If one coordinate is zero, then the point is located on the x-axis or the y-axis. If y coordinate is zero, then the point is located on the y-axis. If both coordinates are zero, then the point represents the origin.

8 0

3 years ago

X-y=6 2x-3z=16 2y+z=4 Solve the systems of equations.

Neko [114]

Answer:

Given System of equation:

x-y =6 .....,[1]

2x-3z = 16 ......[2]

2y+z = 4 .......[3]

Rewrite the equation [1] as

y = x - 6 .......[4]

Substitute the value of [4] in [3], we get

$2(x-6)+z = 4$

Using distributive property on LHS ( i.e, $a \cdot (b+c) =a \cdot b+ b \cdot c$ )

then, we have

2x - 12 +z =4

Add 12 to both sides of an equation:

2x-12+z+12=4+12

Simplify:

2x +z = 16 .......[5]

On substituting equation [2] in [5] we get;

2x+z=2x -3z

z = -3z

Add 3z both sides of an equation:

z+3z = -3z+3z

4z = 0

Simplify:

z = 0

Substitute the value of z = 0 in [2] to solve for x;

$2x-3(0) = 16$

2x = 16

Divide by 2 both sides of an equation:

$\frac{2x}{2} =\frac{16}{2}$

Simplify:

x= 8

Substitute the value of x =8 in equation [4] to solve for y;

y = 8-6 = 2

y = 2

Therefore, the solution for the given system of equation is; x = 8 , y = 2 and z =0

5 0

3 years ago

What value of n makes the equation true? (2x^9y^n)(4x^2y^10)=8x^11y^20

Komok [63]

Hi there! The answer is n = 10.

$(2 {x}^{9} {y}^{n} )(4 {x}^{2} {y}^{10} ) = 8 {x}^{11} {y}^{20}$

As you see at the powers of x, we need to add the exponents of the power we when multiply them.
${x}^{9} \times {x}^{2} = {x}^{9 + 2} = x {}^{11}$

The powers of y work the same way.
${y}^{n} \times {y}^{10} = {y}^{n + 10} = {y}^{20}$

Hence, n = 10, since
${y}^{10 + 10} = {y}^{20}$

8 0

3 years ago

Is 1.75 larger than 6.2

Dominik [7]

No 6.2 could be put as 6.20 and 1.75

6 is bigger than 1

8 0

3 years ago

Read 2 more answers