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il63 [147K]
2 years ago
12

Find the equation of the circle whose center is at (2,3) and radius is 6

Mathematics
1 answer:
Advocard [28]2 years ago
7 0

Answer:

Step-by-step explanation:

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11 more than a number q is negative 15 ; Solve for the variable.
Cloud [144]

Answer:

q = -26

Step-by-step explanation:

Set up an equation where q is the unknown number:

q + 11 = -15

Solve for q by subtracting 11 from both sides:

q + 11 = -15

q = -26

4 0
2 years ago
A part selected for testing is equally likely to have been produced on any one of six cutting tools.
JulijaS [17]

Answer:

a)S= (1,2,3,4,5,6)

b) 1/6 or 16.67%

c) 1/3 or 33.33%

d) 5/6 or 83.33%

Step-by-step explanation:

a) Sample space is the list of all the possible sample in this its either tool 1,2,3,4,5 or 6

b) probability of selecting from tool 1 is equally likely to picking from any cutting tool therefore it

1/6 or 16.7%

c) probability of selecting from any tool is 16.7% so probability of selecting from 3 or 5 = 16.67% + 16.67% = 33.3%

d) if the part is not from tool 4 then its from tool 1, or tool 2 or tool 3 or tool 5 or tool 6 therefore its

1/6+1/6+1/6+1/6+1/6 = 5/6 or 8.33%

8 0
3 years ago
Q8 Q18.) How many adults and students should attend to raise the maximum amount of​ money?
Kruka [31]
160 adults and 80 students seems to be correct
5 0
3 years ago
Read 2 more answers
A sugar box manufacturing company is accurately making 100 g packets. Suppose a business researcher randomly selects 100 boxes,
creativ13 [48]

Answer:

...........................................................................

6 0
3 years ago
At most, how many unique roots will a third-degree polynomial function have?
beks73 [17]

The fundamental theorem of algebra states that a polynomial with degree n has at most n solutions. The "at most" depends on the fact that the solutions might not all be real number.

In fact, if you use complex number, then a polynomial with degree n has exactly n roots.

So, in particular, a third-degree polynomial can have at most 3 roots.

In fact, in general, if the polynomial p(x) has solutions x_1,\ x_2,\ldots x_n, then you can factor it as

p(x) = (x-x_1)(x-x_2)\ldots (x-x_n)

So, a third-degree polynomial can't have 4 (or more) solutions, because otherwise you could write it as

p(x)=(x-x_1)(x-x_2)(x-x_3)(x-x_4)

But this is a fourth-degree polynomial.

7 0
3 years ago
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