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3 years ago

Compare the line passing through the points (—2, —9) and (4, 6) to the line given by the equation y = 2/5x - 4

Mathematics

1 answer:

Romashka-Z-Leto [24]3 years ago

6 0

Answer:Option D

Step-by-step explanation:Can't be asked to explain.

Hope it helps :)

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The perimeter of an equilateral triangle is 6 inches more than the perimeter of a square, and the side of the triangle is 4 inch

soldier1979 [14.2K]

Answer:

The length of the side of the triangle is 10 inches.

Step-by-step explanation:

Let p = perimeter of the equilateral triangle

Let P = perimeter of the square

Let s = length of side of the triangle

Let S = length of side of the square

"The perimeter of an equilateral triangle is 6 inches more than the perimeter of a square"

p = P + 6 Equation 1

"the side of the triangle is 4 inches longer than the side of the square"

s = S + 4 Equation 2

We have 2 equations and 4 unknowns. We need two more equations. We use the definition of perimeter to get the other two equations.

For an equilateral triangle,

p = 3s Equation 3

For a square,

P = 4S Equation 4

Substitute p and P of Equation 1 with equations 3 and 4. Then write equation 2.

3s + 4S = 6

s = S + 4

Now we have a system of 2 equations in 2 unknowns. We can solve for s and S. We can use the substitution method. Solve the second equation for S.

S = 4 - s

Substitute S = 4 - s into equation 3s + 4S = 6.

3s + 4(4 - s) = 6

3s + 16 - 4s = 6

-s = -10

s = 10

Answer: The length of the side of the triangle is 10 inches.

7 0

3 years ago

The table shows x-values going in and y-values coming out. The function being used is

strojnjashka [21]

F(X) = 2x-8

I hope it helped you

3 0

3 years ago

Read 2 more answers

A square has a side length of 3 inches. What is the length of the diagonal distance across the square ?

astraxan [27]

Answer:

4.243

Step-by-step explanation:

To calculate the diagonal of a square, multiply the length of the side by the square root of 2:

d = a√2

d = 3√2=4.24264068

7 0

3 years ago

A year on mars is 1.88 times as long as a year n Earth.An Earth year lasts 365.3 days.Find the legth of a year on mars.

Kazeer [188]

I got 686.764 days for a year on Mars your welcome

8 0

3 years ago

14, 16, and 20 using elimination method showing work. Thanks so much

Nady [450]

14) x=0, y=3, z=-2

Solution Set (0,3,-2)

16) x=1, y=1 and z=1

Solution set = (1,1,1)

20) x = -263/31, y=164/31 ,z=122/31

Solution set (-263/31, 164/31 ,122/31)

Step-by-step explanation:

14)

$x-y+2z=-7\\y+z=1\\x=2y+3z$

Rearranging and solving:

$x-y+2z=-7\,\,\,eq(1)\\y+z=1\,\,\,eq(2)\\x-2y-3z=0\,\,\,eq(3)$

Eliminate y:

Adding eq(1) and eq(2)

$x-y+2z=-7\,\,\,eq(1)\\ 0x+y+z=1\,\,\,eq(2)\\-------\\x+3z=-6\,\,\,eq(4)$

Multiply eq(2) with 2 and add with eq(3)

$0x+2y+2z=2\,\,\,eq(2)\\\\x-2y-3z=0\,\,\,eq(3)\\--------\\x-z=2\,\,\,eq(5)$

Eliminate x:

Subtract eq(4) and eq(5)

$x+3z=-6\,\,\,eq(4)\\x-z=2\,\,\,eq(5)\\-\,\,\,+\,\,\,\,\,\,-\\---------\\4z=-8\\z= -2$

So, value of z = -2

Now putting value of z in eq(2)

$y+z=1\\y+(-2)=1\\y-2=1\\y=1+2\\y=3$

So, value of y = 3

Now, putting value of z and y in eq(1)

$x-y+2z=-7\\x-(3)+2(-2)=-7\\x-3-4=-7\\x-7=-7\\x=-7+7\\x=0$

So, value of x = 0

So, x=0, y=3, z=-2

S.S(0,3,-2)

16)

$3x-y+z=3\\\x+y+2z=4\\x+2y+z=4$

Let:

$3x-y+z=3\,\,\,eq(1)\\x+y+2z=4\,\,\,eq(2)\\x+2y+z=4\,\,\,eq(3)$

Eliminating y:

Adding eq(1) and (2)

$3x-y+z=3\,\,\,eq(1)\\x+y+2z=4\,\,\,eq(2)\\---------\\4x+3z=7\,\,\,eq(4)$

Multiply eq(1) by 2 and add with eq(3)

$6x-2y+2z=6\,\,\,eq(1)\\x+2y+z=4\,\,\,eq(3)\\---------\\7x+3z=10\,\,\,eq(5)$

Now eliminating z in eq(4) and eq(5) to find value of x

Subtracting eq(4) and eq(5)

$4x+3z=7\,\,\,eq(4)\\7x+3z=10\,\,\,eq(5)\\-\,\,\,-\,\,\,\,\,\,\,\,\,\,-\\-----------\\-3x=-3\\x=-3/-3\\x=1$

So, value of x = 1

Putting value of x in eq(4) to find value of x:

$4x+3z=7\\4(1)+3z=7\\4+3z=7\\3z=7-4\\z=3/3\\z=1$

So, value of z = 1

Putting value of x and z in eq(2) to find value of y:

$x+y+2z=4\\1+y+2(1)=4\\1+y+2=4\\y+3=4\\y=4-3\\y=1$

So, x=1, y=1 and z=1

Solution set = (1,1,1)

20)

$x+4y-5z=-7\\3x+2y+2z=-7\\2x+y+5z=8$

Let:

$x+4y-5z=-7\,\,\,eq(1)\\3x+2y+2z=-7\,\,\,eq(2)\\2x+y+5z=8\,\,\,eq(3)$

Solving:

Eliminating z :

Adding eq(1) and eq(3)

$x+4y-5z=-7\,\,\,eq(1)\\2x+y+5z=8\,\,\,eq(3)\\---------\\3x+5y=1\,\,\,eq(4)$

Multiply eq(1) with 2 and eq(2) with 5 and add:

$2x+8y-10z=-14\,\,\,eq(1)\\15x+10y+10z=-35\,\,\,eq(2)\\----------\\17x+18y=-49\,\,\,eq(5)$

Eliminate y:

Multiply eq(4) with 18 and eq(5) with 5 and subtract:

$54x+90y=18\\85x+90y=-245\\-\,\,\,-\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+\\-------\\-31x=158\\x=-\frac{263}{31}$

So, value of x = -263/31

Putting value of x in eq(4)

$3x+5y=1\\3(-\frac{263}{31})+5y=1\\-\frac{789}{31}+5y=1 \\5y=1+\frac{789}{31}\\5y=\frac{820}{31}\\y=\frac{820}{31*5}\\y=\frac{164}{31}$

Now putting x = -263/31 and y=164/31 in eq(1) and finding z:

We get z=122/31

So, x = -263/31, y=164/31 ,z=122/31

Solution set (-263/31, 164/31 ,122/31)

Keywords: Solving system of Equations

Learn more about Solving system of Equations at:

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4 0

3 years ago