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3 years ago

Solve x:(x+4)/3=2 Will give points

2 answers:

8 0

$\frac{x + 4}{3} = 2 \\ \\ \\ \text{ Multiply by 3 on both sides, }\\ \\ \\ = > \frac{x + 4}{3} \times 3 = 2 \times 3 \\ \\ \\ \\ = > x + 4 = 6 \\ \\ \\ = > x = 6 - 4 \\ \\ \\ = > x = 2$

Hence, value of x in the given equation is 2

Oksana_A [137]3 years ago

6 0

(X+4)/3=2
-4. -4
X/3=(-2)
x3. x3
X=6

Step one: isolate x
- take away 4. What you do to one side, you must do to the other. This leaves you with [x/3=(-2)] because 2-4=(-2)
- multiply by 3. This gets rid of the three so x is by itself. Once again, what you do to one side of the equal sign, you must do to the other. It leaves you with [x=6] because 3x(-2)=6

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If the length of diagonal of a square is 4√2 cm, find it's length, perimeter and area.

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Answer:

As Per Provided Information

Length of diagonal of square is 4√2 cm

We have been asked to find the length , perimeter and area of square .

First let's calculate the side of square .

Using Formulae

$\boxed{\bf \:Diagonal_{(Square)} \: = side \sqrt{2}}$

On substituting the value in above formula we obtain

$\qquad\longrightarrow\sf \:4 \sqrt{2} = side \sqrt{2} \\ \\ \\ \qquad\longrightarrow\sf \:4 \cancel{\sqrt{2}} = side \cancel{ \sqrt{2}} \\ \\ \\ \qquad\longrightarrow\sf \:side \: = 4 \: cm$

Therefore,

Length of its side is 4 cm.

Finding the perimeter of square.

$\boxed{\bf \: Perimeter_{(Square)} = 4 \times side}$

Substituting the value we obtain

$\qquad\longrightarrow\sf \:Perimeter_{(Square)} \: = 4 \times 4 \\ \\ \\ \qquad\longrightarrow\sf \:Perimeter_{(Square)} = 16 \: cm$

Therefore,

Perimeter of square is 16 cm .

Finding the area of square .

$\boxed{\bf \: Area_{(Square)} = {side}^{2}}$

Substituting the value we get

$\qquad\longrightarrow\sf \:Area_{(Square)} \: = {4}^{2} \\ \\ \\ \qquad\longrightarrow\sf \:Area_{(Square)} = 16 \: {cm}^{2}$

Therefore,

Area of square is 16 cm².

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Which of the following is an improper integral?

guapka [62]

Answer:

A) $\displaystyle \int\limits^3_0 {\frac{x + 1}{3x - 2}} \, dx$

General Formulas and Concepts:

Calculus

Discontinuities

Removable (Hole)
Jump
Infinite (Asymptote)

Integration

Integrals
Definite Integrals
Integration Constant C
Improper Integrals

Step-by-step explanation:

Let's define our answer choices:

A) $\displaystyle \int\limits^3_0 {\frac{x + 1}{3x - 2}} \, dx$

B) $\displaystyle \int\limits^3_1 {\frac{x + 1}{3x - 2}} \, dx$

C) $\displaystyle \int\limits^0_{-1} {\frac{x + 1}{3x - 2}} \, dx$

D) None of these

We can see that we would have a infinite discontinuity if x = 2/3, as it would make the denominator 0 and we cannot divide by 0. Therefore, any interval that includes the value 2/3 would have to be rewritten and evaluated as an improper integral.

Of all the answer choices, we can see that A's bounds of integration (interval) includes x = 2/3.

∴ our answer is A.

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Integration

Book: College Calculus 10e

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