 1 + 0 × 1/10 + 6 × 1/100

D) 2 × 100 + 6 × 1/100

Step-by-step explanation:

Given:

Number is

200.06

For option A)

2 × 100 + 0 × 1 + 0 × 1/10 + 6 × 1/100

= 200 + 0 + 0 + 0.06

= 200.06

Which is CORRECT.

For option B)

2 × 100 + 0 × 1 + 0 × 1/100 + 6 × 1/1,000

= 200 + 0 + 0 + 0.006

=200.006

Which is INCORRECT.

For option C)

2 × 100 + 6 × 1/10

= 200 + 0.6

= 200.6

Which is INCORRECT.

For option D)

2 × 100 + 6 × 1/100

= 200 + 0.06

= 200.06

Which is CORRECT.

Therefore,

Correct options are

A) 2 × 100 + 0 × 1 + 0 × 1/10 + 6 × 1/100

D) 2 × 100 + 6 × 1/100

5 0

2 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> $sin^2(3 x)$

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

2 years ago

If R= {(1,2), (3,4), (5,6), (8,9)} find the domain and range of R a. Find value of X and Y.

podryga [215]

<h2>The Domain and Range of a Set of Ordered Pairs</h2><h3>Answer:</h3>

Domain: $x \in \{ 1, 3, 5, 8 \}$

Range: $y \in \{ 2, 4, 6, 9 \}$

<h3>Step-by-step explanation:</h3>

We are given the set of ordered pairs: $R$ . The Domain is just the set containing all the $x$ values of each ordered pair of $R$ . Recall that in the ordered pair $(a,b)$ , $a$ is the $x$ value. In set $R$ , we can see that the $x$ values are $\{ 1, 3, 5, 8 \}$ .

For the Range, it is just the set containing all the $y$ values of each ordered pair of $R$ . Recall that in the ordered pair $(a,b)$ , $b$ is the $y$ value. In set $R$ , we can see that the $y$ values are $\{ 2, 4, 6, 9 \}$ .

7 0

2 years ago