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Shalnov [3]
3 years ago
14

If B=3x-1 and C=2x^2+x+9, find an expression that equals 2B+2C in standard form.

Mathematics
1 answer:
Anestetic [448]3 years ago
3 0

Answer:

4x^2 +8x + 16

Step-by-step explanation:

2 (3x - 1) + 2 (2x^2 + x + 9)

6x - 2 + 4x^2 +2x + 18

4x^2 + 8x +16

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Data consistent with summary quantities in the article "Effects of Fast-Food Consumption on Energy Intake and Diet Quality Among
Olegator [25]

Answer:

a) X[bar]₁=  1839.20 cal

b) X[bar]₂= 1779.07 cal

c) S₁= 386.35 cal

Step-by-step explanation:

Hello!

You have two independent samples,

Sample 1: n₁= 15 children that did not eat fast food.

Sample 2: n₂= 15 children that ate fast food.

The study variables are:

X₁: Calorie consumption of a kid that does not eat fast food in one day.

X₂: Calorie consumprion of a kid that eats fast food in one day.

a)

The point estimate of the population mean is the sample mean

X[bar]₁= (∑X₁/n₁) = (27588/15)= 1839.20 cal

b)

X[bar]₂= (∑X₂/n₂)= (26686/15)= 1779.07 cal

c)

To calculate the sample standard deiation, you have to calculate the sample variance first:

S₁²= \frac{1}{n_1-1}[∑X₁² - (( ∑X₁)²/n₁)]= \frac{1}{14} + [52829538 - (\frac{27588^{2} }{15}) = 149263.4571 cal²

S₁= 386.35 cal

I hope it helps!

7 0
3 years ago
A wall in Maria’s bedroom is in the shape of a trapezoid. The wall can be divided into a rectangle and a triangle. Using the 45°
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Using the 45°-45°-90° triangle theorem, find the value of h, the height of the wall. C. 13 ft

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What quadrants is csc less than 0.
adelina 88 [10]

quadrant IV

Sure hope this helped you, have a good day! =)

6 0
2 years ago
Estimate numbers 1-4 from the tens place value.<br> 1. 89<br> 2. 64<br> 3. 985<br> 4. 467
zhenek [66]

\huge\text{Hey there!}

\huge\text{We are not sure what you want us to}\\\huge\text{round to so I will...}\huge\boxed{\downarrow}\\\\\\\huge\boxed{\mathsf{original \ number\rightarrow tens\rightarrow hundreds/thousands}}

\huge\boxed{1.) \ \mathsf{89} \rightarrow \bold{90}\rightarrow\bf 100}}\\\huge\boxed{2.) \mathsf{\ 64\rightarrow \bold{6} \rightarrow \bf 100}}\\\huge\boxed{3.) \mathsf{ \ 985 \rightarrow \bold{990}\rightarrow \bf 1,000}}\\\huge\boxed{4.)\mathsf{\ 467\rightarrow\bold{470}\rightarrow \bf 500}}}

\huge\textsf{Reminder: If you have \bf 0 - 4 you are going}\\\huge\textsf{\bf downward \& 5 - 9 you are going upward}

\huge\text{Good luck on your assignment \& enjoy your day!}

~\huge\boxed{\frak{Amphitrite1040:)}}

5 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
2 years ago
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