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3 years ago

If I have 18 pies and eat 13.67 percent of each one how much pie have I eaten?

Mathematics

1 answer:

PilotLPTM [1.2K]3 years ago

7 0

Answer:

2.46 pies (round if asked)

Step-by-step explanation:

You are eating 13.67% of each pie, so you will multiply 18 × 0.1367 = 2.46

You might be interested in

How can Kendra determine if the function is actually linear?

pashok25 [27]

Answer:

D. She can check to see if the rate of change between the first two ordered pairs is the same as the rate of change between the first and last ordered pairs.

Step-by-step explanation:

Find the rate of change between first two ordered pairs and the second two ordered pairs:

1. Points (2,4) and (3,9). Rate of change:

$\dfrac{y_2-y_1}{x_2-x_1}=\dfrac{9-4}{3-2}=\dfrac{5}{1}=5$

2. Points (3,9) and (4,16). Rate of change:

$\dfrac{y_2-y_1}{x_2-x_1}=\dfrac{16-9}{4-3}=\dfrac{7}{1}=7$

The rate of change for the linear function must the same for each two points on the graph of the function. In this case, the reate of change differs, so this function is not linear and correct option is D.

8 0

3 years ago

Read 2 more answers

Look at pic plz help

jasenka [17]

Answer:

Mochi

Step-by-step explanation:

Mochi

Each week theres a increase of 4 in Mochi's weight. So you will add 4 twice to get to the 5th week.

4th week : 17

5th week : 21

Kappa

In 4 weeks Kappa weighed 14, which is less that Mochi at the pace Kappa is going at.

on the 3rd week Kappa weighed 11

and 2nd week Kappa weighed 5

and 1st week Kappa weighed 3

Looking at the graph, we can give a rough estimate on the 5th week it's going to be 16 or 17 either way making Mochi heavier.

8 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:

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

3 years ago

Kristin bought 8 books for a total of $238.60. Math books cost $29.60 and english books cost $29.90. How many of each type of bo

Olegator [25]

5 math books and six english books

6 0

2 years ago

I NEED HELP ASAP PLZZZZ LOOK AT THE PIC

umka2103 [35]

Answer:

The red line is g(x)

Step-by-step explanation:

* To solve this problem you must to know some fact about

transformation

- If the function f(x) translated horizontally to the left

by h units, then the new function g(x) = f(x + h)

- If the function f(x) translated vertically up

by k units, then the new function g(x) = f(x) + k

- If the function f(x) translated vertically down

by k units, then the new function g(x) = f(x) – k

- A vertical stretching is the stretching of the graph away from

the x-axis

• if k > 1, the graph of y = k•f (x) is the graph of f (x) vertically

stretched by multiplying each of its y-coordinates by k.

- A vertical compression is the squeezing of the graph toward

the x-axis.

• if 0 < k < 1 (a fraction), the graph is f (x) vertically compressed

by multiplying each of its y-coordinates by k.

• if k should be negative, the vertical stretch or compress is

followed by a reflection across the x-axis.

* Now lets solve the problem

- g(x) = -1/2f(x + 2)

# x + 2 ⇒ means the graph moved 2 units to the left

# -1/2 ⇒ means the graph has vertical compressed followed

by a reflection across the x-axis, then multiply

each y-coordinate by -1/2

- The graph of f(x) intersects x-axis at point (3 , 0)

∴ The point will be (1 , 0) on g(x)

- The graph of f(x) intersects y-axis at point (0 , -6)

∴ The point will be (-2 , 3) on g(x)

* Look to the graph

# The red line is g(x)

# The blue line is f(x)

- To check your answer

# The equation of f(x) is f(x) = 2x - 6

∵ g(x) = -1/2f(x + 2)

∴ g(x) = -1/2[2(x + 2) - 6] = -1/2[2x + 4 - 6] = -1/2[2x - 2] = -x + 1

* g(x) intersect x-axis at point (1 , 0) and y-axis at point (0 , 1)

3 0

3 years ago