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2 years ago

UGHHHHHHHHHHHHHHHHHHHHHHHHH HELPPPPPPPPPPPPPPPPPPPPPPPPPPPPPP MEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEE

Mathematics

2 answers:

LekaFEV [45]2 years ago

5 0

Answer:

1/ 6 of 54find

So now that we've converted 54 into a fraction, to work out the answer, we put the fraction 1/6 side by side with our new fraction, 54/1 so that we can multiply those two fractions.

That's right, all you need to do is convert the whole number to a fraction and then multiply the numerators and denominators. Let's take a look:

1 x 54

6 x 1

In this case, our new fraction can actually be simplified down further. To do that, we need to find the greatest common factor of both numbers.

Step-by-step explanation:

please mark me as brainlist

dimaraw [331]2 years ago

4 0

Answer:

11/25

Step-by-step explanation:

44% can be written as a Fraction as 44/100. This fraction can ultimately be simplified to 22/50 before being finally reduced again to 11/25.

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coldgirl [10]

Division word problems often ask you to split something evenly. You should solve this problem by dividing the total amount of ice cream by the amount you will serve each guest. After converting to fractions, 4.5 divided by .75. You will have enough to serve six guests, and no ice cream left over.

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3 years ago

A rectangular yard has dimensions 10 meters by 8 meters. In the middle of the yard is a rectangular pond of 8 square meters, lea

Nikolay [14]

hahahaha i had this same question in my grade

8 0

3 years ago

Suppose you have a right triangle with congruent legs and a hypotenuse that measure (12sqrt(5))/5 What is the length of the smal

anastassius [24]

The length of the smaller leg is 3.79

<h3>How to determine the length of the smaller leg?</h3>

Represent the smaller leg with x.

So, we have:

$x^2 + x^2 = ((12\sqrt5)/5)^2$ -- Pythagoras theorem

This gives

2x^2 = 144/5

Divide by 2

x^2 = 72/5

This gives

x^2 = 14.4

Take the square root

x = 3.79

Hence, the length of the smaller leg is 3.79

Read more about right triangles at

brainly.com/question/6322314

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4 0

2 years ago

Rita drank the following ounces of milk during the last week

Andre45 [30]

Answer:

Mean = 14.14

Mode = 14

Median = 14

Range = 16

Step-by-step explanation:

Given Data Set : 18 , 14 , 10 , 22, 14 , 15 and 6

Identifying the mean

Mean = Sum of given data values / total # of given values

Here sum of given data values = 18 + 14 + 10 + 22 + 14 + 15 + 6 = 99

And there are 7 given data values

So mean = 99 / 7 =

14.14

( approximately )

Identifying the mode

The mode is the number that appears most often

14 appears twice, which is the most times out of any other value

Hence, the mode is

Identifying the median

The median is the middle number in terms of value

To make this easier, its best to order the given values from least to greatest
Doing so we acquire 6 , 10 , 14 , 14 , 15 , 18 , 22

Then you go to the middle of the listed terms.

Doing so we would end up with 14 being in the middle

Hence the median is

Identifying the range

Range = largest given value - smallest given value

Here the most rita drank was 22 ounces and the least amount of milk she drank was 6 ounces

So the range would be 22 - 6 =

ounces

And we are done!

3 0

2 years ago

What is the Laplace Transform of 7t^3 using the definition (and not the shortcut method)

Leokris [45]

Answer:

Step-by-step explanation:

By definition of Laplace transform we have

L{f(t)} = $L{{f(t)}}=\int_{0}^{\infty }e^{-st}f(t)dt\\\\Given\\f(t)=7t^{3}\\\\\therefore L[7t^{3}]=\int_{0}^{\infty }e^{-st}7t^{3}dt\\\\$

Now to solve the integral on the right hand side we shall use Integration by parts Taking $7t^{3}$ as first function thus we have

$\int_{0}^{\infty }e^{-st}7t^{3}dt=7\int_{0}^{\infty }e^{-st}t^{3}dt\\\\= [t^3\int e^{-st} ]_{0}^{\infty}-\int_{0}^{\infty }[(3t^2)\int e^{-st}dt]dt\\\\=0-\int_{0}^{\infty }\frac{3t^{2}}{-s}e^{-st}dt\\\\=\int_{0}^{\infty }\frac{3t^{2}}{s}e^{-st}dt\\\\$

Again repeating the same procedure we get

$=0-\int_{0}^{\infty }\frac{3t^{2}}{-s}e^{-st}dt\\\\=\int_{0}^{\infty }\frac{3t^{2}}{s}e^{-st}dt\\\\\int_{0}^{\infty }\frac{3t^{2}}{s}e^{-st}dt= \frac{3}{s}[t^2\int e^{-st} ]_{0}^{\infty}-\int_{0}^{\infty }[(t^2)\int e^{-st}dt]dt\\\\=\frac{3}{s}[0-\int_{0}^{\infty }\frac{2t^{1}}{-s}e^{-st}dt]\\\\=\frac{3\times 2}{s^{2}}[\int_{0}^{\infty }te^{-st}dt]\\\\$

Again repeating the same procedure we get

$\frac{3\times 2}{s^2}[\int_{0}^{\infty }te^{-st}dt]= \frac{3\times 2}{s^{2}}[t\int e^{-st} ]_{0}^{\infty}-\int_{0}^{\infty }[(t)\int e^{-st}dt]dt\\\\=\frac{3\times 2}{s^2}[0-\int_{0}^{\infty }\frac{1}{-s}e^{-st}dt]\\\\=\frac{3\times 2}{s^{3}}[\int_{0}^{\infty }e^{-st}dt]\\\\$

Now solving this integral we have

$\int_{0}^{\infty }e^{-st}dt=\frac{1}{-s}[\frac{1}{e^\infty }-\frac{1}{1}]\\\\\int_{0}^{\infty }e^{-st}dt=\frac{1}{s}$

Thus we have

$L[7t^{3}]=\frac{7\times 3\times 2}{s^4}$

where s is any complex parameter

5 0

3 years ago