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3 years ago

What is the approximate area of the figure ?

Mathematics

1 answer:

Strike441 [17]3 years ago

6 0

Answer:

89.6 square feet

Step-by-step explanation:

The total area is that of a 5×14 rectangle and a circle with a radius of 2.5 feet:

A = bh +πr^2

= (5 ft)(17 f) +π(2.5 ft)^2

= 70 ft^2 + 6.25π ft^2

≈ (70 +19.6) ft^2 = 89.6 ft^2

The approximate area is 89.6 square feet.

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PLEASE, HELP ASAP!!!!!!!!!! THANKS, THANKS!!!!!!!!!!

Alik [6]

Here's one way to convert the number to the ratio of integers.

Count the number of repeating digits. Call this number n. Here, only 5 has an overline, so there is only one repeating digit. That is, n = 1.
Multiply the number by 10^(-n) and subtract that from the original number. Call the result p. Here, doing this gives $p=3.25\overline{5}-0.32\overline{5}=3.23$
Compute the value q = 1 - 10^(-n). The result of step 2 is q times the original number. Here, q = 1 - 10^-1 = 0.9.
Create the fraction p/q. Note that both p and q have a finite number of decimal digits, so this is easily converted to the ratio of integers, by moving the decimal point an equal number of places in the numerator and denominator. Reduce the resulting fraction as may be required. This is the fraction you're looking for. Here, p/q = 3.23/0.9 = 323/90. No reduction of this fraction is possible.

The fraction you're looking for is ...

$3.2\overline{5}=\dfrac{323}{90}$

3 0

3 years ago

Approximate the value of 44 (Square root) to the nearest hundreth

Svetlanka [38]

I assume that calculators are not allowed, perhaps not even paper and pencil.

With paper and pencil, there is an algorithm to find the square-root of any number accurate to as many digits as we wish. Please post if that is the case.

Without calculator, pen and pencil, we resort to mental arithmetic. Someone with a little practice can memorize or calculate the square of all numbers up to 100, or as many as one can.

Here we can take advantage of exactly the property suggested by Xisca... that is, sqrt(44)=2sqrt(11), which reduces the calculation to more accuracy with a lesser knowledge of the squares of numbers.

As a first approximation, we know that 33*33=1089, or 3.3*3.3=10.89, which is already very close to 11.

Thus a first approximation to sqrt(44)=2sqrt(11)=2*3.3=6.6 (approximately).

To refine the calculation of sqrt(11), we use Newton's approximation, namely

a better approximation is given by 3.3 + (11 - 3.3^2) / (2*3.3) = 3.3+0.11/6.6 = 3.3+1/60

which gives

sqrt(44) = 2(3.3+1/60) = 6.6+1/30 = 6.6333,

actually accurate almost to the nearest 1/10000.

All of the above can be done mentally

For those who need help with rounding to the nearest hundredth (0.01), we drop digits AFTER the 2nd decimal, namely *.**33. If the first digit dropped is 5 or more, add one to the last digit retained, otherwise, retain only the two digits after the decimal. In this case, the first digit dropped is 3, so it will simply be dropped. The final answer is then 6.63, to the nearest hundredth.

5 0

3 years ago

Read 2 more answers

Sketch the graph of each line y=-9x-5

ycow [4]

I can’t answer it without a picture of the graph

5 0

3 years ago

2.11 A concert is performed for a crowd of 118

never [62]

Answer:

Number of Students who attended = 22

Number of Adults who attended = 34

Number of Children who attended = 62

Step-by-step explanation:

Total crowd at concert = 118

Cost of each adult's ticket = $20

Cost of each student's ticket = $16

Cost of each children's ticket = $11.50

The total revenue of the concert = $1745.00

Now, Let the number of students = k

So, the number of children = k + 40

So, number of adults = 118 - {k + (k + 40)}

= 118- 2k -40 = 78 -2k

So, the total revenue = Total revenue from {adults + students + children}

or, $1745.00 = (78 -2k)(20) + k(16) +(k+40)(11.50)

here, solving for the value of k :

12.5k = 2020 - 1745

or k = 275\12.5 = 22

So, the Number of students who attended = 22

Number of Adults who attended = 78 - 2k = 78 - 44 = 34

Number of children who attended = k + 40 = 22 + 40 = 62

5 0

3 years ago

Which of the following matrix addition problems are possible?

sergejj [24]

Answer:

Option A

Step-by-step explanation:

When adding two matrices, they should be of the same sizes.

Option A:

Matrix $\left[\begin{array}{c}2&3\end{array}\right]$ has 2 rows and 1 column.

Matrix $\left[\begin{array}{c}3&4\end{array}\right]$ has 2 rows and 1 column too.

So, these matrices can be added.

The result will be

$\left[\begin{array}{c}2+3&3+4\end{array}\right]=\left[\begin{array}{c}5&7\end{array}\right]$

7 0

3 years ago

Read 2 more answers