Solve the equation –(–7y

Mekhanik [1.2K]

Answer:

Mass
Surface area
Volume

Step-by-step explanation:

The size of a surface. The amount of space inside the boundary of a flat (2-dimensional) object such as a triangle or circle, or surface of a solid (3-dimensional) object

Mass

A property of a physical body and a measure of its resistance to acceleration when a net force is applied.

Surface

The area is the sum of the areas of all faces (or surfaces) on a 3D shape. A cuboid has 6 rectangular faces. To find the surface area of a cuboid, add the areas of all 6 faces. We can also label the length (l), width (w), and height (h) of the prism and use the formula, SA=2lw+2lh+2hw, to find the surface area.

Volume

The Measure of the amount of space that a substance or an object takes up.

8 0

2 years ago

1/2x+y=10 solving literal equation

hram777 [196]

Step-by-step explanation:

1 x-y = 10

2

multiply by 2

2* (1 x+y) =10

2

x+y=20

5 0

2 years ago

❊ Simplify :

DiKsa [7]

Answer:

See Below.

Step-by-step explanation:

Problem 1)

We want to simplify:

$\displaystyle \frac{a+2}{a^2+a-2}+\frac{3}{a^2-1}$

First, let's factor the denominators of each term. For the second term, we can use the difference of two squares. Hence:

$\displaystyle =\frac{a+2}{(a+2)(a-1)}+\frac{3}{(a+1)(a-1)}$

Now, create a common denominator. To do this, we can multiply the first term by (a + 1) and the second term by (a + 2). Hence:

$\displaystyle =\frac{(a+2)(a+1)}{(a+2)(a-1)(a+1)}+\frac{3(a+2)}{(a+2)(a-1)(a+1)}$

Add the fractions:

$\displaystyle =\frac{(a+2)(a+1)+3(a+2)}{(a+2)(a-1)(a+1)}$

Factor:

$\displaystyle =\frac{(a+2)((a+1)+3)}{(a+2)(a-1)(a+1)}$

Simplify:

$\displaystyle =\frac{a+4}{(a-1)(a+1)}$

We can expand. Therefore:

$\displaystyle =\frac{a+4}{a^2-1}$

Problem 2)

We want to simplify:

$\displaystyle \frac{1}{(a-b)(b-c)}+\frac{1}{(c-b)(a-c)}$

Again, let's create a common denominator. First, let's factor out a negative from the second term:

$\displaystyle \begin{aligned} \displaystyle &= \frac{1}{(a-b)(b-c)}+\frac{1}{(-(b-c))(a-c)}\\\\&=\displaystyle \frac{1}{(a-b)(b-c)}-\frac{1}{(b-c)(a-c)}\\\end{aligned}$