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3 years ago

A store sells ice cream with assorted toppings. They charge $3.00 for an ice cream, plus 50 cents

Mathematics

1 answer:

Nostrana [21]3 years ago

3 0

Answer:

$5.00

Step-by-step explanation:

$3.00 for ice cream

+ $2.00 for the 4 ounces of toppings

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Pls guys help me pls :(

Olenka [21]

Answer:good luck!

Step-by-step explanation:

3 0

4 years ago

Use the definition of a Taylor series to find the first four nonzero terms of the series for f(x) centered at the given value of

Black_prince [1.1K]

Answer:

The first four nonzero terms of the Taylor series of $\frac{7}{x + 1}$ around $a=2$ are:

$f\left(x\right)\approx P\left(x\right) = \frac{7}{3}- \frac{7}{9}\left(x-2\right)+\frac{7}{27}\left(x-2\right)^{2}- \frac{7}{81}\left(x-2\right)^{3}+\frac{7}{243}\left(x-2\right)^{4}$

Step-by-step explanation:

The Taylor series of the function f at a (or about a or centered at a) is given by

$f\left(x\right)=\sum\limits_{k=0}^{\infty}\frac{f^{(k)}\left(a\right)}{k!}\left(x-a\right)^k$

To find the first four nonzero terms of the Taylor series of $\frac{7}{x + 1}$ around $a=2$ you must:

In our case,

$f\left(x\right) \approx P\left(x\right) = \sum\limits_{k=0}^{n}\frac{f^{(k)}\left(a\right)}{k!}\left(x-a\right)^k=\sum\limits_{k=0}^{4}\frac{f^{(k)}\left(a\right)}{k!}\left(x-a\right)^k$

So, what we need to do to get the desired polynomial is to calculate the derivatives, evaluate them at the given point, and plug the results into the given formula.

$f^{(0)}\left(x\right)=f\left(x\right)=\frac{7}{x + 1}$

Evaluate the function at the point: $f\left(2\right)=\frac{7}{3}$

$f^{(1)}\left(x\right)=\left(f^{(0)}\left(x\right)\right)^{\prime}=\left(\frac{7}{x + 1}\right)^{\prime}=- \frac{7}{\left(x + 1\right)^{2}}$

Evaluate the function at the point: $\left(f\left(2\right)\right)^{\prime }=- \frac{7}{9}$

$f^{(2)}\left(x\right)=\left(f^{(1)}\left(x\right)\right)^{\prime}=\left(- \frac{7}{\left(x + 1\right)^{2}}\right)^{\prime}=\frac{14}{\left(x + 1\right)^{3}}$

Evaluate the function at the point: $\left(f\left(2\right)\right)^{\prime \prime }=\frac{14}{27}$

$f^{(3)}\left(x\right)=\left(f^{(2)}\left(x\right)\right)^{\prime}=\left(\frac{14}{\left(x + 1\right)^{3}}\right)^{\prime}=- \frac{42}{\left(x + 1\right)^{4}}$

Evaluate the function at the point: $\left(f\left(2\right)\right)^{\prime \prime \prime }=- \frac{14}{27}$

$f^{(4)}\left(x\right)=\left(f^{(3)}\left(x\right)\right)^{\prime}=\left(- \frac{42}{\left(x + 1\right)^{4}}\right)^{\prime}=\frac{168}{\left(x + 1\right)^{5}}$

Evaluate the function at the point: $\left(f\left(2\right)\right)^{\prime \prime \prime \prime }=\frac{56}{81}$

Apply the Taylor series definition:

$f\left(x\right)\approx\frac{\frac{7}{3}}{0!}\left(x-\left(2\right)\right)^{0}+\frac{- \frac{7}{9}}{1!}\left(x-\left(2\right)\right)^{1}+\frac{\frac{14}{27}}{2!}\left(x-\left(2\right)\right)^{2}+\frac{- \frac{14}{27}}{3!}\left(x-\left(2\right)\right)^{3}+\frac{\frac{56}{81}}{4!}\left(x-\left(2\right)\right)^{4}$

The first four nonzero terms of the Taylor series of $\frac{7}{x + 1}$ around $a=2$ are:

$f\left(x\right)\approx P\left(x\right) = \frac{7}{3}- \frac{7}{9}\left(x-2\right)+\frac{7}{27}\left(x-2\right)^{2}- \frac{7}{81}\left(x-2\right)^{3}+\frac{7}{243}\left(x-2\right)^{4}$

8 0

3 years ago

40. A triangle is placed inside of a circle of radius 3x. The base of the

Minchanka [31]

Answer:

P=a+b+c. To find the area of a triangle, we need to know its base and height. ... Name. Choose a variable to represent it. Let x= the measure of the angle. Step 4.

Step-by-step explanation:

4 0

3 years ago

Can u help me with number 7

vladimir2022 [97]

Step-by-step explanation:

This triangle is acute! ✅

It is also scalene.

see the picture below

7 0

2 years ago

I will give brainiest

OverLord2011 [107]

Answer:

1,Let us suppose that the height of the ladder is h and the horizontal distance of the ladder from the wall is d, then we use these two important trigonometric functions; the sine and the cosine.

h = sin 78 * 8 m and d = cos 78 * 8m where sin 78 = 0.9781 and cos 78 = 0.2079

With the given values, I expect you to do the two multiplications yourself. You will obtain a near to 8 m answer for the height h and a near to 1.7 m answer for the distance d.

2, The ship is 15 km from the port.

3,30m The ship must travel 15.6km 10 km 8. A rectangular field is 40 m long and 30 m wide. Carl walks from one corner of the field to the opposite corner along the edge of the field. Jade walks across the field diagonally to arrive at the same corner. Tip How much shorter is Jade's shortcut? Show your work. Sketch a diagram Carl-40m ade - b?= C² first. 40m. Jade = a't 62=(3 30' + 402 c 900 + 1,600 e? C² =2,5 U0 %3D C=/2500 C=50 10m shorter. jade shortcut is 172

Step-by-step explanation:

4 0

3 years ago