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2 years ago

E=3.8cos(110πt). Determine the period.

Mathematics

1 answer:

kotykmax [81]2 years ago

6 0

Check the picture below.

$f(t)=3.8\cos(\stackrel{B}{110\pi } t)~\hfill \stackrel{\textit{\Large Period}}{\cfrac{2\pi }{110\pi }\implies \cfrac{1}{55}}$

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Julian walked 6/10 of a mile to his friends house and another 35/100 mile to the store. He walked 1/4 of a mile back home. Julia

AveGali [126]

The statement, "Julian's sister said he walked 1/5 mile" cannot be agreed because Julian totally walked $1\frac{1}{5} \text{ or } \frac{6}{5}$ miles.

<u>Solution:</u>

Given that,

Julian walked 6/10 of a mile to his friends house
Another 35/100 mile to the store
He walked 1/4 of a mile back home

To find total distance walked by Julian we have to add the above stated values. That is, $\frac{6}{10} +\frac{35}{100} +\frac{1}{4}$

Factors of 10 = $5\times2$

Factors of 100 = $5\times2\times5\times2$

Factors of 4 = $2\times2$

Therefore, the least common factor of 10, 100 and 4 is 100. With like denominators we can operate on just the numerators,

$\frac{6\times10}{10\times10} +\frac{35\times1}{100\times1} +\frac{1\times25}{4\times1}\rightarrow\frac{60+35+25}{100}\rightarrow\frac{120}{100}$

$\Rightarrow\frac{120}{100}\rightarrow\frac{6}{5}$

Which can also be written as $1\frac{1}{5}$ .

So, from the above calculation it can be said that Julian walked $1\frac{1}{5} \text{ miles }$ .

3 0

3 years ago

Solve the following word problem using any method. Show your work.

VLD [36.1K]

Let's set up two equations:
x+y=20 We know she brought total of 20, some of each so x=small y=big
20x+ 30y=450

x+y= 20
-y -y
------------
x= 20 -y

substitute into the other equation:
20 (20-y) + 30y = 450
400 - 20y + 30y = 450
400+ 10 y = 450
10 y = 50
y= 5

So she brought 5 big cases. Then substitute the value of y into one of the equations.

5+ x = 20
x= 15

So we bought 15 small cases and 5 big cases.

Hope this helps! Please give me the brainliest answer if you like it! If you have further questions, please leave a comment or add me as a friend!

5 0

2 years ago

Read 2 more answers

Helen's house is located on a rectangular lot that is1 1-8 miles by 9/10 mile.Estimate the distance around the lot

klemol [59]

Answer: The distance around the lot is given by

$4\frac{1}{20}\ miles$

Step-by-step explanation:

Since we have given that

Length of rectangular lot is given by

$1\frac{1}{8}\\\\=\frac{9}{8}\ miles$

Width of rectangular lot is given by

$\frac{9}{10}\ mile$

We need to find the distance around the lot.

As we know the formula for "Perimeter of rectangle":

$Perimeter=2(Length+Width)\\\\Perimeter=2(\frac{9}{8}+\frac{9}{10})\\\\Perimeter=2(\frac{45+36}{40})\\\\Perimeter=2\times \frac{81}{40}\\\\Perimeter=\frac{81}{20}\\\\Perimeter=4\frac{1}{20}\ miles$