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2 years ago

Marcus is going to start saving $20 every month to buy

Mathematics

1 answer:

algol132 years ago

7 0

Answer:

y = $10 + I

where I = PRT/100

P = $10, R = $20 T = x

Step-by-step explanation:

I represents the Simple Interest formula

P is the Principal or the Money used to start the savings account.

R represents the rate which in this case is $20 let month

T is the time expended which in this case is x

Hope this helps!!

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A dog is on a 5 ft leash staked to the ground. If he can rotate around the stake, forming a circle, what is the area of the gras

Zanzabum

Answer:

78.5 ft²

Step-by-step explanation:

In this situation, the dog's 5 ft leash is the radius of the circle that he forms.

So, the radius is 5.

Use the area formula, A = $\pi$ r². Plug in 5 as r, and 3.14 as $\pi$

A = $\pi$ r²

A = (3.14)(5²)

A = (3.14)(25)

A = 78.5

So, the area of the grass he can cover is 78.5 ft²

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2 years ago

Help me solve this question please

Vanyuwa [196]

Answer:

Step-by-step explanation:

3 0

2 years ago

6th grade math help me please ! :))

polet [3.4K]

Cost of 32 ounces yoghurt = $1.62

Cost of 1 ounce = $1.62/32 = $0.05 (approx)

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2 years ago

A pattern follows the rule "Starting with three, every consecutive line has 2 less than twice the previous line."

SVEN [57.7K]

"Starting with three, every consecutive line has 2 less than twice the previous line."

this statement means that
your staring line has 3 marbles. You multiply the 3 marblesby 2 so
3x2=6
And then you minus it by 2
6-2=4

which means that you'll get 4 marbles for the next line.

So to get your 6th line, you count how many marbles is on the 5th line but since your diagram doesn't have the 5th line you have to figure out the 5th line by counting how.many marbles is on the 4th line.

4th line = 10 marbles

10×2=20
20-2=18

5th line = 18

18×2=36
36-2=34

So the 6th line has 34 marbles.

8 0

3 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:

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

Like in the linked problem, the limit is 0 so the series for $f(x)$ converges everywhere.

7 0

2 years ago