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3 years ago

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.

The point estimate of the population mean is the sample mean

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

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

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!

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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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2 years ago

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What quadrants is csc less than 0.

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quadrant IV

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

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2 years ago

Estimate numbers 1-4 from the tens place value. 1. 89 2. 64 3. 985 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:

These are the steps to find the Taylor series for the function $sin^2(3 x)$

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:

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

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

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

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

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