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Lelu [443]
2 years ago
9

If a number is divisible by both 2 and 3 then we can say the number is divisible by A. 5. B. 2. C. 4. D. 6.

Mathematics
2 answers:
Elza [17]2 years ago
7 0
I believe it is d 6. you can divide 6 by both 3 and 2.
Komok [63]2 years ago
7 0
D. 6 because if one number is divisible by another number's factors, then the first number is divisible by the second number. for example, take the numbers 18 and 6.  the factors of 6 are 2 and 3. 18 is divisible by both 2 and 3, therefore it has to be divisible by 6.<span />
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The length of a rectangle is 3 inches greater than the width.
kati45 [8]

Answer:

A - x (x+3) = Area of the rectangle

B- 28 inches squared

Step-by-step explanation:

B- using the formula from A

x = 4

x + 3 = 4 +3 = 7

4 times 7 is 28

4 0
3 years ago
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Write the equation of the lines resented below in slope-intercept form
scoray [572]

Slope intercept form: y = mx + b

mx = slope

b = y-intercept

We know the y intercept is 0, so nothing will be written there.

To find the slope of this line, we can use the slope formula.

\frac{y.2 - y.1}{x.2 - x.1} = m

We'll use the points (1, 0) and (3, 1) to find the slope.

Now we can just plug these values into the equation to find the slope.

1 - 0 / 3 - 1

1 / 2

The slope of the line is 1/2, or 0.5.

The slope-intercept form of this line can be written as:

y = 0.5x

6 0
3 years ago
What are the types of roots of the equation below?<br> - 81=0
Tju [1.3M]

Option B, that is Two Complex and Two Real which are x + 3, x - 3, x + 3i and x - 3i, are the types of roots of the equation x⁴ - 81 = 0. This can be obtained by finding root of the equation using algebraic identity.    

<h3>What are the types of roots of the equation below?</h3>

Here in the question it is given that,

  • the equation x⁴ - 81 = 0

By using algebraic identity, (a + b)(a - b) = a² - b², we get,  

⇒ x⁴ - 81 = 0                      

⇒ (x² +  9)(x² - 9) = 0

⇒ (x² + 9)(x² - 9) = 0

  1. (x² -  9) = (x² - 3²) = (x - 3)(x + 3) [using algebraic identity, (a + b)(a - b) = a² - b²]
  2. x² + 9 = 0 ⇒ x² = -9 ⇒ x = √-9 ⇒ x= √-1√9 ⇒x = ± 3i

⇒ (x² + 9) = (x - 3i)(x + 3i)

Now the equation becomes,

[(x - 3)(x + 3)][(x - 3i)(x + 3i)] = 0

Therefore x + 3, x - 3, x + 3i and x - 3i are the roots of the equation

To check whether the roots are correct multiply the roots with each other,

⇒ [(x - 3)(x + 3)][(x - 3i)(x + 3i)] = 0

⇒ [x² - 3x + 3x - 9][x² - 3xi + 3xi - 9i²] = 0

⇒ (x² +0x - 9)(x² +0xi - 9(- 1)) = 0

⇒ (x² - 9)(x² + 9) = 0

⇒ x⁴ - 9x² + 9x² - 81 = 0

⇒ x⁴ - 81 = 0

Hence Option B, that is Two Complex and Two Real which are x + 3, x - 3, x + 3i and x - 3i, are the types of roots of the equation x⁴ - 81 = 0.

Disclaimer: The question was given incomplete on the portal. Here is the complete question.

Question: What are the types of roots of the equation below?

x⁴ - 81 = 0

A) Four Complex

B) Two Complex and Two Real

C) Four Real

Learn more about roots of equation here:

brainly.com/question/26926523

#SPJ9

5 0
1 year ago
What is the value of y in the solution to the system of equations?
Yuliya22 [10]

Answer: y = 8

Step-by-step explanation:

8 0
2 years ago
Read 2 more answers
Let f(x)=5x3−60x+5 input the interval(s) on which f is increasing. (-inf,-2)u(2,inf) input the interval(s) on which f is decreas
o-na [289]
Answers:

(a) f is increasing at (-\infty,-2) \cup (2,\infty).

(b) f is decreasing at (-2,2).

(c) f is concave up at (2, \infty)

(d) f is concave down at (-\infty, 2)

Explanations:

(a) f is increasing when the derivative is positive. So, we find values of x such that the derivative is positive. Note that

f'(x) = 15x^2 - 60&#10;

So,

&#10;f'(x) \ \textgreater \  0&#10;\\&#10;\\ \Leftrightarrow 15x^2 - 60 \ \textgreater \  0&#10;\\&#10;\\ \Leftrightarrow 15(x - 2)(x + 2) \ \textgreater \  0&#10;\\&#10;\\ \Leftrightarrow \boxed{(x - 2)(x + 2) \ \textgreater \  0} \text{   (1)}

The zeroes of (x - 2)(x + 2) are 2 and -2. So we can obtain sign of (x - 2)(x + 2) by considering the following possible values of x:

-->> x < -2
-->> -2 < x < 2
--->> x > 2

If x < -2, then (x - 2) and (x + 2) are both negative. Thus, (x - 2)(x + 2) > 0.

If -2 < x < 2, then x + 2 is positive but x - 2 is negative. So, (x - 2)(x + 2) < 0.
 If x > 2, then (x - 2) and (x + 2) are both positive. Thus, (x - 2)(x + 2) > 0.

So, (x - 2)(x + 2) is positive when x < -2 or x > 2. Since

f'(x) \ \textgreater \  0 \Leftrightarrow (x - 2)(x + 2)  \ \textgreater \  0

Thus, f'(x) > 0 only when x < -2 or x > 2. Hence f is increasing at (-\infty,-2) \cup (2,\infty).

(b) f is decreasing only when the derivative of f is negative. Since

f'(x) = 15x^2 - 60

Using the similar computation in (a), 

f'(x) \ \textless \  \ 0 \\ \\ \Leftrightarrow 15x^2 - 60 \ \textless \  0 \\ \\ \Leftrightarrow 15(x - 2)(x + 2) \ \ \textless \  0 \\ \\ \Leftrightarrow \boxed{(x - 2)(x + 2) \ \textless \  0} \text{ (2)}

Based on the computation in (a), (x - 2)(x + 2) < 0 only when -2 < x < 2.

Thus, f'(x) < 0 if and only if -2 < x < 2. Hence f is decreasing at (-2, 2)

(c) f is concave up if and only if the second derivative of f is positive. Note that

f''(x) = 30x - 60

Since,

f''(x) \ \textgreater \  0&#10;\\&#10;\\ \Leftrightarrow 30x - 60 \ \textgreater \  0&#10;\\&#10;\\ \Leftrightarrow 30(x - 2) \ \textgreater \  0&#10;\\&#10;\\ \Leftrightarrow x - 2 \ \textgreater \  0&#10;\\&#10;\\ \Leftrightarrow \boxed{x \ \textgreater \  2}

Therefore, f is concave up at (2, \infty).

(d) Note that f is concave down if and only if the second derivative of f is negative. Since,

f''(x) = 30x - 60

Using the similar computation in (c), 

f''(x) \ \textless \  0 &#10;\\ \\ \Leftrightarrow 30x - 60 \ \textless \  0 &#10;\\ \\ \Leftrightarrow 30(x - 2) \ \textless \  0 &#10;\\ \\ \Leftrightarrow x - 2 \ \textless \  0 &#10;\\ \\ \Leftrightarrow \boxed{x \ \textless \  2}

Therefore, f is concave down at (-\infty, 2).
3 0
3 years ago
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