All categories

4 years ago

This quarter circle has a radius of 7 centimeters what is the area of this shape?

Mathematics

1 answer:

IgorLugansk [536]4 years ago

8 0

Answer:

153.94

Step-by-step explanation:

I literally just typed in "how to find area of a circle with given radius" to refresh my memory lol. But then all I had to do was type in the radius and it gave me the answer: 153.94

However, if you need the actual steps:

Use the formula: A = πr^2

A = π7^2

A = 49π

A = 153.938 which rounds to 153.94 :)

You might be interested in

15 miles from home after 20 minutes and 36 miles from home after 48 minutes. Proportional?

elena-s [515]

Answer: yes, they are proportional

Step-by-step explanation:

15/20 = 36/48 (cross multiplication)

15 x 48 = 720

36 x 20 = 720

6 0

3 years ago

Let y = (2 6) and u = (7 1). Write y as the sum of a vector in Span(u) and a vector orthogonal to .

Minchanka [31]

The question is missing. Here is the complete question.

Let y = $\left[\begin{array}{ccc}2\\6\end{array}\right]$ and u = $\left[\begin{array}{ccc}7\\1\end{array}\right]$ . Write y as the sum of a vector in Span(u) and a vector orthogonal to u.

Answer: y = $\left[\begin{array}{ccc}\frac{21}{10} \\ \frac{3}{10} \end{array}\right] + \left[\begin{array}{ccc}\frac{-1}{10}\\ \frac{57}{10} \end{array}\right]$

Step-by-step explanation: The sum of vectors is given by

y = $y_{1}$ + z

where $y_{1}$ is in Span(u);

vector z is orthogonal to it;

First you have to compute the orthogonal projection $y_{1}$ of y:

$y_{1}$ = proj y = $\frac{y.u}{u.u}.u$

Calculating orthogonal projection:

$\left[\begin{array}{c}2\\6\end{array}\right]$ . $\left[\begin{array}{c}7\\1\end{array}\right]$ = $\left[\begin{array}{c}9\\6\end{array}\right]$

$\left[\begin{array}{c}7\\1\end{array}\right]$ . $\left[\begin{array}{c}7\\1\end{array}\right]$ = $\left[\begin{array}{c}49\\1\end{array}\right]$

$y_{1} = \frac{9+6}{49+1}.u$

$y_{1} = \frac{15}{50}.u$

$y_{1} = \frac{3}{10}.u$

$y_{1} = \frac{3}{10}.\left[\begin{array}{c}7\\1\end{array}\right]$

$y_{1} = \left[\begin{array}{c}\frac{21}{10} \\\frac{3}{10} \end{array}\right]$

Calculating vector z:

z = y - $y_{1}$

z = $\left[\begin{array}{c}2\\6\end{array}\right] - \left[\begin{array}{c}\frac{21}{10} \\\frac{3}{10} \end{array}\right]$

z = $\left[\begin{array}{c}\frac{-1}{10} \\\frac{57}{10} \end{array}\right]$

Writing y as the sum:

$y = \left[\begin{array}{c}\frac{21}{10} \\\frac{3}{10} \end{array}\right] + \left[\begin{array}{c}\frac{-1}{10} \\\frac{57}{10} \end{array}\right]$

5 0

3 years ago

PLEASEEEEEEEE HELPPPPPPPPPPPPPPPPPPPPPPPP MEEEEEEEEEEEEEEEEEEEE

spayn [35]

Answer:

First, find the multiplier ( 100-18/100)

=0.82

364000*0.82

=298480 is the new value after the years

7 0

3 years ago

James' grades on his last five tests were 93, 95, 89, 95, and 100. What does he need to make on his last test to have a mean (av

leonid [27]

Answer: 98

Step-by-step explanation:

First find the mean of the numbers you currently have.

93 + 95 + 89 + 95 + 100

and divide by 5 to get 94.4.

Now you need to find the extra number.

The number has to be higher than 94. I know there is an easier way, but I started by adding 100 and kept going down til I was at 98.

93 + 95 + 89 + 95 + 100 + 98

divided by 6 is 95

8 0

3 years ago

Two particles are projected simultaneously towards each other from opposite ends of a straight horizontal tube. One particle tra

zysi [14]

This shows that it will take the particle 23 secs to meet.

The distance travelled by each particle is 50cm

If two particles are projected simultaneously towards each other from opposite ends of a straight horizontal tube,

The first particle distances are 97cm, 93cm, 89cm... every 2secs

This form an arithmetic sequence and the nth term is expressed as:

Tn = a + (n - 1)d

a is the first term = 97

d is the common difference = 4

Substitute

Tn = 97+(n - 1)(4)

Tn = 97 + 4n -4

Tn = 4n + 93

Similarly for the distances 49 cm, 47 cm, 45cm .....

This also forms an arithmetic sequence and the nth term is expressed as:

Tn = a + (n - 1)d

a is the first term = 49

d is the common difference = 2

Substitute

Tn = 49+(n - 1)(2)

Tn = 49 + 2n -2

Tn = 2n + 47

For the particles to meet, this means that 4n + 93 = 2n + 47

4n - 2n = 47 - 93

2n = -46

n = -23

This shows that it will take the particle 23 secs to meet.

To get the distance travelled by each particle

Tn = 4n + 93

T2 = 4(2) + 93

T2 = 101

For the other sequence;

Tn = 2n + 47

T2 = 2(2) + 47

T2 = 51

Taking the difference;

D = 101 - 51 = 50cm

Hence the distance travelled by each particle is 50cm

Learn more here: brainly.com/question/13989292

3 0

3 years ago