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nalin [4]
4 years ago
5

How many different lock combinations are possible on a bike lock with 4 rotating discs that go from 0-9

Mathematics
1 answer:
AlexFokin [52]4 years ago
3 0
You would multiply 9x___ whatever the other number is witch is 4
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QUICK<br><br> Find the area of the shaded region. Use 3.14 for pi.
ivolga24 [154]

Answer:

The shaded region is 353.25m^2.

Step-by-step explanation:

The area of a circle is defined by the multiplication of the square of its radius, or half its diameter, by pi, or simply A=\pi r^2.

15m is the radius of the large circle, and also the diameter of the smaller circles.

The area of the shaded region will be the subtraction of the area of the two smaller circles from the area of the larger circle.

As such:

A=\pi(15)^{2}-2(\pi(\frac{15}{2} )^{2})\\\\A=(3.14)(15)^{2}-2(3.14)(\frac{15}{2} )^{2}\\A=706.5-353.25\\A=353.25

4 0
3 years ago
Read 2 more answers
Find the sum of the measures of the interior angles of the figure.
Ivahew [28]

Step-by-step explanation:

Given figure is of HEXAGON.

the sum of the measures of the interior angles of a hexagon

=  \frac{(6 - 2) \times 180 \degree}{6}  \times 6 \\  \\ =  {4 \times 30 \degree} \times 6  \\   \\ = 720 \degree

4 0
4 years ago
The figure shows a parallelogram inside a rectangle outline:
Zolol [24]

Answer:

well this is the answer

Step-by-step explanation:

3 0
3 years ago
A uranium mining town reported population declines of 3.2%, 5.2%, and 4.7% for the three successive five-year periods 1985–89, 1
KATRIN_1 [288]

Answer:

Step-by-step explanation:

Heres the complete question:

A uranium mining town reported population declines of 3.2%, 5.2%, and 4.7% for the three successive five-year periods 1985–89, 1990–94, and 1995–99. If the population at the end of 1999 was 9,320:

How many people lived in the town at the beginning of 1985? (Round your answer to the nearest whole number.)

solution:

Let the population of the town at the beginning of 1985 be P. Then, given that in the first five-year period the population declined by 3.2%, i.e., 0.032, the population of the town at the end of 1989 would be

(1 – 0.032)P = 0.968P.

Again, given that in the second five-year period the population declined by 5.2%, i.e., 0.052, the population of the town at the end of 1994 would be

(1 – 0.052)(0.968P) = 0.948 x 0.968P = 0.917664P.

Finally, given that in the third five-year period the population declined by 4.7%, i.e., 0.047, the population of the town at the end of 1999 would be

(1 – 0.047)(0.917664P) = 0.874533792P.

We are given, 0.874533792P = 9320 or

P = 9320/0.874533792 = 10657.11.

Thus, 10657 people lived in the town at the beginning of 1985

3 0
4 years ago
PLEASE HELP
Aliun [14]

Hello from MrBillDoesMath!

Answer:

(x^4-2) (x^4 -1)


Discussion:


Let u = x^4, then

x^8 - 3x^4 + 2

= u^2 - 3u + 2

= (u -2) ( u -1)

= (x^4-2) (x^4 -1)



Thank you,

MrB

8 0
4 years ago
Read 2 more answers
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