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2 years ago

Solve: |x+2|=3 Can someone explain how to do this

Mathematics

1 answer:

katovenus [111]2 years ago

6 0

Hey there!

ORIGINAL EQUATION:
|x + 2| = 3

TRANSLATE:

x + 2 = 3 or x + 2 = -3

SOLVING for: x + 2 = 3

SUBTRACT 2 to BOTH SIDES

x + 2 - 2 = 3 - 2

CANCEL out: 2 - 2 because it give you 1

KEEP: 3 - 2 because help solve for the current x-value

NEW EQUATION: x = 3 - 2

SIMPLIFY IT!
x = 1

Thus, your answer is: x = 2

/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\

SOLVING for: x + 2 = -3

SUBTRACT 2 to BOTH SIDES

x + 2 - 2 = -3 - 2

CANCEL out: 2 - 2 because it give you 0

KEEP: -3 - 2 because it help solve for the current x-value

NEW EQUATION: x = -3 - 2

SIMPLIFY IT!
x = -5

Thus, your answer is: x = -5

/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\

Therefore, your answer is: x = 1 or x = -5

Good luck on your assignment & enjoy your day!

~Amphitrite1040:)

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3 years ago

Use mathematical induction to prove that for each integer n > 4,5" > 2^2n+1 + 100.

Flura [38]

Answer:

The inequality that you have is $5^{n}>2^{2n+1}+100,\,n>4$ . You can use mathematical induction as follows:

Step-by-step explanation:

For $n=5$ we have:

$5^{5}=3125$

$2^{(2(5)+1)}+100=2148$

Hence, we have that $5^{5}>2^{(2(5)+1)}+100.$

Now suppose that the inequality holds for $n=k$ and let's proof that the same holds for $n=k+1$ . In fact,

$5^{k+1}=5^{k}\cdot 5>(2^{2k+1}+100)\cdot 5.$

Where the last inequality holds by the induction hypothesis.Then,

$5^{k+1}>(2^{2k+1}+100)\cdot (4+1)$

$5^{k+1}>2^{2k+1}\cdot 4+100\cdot 4+2^{2k+1}+100$

$5^{k+1}>2^{2k+3}+100\cdot 4$

$5^{k+1}>2^{2(k+1)+1}+100$

Then, the inequality is True whenever $n>4$ .

3 0

2 years ago

Please help, thank you!

shutvik [7]

65 x6 = 390
80 x 3= 240

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3 0

3 years ago

Read 2 more answers

a box has a length of 15 centimetres , a width of 22 centimetres, and a height of 9 centimetres.what is the surface area of the

zzz [600]

Answer:

1326 cm²

Step-by-step explanation:

1) Identify the equation of the surface area of a rectangular prism

$A = 2(wl + wh + hl)$

2) Input the corresponding values $l=15 cm$ ; $w=22 cm$ ; and $h=9 cm$ into the equation, and solve for the area

$A=2(wl+wh+hl)\\A=2((22)(15)+(22)(9)+(9)(15))\\A=2(330+198+135)\\A=2(663)\\A=1326 cm^{2}$

3) ∴ $A=1326cm^2$

5 0

1 year ago

What is the quotient in simplified form? State any restrictions on the variable? \frac{x^2-16}{x^2+5x+6} /\frac{x^2+5x+4}{x^2-2x

lora16 [44]

$\frac{x^2-16}{x^2+5x+6} / \frac{x^2+5x+4}{x^2-2x-8}$

We can begin by rearranging this into multiplication:

$\frac{x^2-16}{x^2+5x+6} * \frac{x^2-2x-8}{x^2+5x+4}$

Now we can factor the numerators and denominators:

$\frac{(x+4)(x-4)}{(x+3)(x+2)} * \frac{(x-4)(x+2)}{(x+4)(x+1)}$

The factors (x+4) and (x+2) cancel out, leaving us with:

$\frac{(x-4)}{(x+3)} * \frac{(x-4)}{(x+1)}$

Our answer comes out to be:

$\frac{(x-4)^{2} }{(x+3)(x+1)}$ or $\frac{ x^{2} -8x+16}{ x^{2}+4x+3 }$

Based on the numerator of the second fraction (since we used its inverse), the denominators of both, and the factors we canceled out earlier, the restrictions are x ≠ -4, -3, -2, -1, 4

4 0

2 years ago