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3 years ago

ASAP! What is the surface area of the solid?

Mathematics

1 answer:

aliya0001 [1]3 years ago

7 0

Answer:

I think is is 14 square inches.

Step-by-step explanation:

2 plus 2 plus 5 plus 5 = 14

I don't know if this is right.

not times because 2 time 2 times 5 times 5 equals 100. there is no 100 . divide is not working and subtract is not working so add. I hoke this is correct and helps.

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What is 5r^2 = 80 It says solve each equation with the quadratic formula.

katovenus [111]

$5r^2 = 80 \ \ /:5\\ \\r^2=\frac{80}{5}\\ \\r^2=16\\ \\r^2-16=0\\ \\ r^2 - 4^2 =0 \\ \\(r-4)(r+4)=0$

$r-4=0 \ \ \vee \ \ r+4 =0 \\ \\r=4 \ \ \vee \ \ r = -4 \\ \\ \\ a^2-b^2 = (a- b)(a+b)$

5 0

3 years ago

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

I need help now it’s due soon

zalisa [80]

Answer:

Ordered pairs are ( -3, 0) (0, 3)

Step-by-step explanation:

Now when your graphing, go up 3 spots in the y-axis ( up & down) and plot your point now do the same for your x-axis (( left and right), so go over to the left 3 places, and there now connect the 2 points and your done. Hope that helps!

7 0

3 years ago

Helpp!!!!! 98 points !!!

Mnenie [13.5K]

Answer:

they don't know the correct length for ad, and they just guessed

5 0

3 years ago

Read 2 more answers

Find the quotient of these Complex Numbers. (5-i) / (3+2i)

JulsSmile [24]

Let the given complex number