All categories

3 years ago

If the dimensions of the following cylinder are doubled, what factor will the surface area change by?

Mathematics

1 answer:

Vedmedyk [2.9K]3 years ago

7 0

Answer:

probably 2 or 4

Step-by-step explanation:

You might be interested in

The surface area of a rectangular prism is 208 cm. Two of the dimensions are 2cm and 10 cm. Find the measure of the other dimens

kompoz [17]

2 times 10 equals 20

208 divided by 20 equals 10.4

6 0

3 years ago

How to solve for x and y?

BlackZzzverrR [31]

(2x-y)+(3y-x) = 2y+X = 25
25 = 2:1 = 2Y:X
25/3 = 25/3
25/3 * 2 = 50/3 = 2y

x = 25/3
y = 25/3

Hope this helps! :)

4 0

3 years ago

Find the x-coordinates of any relative extrema and inflection point(s) for the function f(x)= 6x^1/3 + 3x^4/3. You must justify

irina [24]

Answer:

x-coordinates of relative extrema = $\frac{-1}{2}$

x-coordinates of the inflexion points are 0, 1

Step-by-step explanation:

$f(x)=6x^{\frac{1}{3}}+3x^{\frac{4}{3}}$

Differentiate with respect to x

$f'(x)=6\left ( \frac{1}{3} \right )x^{\frac{-2}{3}}+3\left ( \frac{4}{3} \right )x^{\frac{1}{3}}=\frac{2}{x^{\frac{2}{3}}}+4x^{\frac{1}{3}}$

$f'(x)=0\Rightarrow \frac{2}{x^{\frac{2}{3}}}+4x^{\frac{1}{3}}=0\Rightarrow x=\frac{-1}{2}$

Differentiate f'(x) with respect to x

$f''(x)=2\left ( \frac{-2}{3} \right )x^{\frac{-5}{3}}+\frac{4}{3}x^{\frac{-2}{3}}=\frac{-4x^{\frac{2}{3}}+4x^{\frac{5}{3}}}{3x^{\frac{2}{3}}x^{\frac{5}{3}}}\\f''(x)=0\Rightarrow \frac{-4x^{\frac{2}{3}}+4x^{\frac{5}{3}}}{3x^{\frac{2}{3}}x^{\frac{5}{3}}}=0\Rightarrow x=1$

At x = $\frac{-1}{2}$ ,

$f''\left ( \frac{-1}{2} \right )=\frac{4\left ( -1+4\left ( \frac{-1}{2} \right ) \right )}{3\left ( \frac{-1}{2} \right )^{\frac{5}{3}}}>0$

We know that if $f''(a)>0$ then x = a is a point of minima.

So, $x=\frac{-1}{2}$ is a point of minima.

For inflexion points:

Inflexion points are the points at which f''(x) = 0 or f''(x) is not defined.

So, x-coordinates of the inflexion points are 0, 1

7 0

3 years ago

Use double angle formulas to find the exact value of sin2x,cos2x, and tan2x

denis-greek [22]

Sin2x = 2sinxcosx;
cos2x = (cosx)^2 - (sinx)^2;
tan2x = (sin2x)/(cos2x);

cosx = 5/13 from formula (sinx)^2 + (cosx)^2 = 1;

=> sin2x = 120/169;
.................................

5 0

3 years ago

Does Anyone know this?

lukranit [14]

Answer:

∠AEC = 139°

Step-by-step explanation:

Since EC bisects ∠BED then ∠BEC = ∠CED = 4x + 1

∠AED = ∠AEB + ∠BEC + ∠CED = 180 ← straight angle

Substitute values into the equation

11x - 12 + 4x + 1 + 4x + 1 = 180, that is

19x - 10 = 180 ( add 10 to both sides )

19x = 190 ( divide both sides by 19 )

x = 10

Hence

∠AEC = ∠AEB + ∠BEC = 11x - 12 + 4x + 1 = 15x - 11, hence

∠AEC = (15 × 10) - 11 = 150 - 11 = 139°

6 0

3 years ago