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Novay_Z [31]
3 years ago
13

A type of green paint us made by mixing 2 cups of yellow with 3.5 cups of blue.

Mathematics
2 answers:
AfilCa [17]3 years ago
6 0
<h2>Answer:</h2>

<u>Use 1.75 cups of blue and 1 cup of yellow</u>

<h2>Step-by-step explanation:</h2>

The given problem can be placed in the category of ratios and proportions. There is a ratio of color mixing which contains the proportion of two colors i.e blue and yellow. When we use 2 cups of yellow with 3.5 cups of yellow then we get green color so if we mix half of their amounts then we can get less or simply half amount of color too. Hence adding 1 cup of yellow and 1.75 cup of blue will give us small amount in result.

grigory [225]3 years ago
4 0
Mix 0.5 cups of yellos and 2 cups of blue
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I’m so confused please help I can’t afford another bad grade
iogann1982 [59]

Answer:

2

Step-by-step explanation:

x / (3y)

Let x = 18 and y = 3

18 / ( 3*3)

Determine the denominator first

18 / ( 9)

Divide

2

8 0
3 years ago
Find positive numbers x and y satisfying the equation xyequals12 such that the sum 4xplusy is as small as possible. Let S be the
WITCHER [35]

Answer:

The objective function in terms of one​ number, x is

S(x) = 4x + (12/x)

The values of x and y that minimum the sum are √3 and 4√3 respectively.

Step-by-step explanation:

Two positive numbers, x and y

x × y = 12

xy = 12

S(x,y) = 4x + y

We plan to minimize the sum subject to the constraint (xy = 12)

We can make y the subject of formula in the constraint equation

y = (12/x)

Substituting into the objective function,

S(x,y) = 4x + y

S(x) = 4x + (12/x)

We can then find the minimum.

At minimum point, (dS/dx) = 0 and (d²S/dx²) > 0

(dS/dx) = 4 - (12/x²) = 0

4 - (12/x²) = 0

(12/x²) = 4

4x² = 12

x = √3

y = 12/√3 = 4√3

To just check if this point is truly a minimum

(d²S/dx²) = (24/x³) = (8/√3) > 0 (minimum point)

4 0
3 years ago
Expand the following using the Binomial Theorem and Pascal’s triangle. (x + 2)6 (x − 4)4 (2x + 3)5 (2x − 3y)4 In the expansion o
ivolga24 [154]
\bf (2x+3)^5\implies &#10;\begin{array}{llll}&#10;term&coefficient&value\\&#10;-----&-----&-----\\&#10;1&&(2x)^5(+3)^0\\&#10;2&+5&(2x)^4(+3)^1\\&#10;3&+10&(2x)^3(+3)^2\\&#10;4&+10&(2x)^2(+3)^3\\&#10;5&+5&(2x)^1(+3)^4\\&#10;6&+1&(2x)^0(+3)^5&#10;\end{array}

as you can see, the terms exponents, for the first term, starts at highest, 5 in this case, then every element it goes down by 1, till it gets to 0

for the second term, starts at 0, and every element it goes up by 1, till it gets to the highest

now, to get the coefficient, they way I get it, is "the product of the current coefficient and the exponent of the first term, divided by the exponent of the second term plus 1"

notice the first coefficient is always 1

so...how did we get 10 for the 3rd element?  well, 5*4/2
how did we get 10 for the fourth element?  well, 10*2/4


\bf (2x-3y)^4\implies &#10;\begin{array}{llll}&#10;term&coefficient&value\\&#10;-----&-----&-----\\&#10;1&&(2x)^4(-3y)^0\\&#10;2&+4&(2x)^3(-3y)^1\\&#10;3&+6&(2x)^2(-3y)^2\\&#10;4&+4&(2x)^1(-3y)^3\\&#10;5&+1&(2x)^0(-3y)^4&#10;\end{array}


\bf (3a+4b)^8\implies &#10;\begin{array}{llll}&#10;term&coefficient&value\\&#10;-----&-----&-----\\&#10;1&&(3a)^8(+4b)^0\\&#10;2&+8&(3a)^7(+4b)^1\\&#10;3&+28&(3a)^6(+4b)^2\\&#10;4&+56&(3a)^5(+4b)^3\\&#10;5&+70&(3a)^4(+4b)^4\\&#10;6&+56&(3a)^3(+4b)^5\\&#10;7&+28&(3a)^2(+4b)^6\\&#10;8&+8&(3a)^1(+4b)^7\\&#10;9&+1&(3a)^0(+4b)^8&#10;\end{array}

and from there, you can simplify the elements of the expansion by combining the coefficients

like for example, the 7th element of (3a+4b)⁸ will then be 1032192a²b⁶
7 0
3 years ago
Read 2 more answers
which of the following is q point slope equation of a line that passes through the point (5,2)and (-1,-6)
Salsk061 [2.6K]

Answer:y - y1 = m(x + x1)

m = (y2 - y1)/(x2 - x1) = (-6 - 2)/(-1 - 5) = -8/(-6) = 4/3

y - 2 = 4/3(x - 5) is a possible answer

y + 6 = 4/3(x + 1) is also a possible answer

Step-by-step explanation:

can i be brainliest

4 0
3 years ago
Which series of transformations will not map figure H onto itself
7nadin3 [17]

Answer:

D

Step-by-step explanation:

Given a square with vertices at points (2,1), (1,2), (2,3) and (3,2).

Consider option A.

1st transformation (x+0,y-2) will map vertices of the square into points

  • (2,1)\rightarrow (2,-1);
  • (1,2)\rightarrow (1,0);
  • (2,3)\rightarrow (2,1);
  • (3,2)\rightarrow (3,0).

2nd transformation = reflection over y = 1 has the rule (x,2-y). So,

  • (2,-1)\rightarrow (2,3);
  • (1,0)\rightarrow (1,2);
  • (2,1)\rightarrow (2,1);
  • (3,0)\rightarrow (3,2)

These points are exactly the vertices of the initial square.

Consider option B.

1st transformation (x+2,y-0) will map vertices of the square into points

  • (2,1)\rightarrow (4,1);
  • (1,2)\rightarrow (3,2);
  • (2,3)\rightarrow (4,3);
  • (3,2)\rightarrow (5,2).

2nd transformation = reflection over x = 3 has the rule (6-x,y). So,

  • (4,1)\rightarrow (2,1);
  • (3,2)\rightarrow (3,2);
  • (4,3)\rightarrow (2,3);
  • (5,2)\rightarrow (1,2)

These points are exactly the vertices of the initial square.

Consider option C.

1st transformation (x+3,y+3) will map vertices of the square into points

  • (2,1)\rightarrow (5,4);
  • (1,2)\rightarrow (4,5);
  • (2,3)\rightarrow (5,6);
  • (3,2)\rightarrow (6,5).

2nd transformation = reflection over y = -x + 7 will map vertices into points

  • (5,4)\rightarrow (3,2);
  • (4,5)\rightarrow (2,3);
  • (5,6)\rightarrow (1,2);
  • (6,5)\rightarrow (2,1)

These points are exactly the vertices of the initial square.

Consider option D.

1st transformation (x-3,y-3) will map vertices of the square into points

  • (2,1)\rightarrow (-1,-2);
  • (1,2)\rightarrow (-2,-1);
  • (2,3)\rightarrow (-1,0);
  • (3,2)\rightarrow (0,-1).

2nd transformation = reflection over y = -x + 2 will map vertices into points

  • (-1,-2)\rightarrow (4,3);
  • (-2,-1)\rightarrow (3,4);
  • (-1,0)\rightarrow (2,3);
  • (0,-1)\rightarrow (3,2)

These points are not the vertices of the initial square.

5 0
3 years ago
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