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3 years ago

6

What is the sum of (7x-6y) and (x + y)

1 answer:

melomori [17]3 years ago

7 0

(7x-6y)+(x+y)
7x+x-6y+y
You want to group like terms together like the two x's and then the two y's

Answer:8x-5y

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If anyone can i help me

natulia [17]

Answer:

x=9.4

x=9.43

Step-by-step explanation:

I used the Pythagorean theorem.

Have a nice day!

and can I maybe have brainliest? Thanks!

3 0

3 years ago

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igomit [66]

Follow the rules of PEMDAS:

42/(15-8)
42/7
6

3 0

3 years ago

Find all points on the x-axis that are 14 units from the point (4, -7).

BaLLatris [955]

Answer:

The points are: (16.12,0),(-8.12,0).

Step-by-step explanation:

Solving a quadratic equation:

Given a second order polynomial expressed by the following equation:

$ax^{2} + bx + c, a\neq0$ .

This polynomial has roots $x_{1}, x_{2}$ such that $ax^{2} + bx + c = a(x - x_{1})*(x - x_{2})$ , given by the following formulas:

$x_{1} = \frac{-b + \sqrt{\bigtriangleup}}{2*a}$

$x_{2} = \frac{-b - \sqrt{\bigtriangleup}}{2*a}$

$\bigtriangleup = b^{2} - 4ac$

Distance between two points:

Suppose we have two points, $(x_1,y_1)$ and $(x_2,y_2)$ . The distance between them is given by:

$D = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Find all points on the x-axis that are 14 units from the point (4, -7).

Being on the x-axis mean that they have y-coordinate equal to 0, so the point is (x,0).

The distance is 14. So

$\sqrt{(x-4)^2+(0-(-7))^2} = 14$

$\sqrt{x^2 - 8x + 16 + 49} = 14$

$\sqrt{x^2 - 8x + 65} = 14$

$(\sqrt{x^2 - 8x + 65})^2 = 14^2$

$x^2 - 8x + 65 - 196 = 0$

$x^2 - 8x - 131 = 0$

So $a = 1, b = -8, c = -131$

$\bigtriangleup = (-8)^{2} - 4(1)(-131) = 588$

$x_{1} = \frac{-(-8) + \sqrt{588}}{2} = 16.12$

$x_{1} = \frac{-(-8) - \sqrt{588}}{2} = -8.12$

The points are: (16.12,0),(-8.12,0).

4 0

3 years ago

How do i rearrange a(x-y)=b(z-x) to make x the subject?

Ket [755]

Answer:here's ur answer

Step-by-step explanation:

4 0

3 years ago

Write the expanded form, for 34.75

ira [324]

Answer:

30

Step-by-step explanation:

6 0

3 years ago

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