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3 years ago

14.What is the solution of the system?

Mathematics

1 answer:

Archy [21]3 years ago

3 0

Answer:

14. x=-2.5, y = -7

15. x=28 y = -20

Step-by-step explanation:

14. Let's solve this system by elimination. Multiply the first equation by -1.

-1*(4x-y)= -1*(- 3)

-4x +y = 3

Then add this to the second equation.

-4x+y = 3

6x-y= - 8

--------------

2x = -5

Divide each side by 2

x = -2.5

We still need to find y

-4x+y =3

-4(-2.5) + y =3

10 +y =3

Subtract 10 from each side.

y = 3-10

y = -7

15.I will again use elimination to solve this system, because using substitution will give me fractions which are harder to work with. I will elimiate the y variable. Multiply the first equation by 11

11( 5x+6y)= 11*20

55x+66y = 220

Multiply the second equation by -6

-6(9x+11y)=32*(-6)

-54x-66y = -192

Add the modified equations together.

55x+66y = 220

-54x-66y = -192

---------------------------

x = 28

We still need to solve for y

5x+6y = 20

5*28 + 6y =20

140 + 6y = 20

Subtract 140 from each side

6y = -120

Divide by 6

y = -20

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4 years ago

How to write this numbers in word 76.68

Ad libitum [116K]

Answer:

seventy-six and sixty-eight hundreths

Step-by-step explanation:

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3 years ago

Read 2 more answers

What are the coordinates of the vertices of the pre-image given? ry= −x ◦ T1, −2(x, y)

nataly862011 [7]

<h3><u>Answer:</u></h3>

Hence, the coordinates of pre-image are:

A(1,6) , B(0,4) , C(3,4) , D(2,6)

<h3><u>Step-by-step explanation:</u></h3>

We have to find the coordinates of the vertices of the pre-image given?

Ry= −x ◦ T_1, −2(x, y)

i.e. we have to find the composition of reflection along the line y=-x and translation with the rule:

(x,y) → (x+1,y-2)

Now we have coordinates of A",B",C" and D" as:

A"(-4,-2)

B"(-2,-1)

C"(-2,-4)

D"(-4,-3)

Now we are asked to find the coordinates of the pre-image.

Also when any point is reflected along y=-x then the point is transformed to:

(x,y) → (-y,-x)

Let A,B,C and D are the points of the pre-image.

Hence, the coordinates of the pre-image are given as:

so, the transformation is given as:

A→A'→A"

B→B'→B"

C→C'→C"

D→D'→D"

Where A',B',C',D' represents the transformation after translation and A",B",C",D" represents the transformation after reflection as well.

The coordinates of A' are (2,4)

B' are (1,2)

C' are (4,2)

and D' are (3,4)

Now, the coordinates of pre-image are given as:

A'(x,y) → A(x-1,y+2)=A(1,6)

B'(x,y) → B(x-1,y+2)=B(0,4)

C'(x,y) → C(x-1,y+2)=C(3,4)

and D'(x,y) → D(x-1,y+2)=D(2,6)

Hence, the coordinates of pre-image are:

A(1,6) , B(0,4) , C(3,4) , D(2,6)

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3 years ago

The 3 by 3 grid below shows nine 1 cm × 1 cm squares and uses 24 cm of wire. What length of wire is required for a similar 20 by

Nady [450]

Answer:

840

Step-by-step explanation:

Hi there!

Usually, this kind of problem asks us the perimeter rather than the area. But when talking on this grid on how many cms are necessary to construct one made up of 9 squares 1cm x1 cm. We need to check one thing: Some line segments will be used twice. So this is not a classical perimeter question.

1) So we either manually and carefully count it avoiding not doing it twice the same common line segments.

Or.

2) By just applying an algorithm, Where n is the <em>nxn</em> grid

$n(n+1)+n(n+1)=Length \:of \:the \:wire$