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kow
5 days ago
11

Compute $14A6_{12} - 5B9_{12}$. Give your answer as a base $12$ integer.

Mathematics
1 answer:
PIT_PIT [9.1K]5 days ago
4 0
one must evaluate 12 minus 5B9 as the base and remember to include the integer
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Syd chooses two different primes, both of which are greater than $10,$ and multiplies them. The resulting product is less than $
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The answer is 10. Between 10 and 35, there are 7 prime numbers: 11, 13, 17, 19, 23, 29, and 31. Multiplying 11 by any of the others yields a product smaller than 350, resulting in 6 products. The product of 13 with anything below 26 will also be less than 350, adding 3 more products. Similarly, the product of 17 with anything below 20 yields 1 additional product. Therefore, the total count of different products under 350 amounts to 10.
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4 days ago
Which statements are correct? Check all that apply.
babunello [8423]

Answer:- 1. True. A compass serves the purpose of drawing circles and arcs in geometric constructions.

2. False. A standard geometric construction necessitates the use of a straightedge, compass, and pencil, among other tools, to accurately draw a geometric figure.

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6 0
26 days ago
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A rectangular schoolyard is to be fenced in using the wall of the school for one side and 150
lawyer [9240]

Answer:

Part 1) The equation is A(x)=150x-2x^2

Part 2) When x=40 m, the area of the schoolyard is A=2,800 m^2

Part 3) The valid domain consists of all real numbers exceeding zero and below 75 meters

Step-by-step explanation:

Part 1) Formulate an expression for A(x)

Let

x -----> the length of the rectangular school yard

y ---> the width of the rectangular school yard

It is known that

The perimeter for the fencing (taking the school wall as one side) is

P=2x+y

P=150\ m

thus

150=2x+y

y=150-2x -----> this is equation A

The area of the rectangular school yard is

A=xy ----> this is equation B

Substituting equation A into equation B yields

A=x(150-2x)

A=150x-2x^2

Change to function notation

A(x)=150x-2x^2

Part 2) What is the area when x=40?

With x equal to 40 m

substitute the value into the expression from Part 1 to determine A

A(40)=150(40)-2(40^2)

A(40)=2,800\ m^2    

Part 3) What would be a suitable domain for A(x) in this scenario?

We understand that

A signifies the area of the rectangular school yard

x characterizes the length of the rectangular school yard

It follows that

A(x)=150x-2x^2

This forms a vertical parabola opening downwards

The vertex indicates a maximum point

The x-coordinate of the vertex corresponds to the length that maximizes the area

The y-coordinate of the vertex denotes the maximum area

The vertex corresponds to (37.5, 2812.5)

Refer to the accompanying figure

Consequently,

The peak area achieved is 2,812.5 m^2

The x-intercepts are located at x=0 m and x=75 m

The domain for A is the range -----> (0, 75)

All real numbers greater than zero and less than 75 meters

5 0
1 month ago
Sal is trying to determine which cell phone and service plan to buy for his mother. The first phone costs $100 and $55 per month
AnnZ [9104]

Answer:

a) The first inequality is 100 + 55x > 150 + 51x;

b) The final inequality results in x > 12.5

c) Sal's mother will need to use the second phone for at least 13 months.

Step-by-step explanation:

a) Let x represent the number of months.

1. The first phone is priced at $100, with a monthly fee of $55 for unlimited use, leading to a total cost of $(100 + 55x) for x months.

2. The second phone costs $150 with a monthly fee of $51 for unlimited use, resulting in a total of $(150 + 51x) for x months.

3. For the second phone to be cheaper, we set up the inequality:

150 + 51x < 100 + 55x

which simplifies to

100 + 55x > 150 + 51x

b) Now solve this:

55x - 51x > 150 - 100

4x > 50

so x > 12.5

c) This means Sal's mother has to retain the second phone for at least 13 months (since x > 12.5).

8 0
1 month ago
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