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3 years ago

This equation is an equation that expresses a relationship among variables. Formulas are examples of this.

Mathematics

2 answers:

noname [10]3 years ago

8 0

Answer:

The answer is Literal Equation. (Im doing the usa test prep)

Step-by-step explanation:

Yuki888 [10]3 years ago

3 0

I believe that it is a Proportional Relationship

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Please help asap !!!!!! what vaule of x makes the equation true 3(x-4)=15

DaniilM [7]

3(x-4) = 15

Distribute the 3:

3x-12 = 15

Add 12 to both sides:

3x = 27

Divide both sides by 3:

X = 9

The answer is D. 9

5 0

3 years ago

What are the like terms in the expression?

Anarel [89]

Since none of the terms have the same variables the like terms would be (A) because they are both constants

5 0

3 years ago

Read 2 more answers

Solve for t in scientific formula d=rt

yarga [219]

The problem is: d=rt
you are solving for r

d=rt
-- --
t t

you divide both sides by r, so your answer is:

d/t=r or r=d/t

I hope this helped! :))

3 0

3 years ago

In a random sample of 70 people, it was found that 44 of them were fans of the New York Yankees. What is the margin of error for

Veronika [31]

The general formula for the margin of error would be:

z * √[p (1-p) ÷ n]

where:
z = values for selected confidence level
p = sample proportion
n = sample size

Since the confidence level is not given, we can only solve for the
<span>√[p (1-p) ÷ n] part.
</span>
p = 44/70
n = 70

√[44/70 (1 - (44/70) ÷ 70]

√[0.6286 (0.3714)] ÷ 70

√0.2335 ÷ 70

√0.0033357 = 0.05775 or 0.058 Choice B.

7 0

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 $(-\infty, 2)$ .

3 0

3 years ago