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2 years ago

How do you calculate the minimised dimensions: length width and height of a rectangular prism with the given volume of 10m^3?

Mathematics

1 answer:

Elan Coil [88]2 years ago

3 0

The length, width and height of the rectangular prism is ∛10

<h3>How to determine the dimensions?</h3>

The volume is given as:

Volume, V = 10m^3

For a minimised volume, we have:

Length = Width = Height

This means that:

V = L^3

So, we have:

L^3 = 10

Take the cube root of both sides

L = ∛10

Hence, the length, width and height of the rectangular prism is ∛10

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2. Let f(x) = 3x² -x+5, find f(x+1)

Anna11 [10]

Answer:

f(x + 1) = 3x² + 5x + 7

Step-by-step explanation:

To find f(x + 1), substitute x = x + 1 into f(x), that is

f(x + 1) = 3(x + 1)² - (x + 1) + 5 ← expand (x + 1)² using FOIL

= 3(x² + 2x + 1) - x - 1 + 5 ← distribute parenthesis by 3

= 3x² + 6x + 3 - x - 1 + 5 ← collect like terms

= 3x² + 5x + 7

8 0

3 years ago

write an equation in slop-intercept form for a line that is perpendicular to the line y= -5x-3 and has a y-intercept of 6.

jenyasd209 [6]

Answer:

The equation of the line would be y = 1/5x + 6

Step-by-step explanation:

First, to find the slope of a perpendicular line, you need to find the opposite and reciprocal of the original slope. Since the original is -5, we first change the sign (which makes it just 5). Now we take that 5 and flip it to get 1/5.

Now that we have the slope as 1/5, we can use that and the intercept in slope-intercept form to get the equation.

y = mx + b

y = 1/5x + 6

6 0

2 years ago

Which of the following would not represent the phrase “the difference of two numbers divided by 5”?

Sophie [7]

It would be (a) because it is now
it

5 0

2 years ago

Find the surface area of the solid generated by revolving the region bounded by the graphs of y = x2, y = 0, x = 0, and x = 2 ab

Nikitich [7]

Answer:

See explanation

Step-by-step explanation:

The surface area of the solid generated by revolving the region bounded by the graphs can be calculated using formula

$SA=2\pi \int\limits^a_b f(x)\sqrt{1+f'^2(x)} \, dx$

If $f(x)=x^2,$ then

$f'(x)=2x$

and

$b=0\\ \\a=2$

Therefore,

$SA=2\pi \int\limits^2_0 x^2\sqrt{1+(2x)^2} \, dx=2\pi \int\limits^2_0 x^2\sqrt{1+4x^2} \, dx$

Apply substitution

$x=\dfrac{1}{2}\tan u\\ \\dx=\dfrac{1}{2}\cdot \dfrac{1}{\cos ^2 u}du$

Then

$SA=2\pi \int\limits^2_0 x^2\sqrt{1+4x^2} \, dx=2\pi \int\limits^{\arctan(4)}_0 \dfrac{1}{4}\tan^2u\sqrt{1+\tan^2u} \, \dfrac{1}{2}\dfrac{1}{\cos^2u}du=\\ \\=\dfrac{\pi}{4}\int\limits^{\arctan(4)}_0 \tan^2u\sec^3udu=\dfrac{\pi}{4}\int\limits^{\arctan(4)}_0(\sec^3u+\sec^5u)du$

Now

$\int\limits^{\arctan(4)}_0 \sec^3udu=2\sqrt{17}+\dfrac{1}{2}\ln (4+\sqrt{17})\\ \\ \int\limits^{\arctan(4)}_0 \sec^5udu=\dfrac{1}{8}(-(2\sqrt{17}+\dfrac{1}{2}\ln(4+\sqrt{17})))+17\sqrt{17}+\dfrac{3}{4}(2\sqrt{17}+\dfrac{1}{2}\ln (4+\sqrt{17}))$

Hence,

$SA=\pi \dfrac{-\ln(4+\sqrt{17})+132\sqrt{17}}{32}$

3 0

2 years ago

If (3,2) is a point on the line whose equation is y = 3x + k, ,<br> determine the value of k.

Svetllana [295]

Answer: -7

Step-by-step explanation: x=3 and y=2

3 0

2 years ago