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3 years ago

Explain why sin(90)=1

1 answer:

4 0

In a right triangle, sin= $\frac{opposite segment}{hypotenuse}$
Because the "opposite segment " is actually the hypotenuse of the triangle we get the fraction $\frac{hypotenuse}{hypotenuse}$ which is 1.

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Which of the following expressions cannot be written as a whole number?

Allisa [31]

9^0 its that answer because all the other number you can make a whole number.

6 0

3 years ago

Read 2 more answers

The mean of a set of eight numbers is 8 and the mean of a different set of twelve numbers is m. Given that the mean if the combi

Basile [38]

Mean=total/the number of values =>total=mean*the number of values
the total of the first set of 8 numbers is : total=8*8=64
the total of the second set of 12 numbers is: 12m
the total of the 20 numbers is: 64+12m
the mean of the 20 number is: (64+12m)/20=5
solve for m: 64+12m=100
12m=36
m=3

8 0

3 years ago

Tacoma's population in 2000 was about 200 thousand, and had been growing by about 9% each year. a. Write a recursive formula for

KIM [24]

Answer:

a) The recurrence formula is $P_n = \frac{109}{100}P_{n-1}$ .

b) The general formula for the population of Tacoma is

$P_n = \left(\frac{109}{100}\right)^nP_{0}$ .

c) In 2016 the approximate population of Tacoma will be 794062 people.

d) The population of Tacoma should exceed the 400000 people by the year 2009.

Step-by-step explanation:

a) We have the population in the year 2000, which is 200 000 people. Let us write $P_0 = 200 000$ . For the population in 2001 we will use $P_1$ , for the population in 2002 we will use $P_2$ , and so on.

In the following year, 2001, the population grow 9% with respect to the previous year. This means that $P_0$ is equal to $P_1$ plus 9% of the population of 2000. Notice that this can be written as

$P_1 = P_0 + (9/100)*P_0 = \left(1-\frac{9}{100}\right)P_0 = \frac{109}{100}P_0$ .

In 2002, we will have the population of 2001, $P_1$ , plus the 9% of $P_1$ . This is

$P_2 = P_1 + (9/100)*P_1 = \left(1-\frac{9}{100}\right)P_1 = \frac{109}{100}P_1$ .

So, it is not difficult to notice that the general recurrence is

$P_n = \frac{109}{100}P_{n-1}$ .

b) In the previous formula we only need to substitute the expression for $P_{n-1}$ :

$P_{n-1} = \frac{109}{100}P_{n-2}$ .

Then,

$P_n = \left(\frac{109}{100}\right)^2P_{n-2}$ .

Repeating the procedure for $P_{n-3}$ we get

$P_n = \left(\frac{109}{100}\right)^3P_{n-3}$ .

But we can do the same operation n times, so

$P_n = \left(\frac{109}{100}\right)^nP_{0}$ .

c) Recall the notation we have used:

$P_{0}$ for 2000, $P_{1}$ for 2001, $P_{2}$ for 2002, and so on. Then, 2016 is $P_{16}$ . So, in order to obtain the approximate population of Tacoma in 2016 is

$P_{16} = \left(\frac{109}{100}\right)^{16}P_{0} = (1.09)^{16}P_0 = 3.97\cdot 200000 \approx 794062$

d) In this case we want to know when $P_n>400000$ , which is equivalent to

$(1.09)^{n}P_0>400000$ .

Substituting the value of $P_0$ , we get

$(1.09)^{n}200000>400000$ .

Simplifying the expression:

$(1.09)^{n}>2$ .

So, we need to find the value of $n$ such that the above inequality holds.

The easiest way to do this is take logarithm in both hands. Then,

$n\ln(1.09)>\ln 2$ .

So, $n>\frac{\ln 2}{\ln(1.09)} = 8.04323172693$ .

So, the population of Tacoma should exceed the 400 000 by the year 2009.

8 0

3 years ago

Read 2 more answers

Now find the sum what 2 5/6 and

salantis [7]

Answer:

$2 \frac{1}{5} + 1 \frac{5}{6} = 2 \frac{6}{30} + 1 \frac{25}{30} \\ \\ \frac{66}{30} + \frac{55}{30} = \frac{121}{30} \\ \\ \\ = 4 \frac{1}{30}$

I hope I helped you^_^

3 0

2 years ago

Please help me with question number 2

Mice21 [21]

Answer:

2. The area of the side walk is approximately 217 m²

3. The distance away from the sprinkler the water can spread is approximately 11 feet

4. The area of the rug is 49.6

Step-by-step explanation:

2. The dimensions of the flower bed and the sidewalks are;

The diameter of the flower bed = 20 meters

The width of the circular side walk, x = 3 meters

Therefore, the diameter of the outer edge of the side walk, D, is given as follows

D = d + 2·x (The width of the side walk is applied to both side of the circular diameter)

∴ D = 20 + 2×3 = 26

The area of the side walk = The area of the sidewalk and the side walk = The area of the flower bed

∴ The area of the side walk, A = π·D²/4 - π·d²/4

∴ A = 3.14 × 26²/4 - 3.14 × 20²/4 = 216.66

By rounding to the nearest whole number, the area of the side walk, A ≈ 217 m²

3. Given that the area formed by the circular pattern, A = 379.94 ft.², we have;

Area of a circle = π·r²

∴ Where 'r' represents how far it can spread, we have;

π·r² = 379.94

r = √(379.94 ft.²/π) ≈ 10.997211 ft.

Therefore, the distance away from the sprinkler the water can spread, r ≈ 11 feet

4. The circumference of the rug = 24.8 meters

The circumference of a circle, C = 2·π·r

Where;

r = The radius of the circle

π = 3.1

∴ For the rug of radius 'r', C = 2·π·r = 24.8

r = 24.8/(2·π) = 12.4/π = 12.4/3.1 = 4

The area = π·r²

∴ The area of the rug = 3.1 × 4² = 49.6.

8 0

2 years ago