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olga_2 [115]
3 years ago
5

Reflect order pair (-2, 5) over y=-x

Mathematics
2 answers:
vesna_86 [32]3 years ago
8 0

When reflecting over y = -x

The x and y values change places and the signs change.

(-2,5) becomes (-5,2)

ivann1987 [24]3 years ago
5 0

Answer:

(-5,2)

Step-by-step explanation:

hope this helps

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Step-by-step explanation:

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Read 2 more answers
Quick Computing Company produces
olganol [36]

Answer:

c(x) = 2x^2 + 6x + 25

Completed question;

Quick Computing Company produces calculators. They have found that the cost, c(x), of making x calculators is a quadratic function in terms of x. The company also discovered that it costs $45 to produce 2 calculators, $81 to produce 4 calculators, and $285 to produce 10 calculators. Derive the function c(x).

Step-by-step explanation:

Given that;

the cost, c(x), of making x calculators is a quadratic function in terms of x.

c(x) = ax^2 + bx + c

Substituting the 3 case scenarios given;

it costs $45 to produce 2 calculators,

45 = a(2^2) + b(2) + c

45 = 4a + 2b +c .......1

$81 to produce 4 calculators,

81 = a(4^2) + b(4) + c

81 = 16a + 4b + c .......2

and $285 to produce 10 calculators.

285 = a(10^2) + b(10) + c

285 = 100a + 10b + c .......3

Solving the simultaneous equation;

Subtracting equation 1 from 2, we have;

36 = 12a + 2b ......4

Subtracting equation 1 from 3

240 = 96a + 8b .......5

Multiply equation 4 by 4

144 = 48a + 8b ......6

Subtracting equation 6 from 5, we have;

96 = 48a

a = 96/48

a = 2

Substituting a = 2 into equation 4;

36 = 12(2) + 2b

36 = 24 + 2b

2b = 36-24 = 12

b = 12/2 = 6

b = 6

Substituting a and b into equation 1;

45 = 4(2) + 2(6) +c

45 = 8 + 12 + c

c = 45 - (8+12)

c = 25

Since a = 2 , b = 6 and c = 25, the quadratic equation for c(x) is ;

c(x) = ax^2 + bx + c

c(x) = 2x^2 + 6x + 25

5 0
3 years ago
A fair 6-faced die is tossed twice. Letting E and F represent the outcomes of the each toss (which are independent), compute the
andreyandreev [35.5K]

Answer:

(a)  \frac{1}{18}            (d)  \frac{1}{9}

(b) \frac{1}{2}               (e) \frac{1}{3}

(c) \frac{4}{9}               (f)  \frac{1}{12}

Step-by-step explanation:

On tossing a 6-face die twice the outcomes of E and F are:

{1, 2, 3, 4, 5 and 6}

And there are total 36 outcomes of the form (E, F).

(a)

The sample space of getting a sum of 11 is: {(5, 6) and (6, 5)}

The probability  of getting a sum of 11 is:

P(Sum 11) =\frac{2}{36} \\=\frac{1}{18}

(b)

The sample space of getting an even sum is:

{(1, 1), (1, 3), (1, 5), (2, 2), (2, 4), (2, 6) (3, 1), (3, 3), (3, 5), (4, 2), (4, 4), (4, 6) (5, 1), (5, 3), (5, 5), (6, 2), (6, 4), (6, 6)}

The probability  of getting an even sum is:

P(Even Sum)=\frac{18}{36}\\ =\frac{1}{2}

(c)

The sample space of getting an odd sum more than 3 is:

{(1, 4), (1, 6), (2, 3), (2, 5), (3, 2), (3, 4), (3, 6), (4, 1), (4, 3), (4, 5), (5, 2), (5, 4),     (5, 6), (6, 1), (6, 3) and (6, 5)}

The probability of getting an odd sum more than 3 is:

<em>P</em> (Odd Sum more than 3) =

=\frac{16}{36}\\=\frac{4}{9}

(d)

The sample space of (E, F) such that E is even and less than 6, and F is odd and greater than 1 is:

{(2, 3), (2, 5), (4, 3) and (4, 5)}

The probability such that E is even and less than 6, and F is odd and greater than 1 is:

P(Even E1) =\frac{4}{36} \\=\frac{1}{9}

(e)

The sample space of E and F such that E is more than 2, and F is less than 4 is:

E = {3, 4, 5 and 6}     F = {1, 2 and 3}

Then the total outcomes of (E, F) will be 12.

The probability such that E is more than 2, and F is less than 4 is:

P(E>2, F

(f)

The sample space of (E, F) such that E is 4 and sum of E and F is odd is:

{(4, 1), (4, 3) and (4, 5)}

The probability such that E is 4 and sum of E and F is odd is:

P(E=4, E+F = Odd)=\frac{3}{36} \\=\frac{1}{12}

6 0
3 years ago
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