What is the approximate circumference of a circle with the diameter of 9

Brandon has a budget of $58 to spend on clothes. The shirts he want to buy are on sale $9 each, and a pair of pants he wants cos

Sauron [17]

58 ≤ ($9 + t ) + ($21 + t )

so, n ≤ ($58 + t ) + ($21 + t )

6 0

3 years ago

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Able to help Q39? Thank you

Alex

A)

y varies inversely as (x-1).

Implies y α 1/(x -1)

   y = k/(x-1). Where k is constant of proportionality.

   y*(x - 1) = k

when y = 12, x = 1/2

12*(1/2 - 1) = k

12*(0.5 - 1) = k

k = 12*(-0.5) = -6.

k = -6.

Equation connecting x and y

y*(x - 1) = k, recall k = -6

y*(x - 1) = -6

b)

when x = a.

y*(x - 1) = -6

y*(a - 1) = -6

y = -6 / (a -1)

when x = 2a.

y*(2a - 1) = -6

y = -6 / (2a -1)

Difference in y is 1.8

-6 / (2a -1) - -6/(a - 1) = 1.8

-6 / (2a -1) + 6/(a - 1) = 1.8

6 / (a -1) - 6 /(2a -1) = 1.8

6( 1/(a-1) - 1/(2a - 1)) = 1.8

1/(a-1) - 1/(2a - 1) = 1.8/6

1/(a-1) - 1/(2a - 1) = 0.3

((2a - 1) - (a - 1)) / ((a-1)(2a -1)) = 0.3

(2a - 1 -a + 1) /((a-1)(2a -1)) = 0.3

a / (2a² - 3a + 1) = 0.3

a/0.3 = 2a² - 3a + 1

10a/3 = 2a² - 3a + 1

2a² - 3a + 1 = 10a/3

2a² - 3a -10a/3 + 1 = 0

2a² - 19a/3 + 1 = 0

6a² - 19a + 3 = 0

This is a quadratic expression which can be factored.

6a² - 18a - a + 3 = 0

6a(a - 3) - 1(a - 3) = 0

(6a - 1)(a - 3) = 0

6a - 1 = 0 a = 1/6

a - 3 = 0   a = 3.

a = 3 or 1/6

Since a is positive integer, a = 3 only.

4 0

3 years ago

1 7/8+1/9= A.0 B.1 C.2 D.2 1/2

vichka [17]

Answer

the answer would be C

Step-by-step explanation:

6 0

3 years ago

Please help meee!!!!

denis-greek [22]

Answer:

Step-by-step explanation:

A) (1+2x)-(2-3x)=

3x 1x = 2x

B) (-3+5x)-(-4x-5)

2x 1x = 1x

C) (x+1)-(-4x-5)

1x 1x x

3 0

3 years ago

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Find set A={1, 2, 6, 10} B={3, 6, 9, 10, 11} C = {1, 2, 4, 7, 11}

PilotLPTM [1.2K]

If U = {1, 2, 3, …, 12} is the universal set, and

A = {1, 2, 6, 10}

B = {3, 6, 9, 10, 11}

C = {1, 2, 4, 7, 11}

then

(1) A U B is the set containing all elements from A and B,

A U B = {1, 2, 3, 6, 9, 10, 11}

(2) A ∩ B is the set of elements that are contained in both A and B,

A ∩ B = {6, 10}

(3) Unfortunately, A ∩ B U C is somewhat ambiguous. It could mean (A ∩ B) U C or A ∩ (B U C ). Then either

(A ∩ B) U C = {6, 10} U {1, 2, 4, 7, 11} = {1, 2, 4, 6, 7, 10, 11}

or

A ∩ (B U C ) = {1, 2, 6, 10} ∩ {1, 2, 3, 4, 6, 7, 9, 10, 11} = {1, 2, 6, 10}

The first interpretation is probably the intended one, since that essentially reads the set operations from left to right.

(4) A' U B is the union of A' and B, where A' is the complement of A, or all elements in U that are not in A. We have

A' = U - A = {3, 4, 5, 7, 8, 9, 11, 12}

and so

A' U B = {3, 4, 5, 6, 7, 8, 9, 10, 11, 12}

(5) We have

A U C = {1, 2, 4, 6, 7, 10, 11}

so that

(A U C )' = U - (A U C ) = {3, 5, 8, 9, 12}

(6) We have

B' = U - B = {1, 2, 4, 5, 7, 8, 12}

and so

A ∩ B' = {1, 2}

(7) Using the complements found in (4) and (6), we have

A' U B' = {1, 2, 3, 4, 5, 7, 8, 9, 11, 12}

Alternatively, we can use the fact that

A' U B' = (A ∩ B)'

and since we know from (2) that A ∩ B = {6, 10}, we end up with the same result,

(A ∩ B)' = U - (A ∩ B) = {1, 2, 3, 4, 5, 7, 8, 9, 11, 12}

(8) We have

A U B U C = {1, 2, 3, 4, 6, 7, 9, 10, 11}

so that

(A U B U C )' = {5, 8, 12}

6 0

3 years ago