Can you please help me do this and explain how you did it so I can look at it for references thanks so much (:

Mathematics

1 answer:

arlik [135]2 years ago

5 0

Answer:

Step-by-step explanation:

By applying Pythagoras theorem in the given right triangles,

(Hypotenuse)² = (leg 1)² + (leg 2)²

a). (123.1)² = x² + (48.7)²

x² = 15153.61 - 2371.69

x² = 12781.92

x = 113.06 cm

b). (117)² = (82)² + x²

x² = 13689 - 6724

x = 83.46 mm

C). (117)² = t² + (10.3)²

t² = 13689 - 106.09

t² = 13582.91

t = 116.55 cm

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andrey2020 [161]

Answer:

If i did the math correctly the answer to number 9 is a.5 and number 10 is c. Vanilla and German chocolate cupcakes represent about 21% of total sales

Step-by-step explanation:

for number 9 you write down each of the numbers with a dot above them. how many dots represents how many times you repeat the number. you then add it together and divide by the amount of numbers you added together. so in this case it would be 65 ÷ 12. this ends up being 5.41 but you'd round down to 5.

for number 10 you write out how many each flavor sold. then you put those numbers over the total cupcakes sold. divide the denominator by 100 and then divide the numerator by that number. then that gives you the fractions of the percent each one was. the german chocolate and the vanilla added together is almost exactly 21% of the sales

6 0

2 years ago

Use the Taylor series you just found for sinc(x) to find the Taylor series for f(x) = (integral from 0 to x) of sinc(t)dt based

Marina CMI [18]

In this question (brainly.com/question/12792658) I derived the Taylor series for $\mathrm{sinc}\,x$ about $x=0$ :

$\mathrm{sinc}\,x=\displaystyle\sum_{n=0}^\infty\frac{(-1)^nx^{2n}}{(2n+1)!}$

Then the Taylor series for

$f(x)=\displaystyle\int_0^x\mathrm{sinc}\,t\,\mathrm dt$

is obtained by integrating the series above:

$f(x)=\displaystyle\int\sum_{n=0}^\infty\frac{(-1)^nx^{2n}}{(2n+1)!}\,\mathrm dx=C+\sum_{n=0}^\infty\frac{(-1)^nx^{2n+1}}{(2n+1)^2(2n)!}$

We have $f(0)=0$ , so $C=0$ and so

$f(x)=\displaystyle\sum_{n=0}^\infty\frac{(-1)^nx^{2n+1}}{(2n+1)^2(2n)!}$

which converges by the ratio test if the following limit is less than 1:

$\displaystyle\lim_{n\to\infty}\left|\frac{\frac{(-1)^{n+1}x^{2n+3}}{(2n+3)^2(2n+2)!}}{\frac{(-1)^nx^{2n+1}}{(2n+1)^2(2n)!}}\right|=|x^2|\lim_{n\to\infty}\frac{(2n+1)^2(2n)!}{(2n+3)^2(2n+2)!}$