Solve: (x+10) - (x-1)

Mathematics

1 answer:

Igoryamba3 years ago

8 0

(x+10) - (x-1)=
(x-x) - (10-(-1)=
0-11=-11

Your answer is -11

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Kipish [7]

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3 years ago

What value of x makes the equation true? 3(4x-5)-4x+1=-6 Please show work

V125BC [204]

Answer:

x = 1

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Equality Properties

Step-by-step explanation:

Step 1: Define Equation

3(4x - 5) - 4x + 1 = -6

Step 2: Solve for x

Distribute 3: 12x - 15 - 4x + 1 = -6
Combine like terms: 8x - 14 = -6
Isolate x term: 8x = 8
Isolate x: x = 1

Step 3: Check

Plug in x into the original equation to verify it's a solution.

Substitute in x: 3(4(1) - 5) - 4(1) + 1 = -6
Multiply: 3(4 - 5) - 4 + 1 = -6
Subtract: 3(-1) - 4 + 1 = -6
Multiply: -3 - 4 + 1 = -6
Subtract: -7 + 1 = -6
Add: -6 = -6

Here we see that -6 does indeed equal -6.

∴ x = 1 is the solution to the equation.

3 0

2 years ago

Suppose a parabola has vertex (6,5) and also passes through the point (7,7). Write the equation of the parabola in vertex form.

jek_recluse [69]

Answer:

Choice B: $y = 2\, (x - 6)^{2} + 5$ .

Step-by-step explanation:

For a parabola with vertex $(h,\, k)$ , the vertex form equation of that parabola in would be:

$\text{$y = a\, (x - h)^{2} + k$ for some constant $a$}$ .

In this question, the vertex is $(6,\, 5)$ , such that $h = 6$ and $k = 5$ . There would exist a constant $a$ such that the equation of this parabola would be:

$y = a\, (x - 6)^{2} + 5$ .

The next step is to find the value of the constant $a$ .

Given that this parabola includes the point $(7,\, 7)$ , $x = 7$ and $y = 7$ would need to satisfy the equation of this parabola, $y = a\, (x - 6)^{2} + 5$ .