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3 years ago

A gazelle can jump over a hundred five feet in a single jump how many yards is that

2 answers:

5 0

You would do 500 ÷ 3 because 3 ft = 1 yard. So, 500 ÷ 3 = 166.6666... Btw, if u have a repeating decimal, just put a line over the first 2 numbers after the decimal. Hoped this helped you!

abruzzese [7]3 years ago

4 0

The answer is 4 yards

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Solve this problem. 4|3x+4|=4x+8

Gelneren [198K]

The correct answer is x= -1. Hope this helps!!

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3 years ago

Is the work shown below correct? Explain your answer.

riadik2000 [5.3K]

(11 + 21) - ( 3 - 100 )
= 32 + 97
= 129

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= 32 - 103
= - 71

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3 years ago

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3 years ago

Read 2 more answers

Form a polynomial function(x) with zeros: -2, multiplicity 1; 1, multiplicity 2; 5, multiplicity 3; and degree 6. Use 1 as the l

Citrus2011 [14]

Answer:

Remember, a number a is a zero of the polynomial p(x) if p(a)=0. And a has multiplicity n if the factor (x-a) appear n times in the factorization of p(x).

1. Since -2 is a zero with multiplicity 1, then (x+2) is a factor of the polynomial.

2. Since 1 is a zero with multiplicity 2, then (x-1) is a factor of the polynomial and appear 2 times.

3. Since 5 is a zero with multiplicity 3, then (x-5) is a factor of the polynomial and appear 3 times.

Then, the polynomial function with the zeros described above is

$p(x)=(x+2)(x-1)^2(x-5)^3= x^6-15x^5+72x^4-78x^3-255x^2+525x-250$

8 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