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Mathematics

2 answers:

Kisachek [45]2 years ago

7 0

Answer:

$8 .9 \sqrt{72.5} = 8.15 \\ \\ 1.2 \sqrt{33.2} \div 27.67 \\ \\ 7.9 \sqrt{74.3 } = 9.41 \\ \\ 3.2 \sqrt{35.5} = 11.09 \\ \\ 3.7 \sqrt{28.1} = 7.59$

Fittoniya [83]2 years ago

4 0

Answer:

8.9 divided by 72.5 = 8.15

1.2 divided by 33.2 = 27.67

7.9 divided by 74.3 = 9.41

3.2 divided by 35.5 = 11.09

3.7 divided by 28.1 = 7.59

hope this helps

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I need help with number 16 please

Sunny_sXe [5.5K]

A) The graph is misleading because it dosent give the number of students, it gives percentages.

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4 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 $(-\infty, 2)$ .

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3 years ago

What is the ratio of clocks to tables? 5 tables 6 clocks 6:5 5:6 5:10 6:10

Anastaziya [24]

6:5 is the answer because you said clock to table Nd there is 6 clock Nd 5 tables

3 0

3 years ago

Read 2 more answers

Write an inequality for each statement.

Pie

Answer:

1. $l \leq 45$ ( in this inequality, the time can be less than or equal to 45, but no more than 45)