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4 years ago

Write in function form X+y=6..?

Mathematics

2 answers:

Ksju [112]4 years ago

6 0

$x+y=6\ \ \ \ /-x\\\\y=-x+6\\\\for\ x=0\to y=-0+6=6\to(0;\ 6)\\\\for\ x=6\to y=-6+6=0\to(6;\ 0)$

Alik [6]4 years ago

5 0

Function form being slope-intercept?

y=-x+6

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When a softball player gets a hit, it can be a single, double, triple, or home run. When Josephine is at bat, the protectity

Fofino [41]

Answer:

The probability of hitting a single or a double is 1/5 or 20% or 20/100 or 0.2

Step-by-step explanation:

In probability, whenever we are to answer an ‘or’ question, we add up the probabilities involved.

The probability of hitting a single is 15%, that is same as 15/100 or just simply 0.15

The probability of hitting a double is 5%, that is simply 5/100 or just simply 0.05

The probability of hitting a single or a double = Probability of hitting a single + Probability of hitting a double = 0.15 + 0.05 = 0.20 or 20/100 or 1/5

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3 years ago

The width if a rectangle is 4ft and the length is unkown. If the area is the expression 12x + 24 what is the unkown length?

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Answer:

Length=6 ft

Step-by-step explanation:

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3 years ago

27.36 divided by 3.14 use compatible numbers to estimate the quotient

stiks02 [169]

Answer:

Step-by-step explanation:

you can round the numbers

27/3= 9

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6 0

4 years ago

Read 2 more answers

What does x stand for in 4+x=3x

Irina18 [472]

x stands for a variable, in which the variable would be replaced by a number(s) that would make the equation true.

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4 years ago

Which series of transformations will not map figure H onto itself

7nadin3 [17]

Answer:

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.

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3 years ago