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4 years ago

8

The minute hand of a clock is 7 inches long. how far does the tip of the minute hand move in 5 minutes? if necessary, round the

answer to two decimal places.

1 answer:

Alina [70]4 years ago

3 0

7/6π" (or 1.17π"). To find the circumference (therefore the distance the hand travels) you do 2rπ (radius is 7in), which is 14π, then divide it by 12 bc the hand is only moving five minutes (which is simplified to 7/6π). You can't simplify π.

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1) Let f(x)=6x+6/x. Find the open intervals on which f is increasing (decreasing). Then determine the x-coordinates of all relat

brilliants [131]

Answer:

1) increasing on (-∞,-1] ∪ [1,∞), decreasing on [-1,0) ∪ (0,1]

$x = -1$ is local maximum, $x = 1$ is local minimum

2) increasing on [1,∞), decreasing on (-∞,0) ∪ (0,1]

$x = 1$ is absolute minimum

3) increasing on (-∞,0] ∪ [8,∞), decreasing on [0,4) ∪ (4,8]

$x = 0$ is local maximum, $x = 8$ is local minimum

4) increasing on [2,∞), decreasing on (-∞,2]

$x = 2$ is absolute minimum

5) increasing on the interval (0,4/9], decreasing on the interval [4/9,∞)

$x = 0$ is local minimum, $x = 4/9$ is absolute maximum

Step-by-step explanation:

To find minima and maxima the of the function, we must take the derivative and equalize it to zero to find the roots.

1) $f(x) = 6x + 6/x$

$f\prime(x) = 6 - 6/x^2 = 0$ and $x \neq 0$

So, the roots are $x = -1$ and $x = 1$

The function is increasing on the interval (-∞,-1] ∪ [1,∞)

The function is decreasing on the interval [-1,0) ∪ (0,1]

$x = -1$ is local maximum, $x = 1$ is local minimum.

2) $f(x)=6-4/x+2/x^2$

$f\prime(x)=4/x^2-4/x^3=0$ and $x \neq 0$

So the root is $x = 1$

The function is increasing on the interval [1,∞)

The function is decreasing on the interval (-∞,0) ∪ (0,1]

$x = 1$ is absolute minimum.

3) $f(x) = 8x^2/(x-4)$

$f\prime(x) = (8x^2-64x)/(x-4)^2=0$ and $x \neq 4$

So the roots are $x = 0$ and $x = 8$

The function is increasing on the interval (-∞,0] ∪ [8,∞)

The function is decreasing on the interval [0,4) ∪ (4,8]

$x = 0$ is local maximum, $x = 8$ is local minimum.

4) $f(x)=6(x-2)^{2/3} +4=0$

$f\prime(x) = 4/(x-2)^{1/3}$ has no solution and $x = 2$ is crtitical point.

The function is increasing on the interval [2,∞)

The function is decreasing on the interval (-∞,2]

$x = 2$ is absolute minimum.

5) $f(x)=8\sqrt x - 6x$ for $x>0$

$f\prime(x) = (4/\sqrt x)-6 = 0$

So the root is $x = 4/9$

The function is increasing on the interval (0,4/9]

The function is decreasing on the interval [4/9,∞)

$x = 0$ is local minimum, $x = 4/9$ is absolute maximum.

5 0

3 years ago

A number line going from negative 3 to positive 5. a closed circle is at negative 1.5. everything to the left of the circle is s

algol13

The inequalities which matches the graph are: x ≥ ₋1.5 and ₋1.5 ≤ x

Given, a number line is moving from ₋3 to ₊5 .

Next a mark is made at ₋1.5 and everything to its left is shaded which means not visible.

When we mark the point and shade the left part of it then we can start applying the inequality expressions.

And from that we can match the applicable inequalities while observing the graph.

For the first inequality ₋1.5 ≥ x.Here,x value ranges from ₋1.5 to ₊5, hence we take this as an inequality expression.
Next, if we consider x ≤ ₋1.5, then here x value will range from ₋1.5 to ₋3. where the region is shaded. Hence this expression doesn't satisfy the graph.
the next expression is ₋1.5 ≤ x. here the value will again range in the shaded area so it is not applicable.
₋1.5 ≥ x, here the values will satisfy the graph.
remaining inequality expressions does not support the graph.

Therefore the only inequalities the graph represents is x ≥ ₋1.5 and ₋1.5 ≤ x

Learn more about "Linear Inequalities" here-

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2 years ago

What is the value of the digit 2 in this number?

antiseptic1488 [7]

The answer is A = 2 tenths
C = .2
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4 years ago

The ratio of a triangle is 2:3:4 the shortest side is 15 inches what is the perimeter

hjlf

<span>The perimeter is the sum of the side lengths, or x+y+z.</span><span> </span><span>The perimeter is 67.5 inches. </span>

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4 years ago

Can someone help me with these geometry questions? I can’t seem to understand them. If u find the answers, can u explain it too?

rusak2 [61]

<em>Question 5</em>

Vertical angles are <u>congruent</u>. Therefore, $3y=\boxed{150}$ . The sum of two supplementary angles equals $\boxed{180^{\circ}}$ .

Therefore, $6x+150=\boxed{180}$

<em>Question 6</em>

$\angle b-14=90-\angle b\\\\2\angle b=104\\\\\angle b=\boxed{52^{\circ}}$

<em>Question 7</em>

<em /> $\angle d+32=180-\angle d\\\\2\angle d=146\\\\\angle d=\boxed{73^{\circ}}$

<em>Question 8</em>

<em /> $7x-80=3x\\\\-80=-4x\\\\x=20\\\\\implies \boxed{m\angle 1=m\angle 3=60^{\circ}}$

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2 years ago