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3 years ago

What are 5 consecutive even integers that add up to 0

2 answers:

7 0

$2n+2n+2+2n+4+2n+6+2n+8=0\\
10n+20=0\\
10n=-20\\
n=-2\\\\
2n=-4\\
2n+2=-2\\
2n+4=0\\
2n+6=2\\
2n+8=4\\\\
\boxed{\{-4,-2,0,2,4\}}$

ale4655 [162]3 years ago

4 0

$2a,2a+2,2a+4,2a+6,2a+8\\\\
2a+2a+2+2a+4+2a+6+2a+8=0\\\\
10a+20= 0\ \ \ \ | subtract\ 20\\\\
10a=-20\ \ \ \ | divide\ by\ 10\\\\
a=-2\\\\
Numers\ are: \ -2,0,2,4,6.$

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How do you do 5+5?

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Answer:

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Problem 4:

Kipish [7]

Answer:

12cm and 16cm

Step-by-step explanation:

The hypotenuse of the right angle triangle = 20cm

let the other two sides be x and y;

The difference;

x = y - 4

Problem:

Find x and y;

Solution:

According to the pythagoras theorem;

x² + y² = 20² ------ i

x² + y² = 400 ----- i

and x - y = 4 ---- ii

So; x = 4 + y

Now input the value into equation (i);

(y + 4)² + y² = 400

(y+4)(y+4) + y² = 400

y² + y² + 4y + 4y + 16 = 400

2y² + 8y + 16 = 400

2y² + 8y + 16 -400 = 0

2y² + 8y - 384 = 0;

y² + 4y - 192 = 0

Factorize the equation;

y² + 16y - 12y - 192 = 0

y(y + 16) - 12(y + 16) = 0

(y-12)(y + 16) = 0

y -12 = 0 or y+ 16 = 0

y = 12 or -16

It is not realistic for the length of a body to be a negative value, so y= 12;

since;

x - y = 4,

x = 4 + 12

x = 16

5 0

3 years ago

Find the number of real number solutions for the equation. x2 – 18 = 0

madam [21]

x ^ 2 - 18 = 0
For this case, the first thing you should do is rewrite the expression:
x ^ 2 - 18 = 0
x ^ 2 = 18
The solutions for this case are:
x1 = root (18)
x2 = -raiz (18)
Therefore the answer is:
2 solutions of real numbers.
Answer:
2

6 0

3 years ago

Read 2 more answers

Xto the 2nd power - 49 = 0

AnnZ [28]

Answer:

7 to the second power... because 7 times is 49 and 49-49 is 0... so 7 to the second power

6 0

3 years ago

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
$

So,

$
f'(x) \ \textgreater \ 0
\\
\\ \Leftrightarrow 15x^2 - 60 \ \textgreater \ 0
\\
\\ \Leftrightarrow 15(x - 2)(x + 2) \ \textgreater \ 0
\\
\\ \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,

$f''(x) \ \textgreater \ 0
\\
\\ \Leftrightarrow 30x - 60 \ \textgreater \ 0
\\
\\ \Leftrightarrow 30(x - 2) \ \textgreater \ 0
\\
\\ \Leftrightarrow x - 2 \ \textgreater \ 0
\\
\\ \Leftrightarrow \boxed{x \ \textgreater \ 2}$

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),

$f''(x) \ \textless \ 0 
\\ \\ \Leftrightarrow 30x - 60 \ \textless \ 0 
\\ \\ \Leftrightarrow 30(x - 2) \ \textless \ 0 
\\ \\ \Leftrightarrow x - 2 \ \textless \ 0 
\\ \\ \Leftrightarrow \boxed{x \ \textless \ 2}$

Therefore, f is concave down at $(-\infty, 2)$ .

3 0

3 years ago