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3 years ago

NEED ANSWERED ASAP!!

Mathematics

2 answers:

kenny6666 [7]3 years ago

8 0

The correct one is option 1.sphere

Sauron [17]3 years ago

4 0

Answer:

2nd , 4th and 5th option are correct that is Cube, Triangular prism and Square pyramid can not be 3-Dimensional Figure.

Step-by-step explanation:

Given: A 3-Dimensional Figure.

A circular cross sectional plane is obtained when a plan horizontally intersect with the 3-Dimensional Solid figure.

To find: From given Solid figures which can to be our required figure.

Since we get a circular cross sectional plane figure. which means Surface on the side of the 3-Dimensional figure must be curved.

So, from given options Only sphere and cone has curved surface in sides.

Therefore, 2nd , 4th and 5th option are correct that is Cube, Triangular prism and Square pyramid can not be 3-Dimensional Figure.

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Let f(x) = 1/x^2 (a) Use the definition of the derivatve to find f'(x). (b) Find the equation of the tangent line at x=2

Verdich [7]

Answer:

(a) $f'(x)=-\frac{2}{x^3}$

(b) $y=-0.25x+0.75$

Step-by-step explanation:

The given function is

$f(x)=\frac{1}{x^2}$ .... (1)

According to the first principle of the derivative,

$f'(x)=lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$

$f'(x)=lim_{h\rightarrow 0}\frac{\frac{1}{(x+h)^2}-\frac{1}{x^2}}{h}$

$f'(x)=lim_{h\rightarrow 0}\frac{\frac{x^2-(x+h)^2}{x^2(x+h)^2}}{h}$

$f'(x)=lim_{h\rightarrow 0}\frac{x^2-x^2-2xh-h^2}{hx^2(x+h)^2}$

$f'(x)=lim_{h\rightarrow 0}\frac{-2xh-h^2}{hx^2(x+h)^2}$

$f'(x)=lim_{h\rightarrow 0}\frac{-h(2x+h)}{hx^2(x+h)^2}$

Cancel out common factors.

$f'(x)=lim_{h\rightarrow 0}\frac{-(2x+h)}{x^2(x+h)^2}$

By applying limit, we get

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

$f'(x)=\frac{-2x)}{x^4}$

$f'(x)=\frac{-2)}{x^3}$ .... (2)

Therefore $f'(x)=-\frac{2}{x^3}$ .

(b)

Put x=2, to find the y-coordinate of point of tangency.

$f(x)=\frac{1}{2^2}=\frac{1}{4}=0.25$

The coordinates of point of tangency are (2,0.25).

The slope of tangent at x=2 is

$m=(\frac{dy}{dx})_{x=2}=f'(x)_{x=2}$

Substitute x=2 in equation 2.

$f'(2)=\frac{-2}{(2)^3}=\frac{-2}{8}=\frac{-1}{4}=-0.25$

The slope of the tangent line at x=2 is -0.25.

The slope of tangent is -0.25 and the tangent passes through the point (2,0.25).

Using point slope form the equation of tangent is

$y-y_1=m(x-x_1)$

$y-0.25=-0.25(x-2)$

$y-0.25=-0.25x+0.5$

$y=-0.25x+0.5+0.25$

$y=-0.25x+0.75$

Therefore the equation of the tangent line at x=2 is y=-0.25x+0.75.

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Arturiano [62]

Answer:

The value of A is 5

Step-by-step explanation:

- The number is divisible by 3 if the sum of its digits is a number

divisible by 3

- Ex: 126 is divisible by 3 because the sum of its digits = 1 + 2 + 3 = 6

and 6 is divisible by 3

- The number is divisible by 5 if its ones digit is zero or 5

- Ex: 675 is divisible by 5 because its ones digit is 5

890 is divisible by 5 because its ones digit is 0

- We are looking for the value of A in the 4-digit number 3A5A which

makes the number divisible by both 3 and 5

∵ A is in the ones position

∴ A must be zero or 5

- Let us try A = 0

∵ A = 0

∴ The number is 3050

∵ The sum of the digits of the number = 3 + 0 + 5 + 0 = 8

∵ 8 is not divisible by 3

∴ 3050 is not divisible by both 3 and 5

∴ A can not be zero

- Let us try A = 5

∵ A = 5

∴ The number is 3555

∵ The sum of the digits of the number = 3 + 5 + 5 + 5 = 18

∵ 18 is divisible by 3

∴ 3555 is divisible by both 3 and 5

∴ A must be equal 5

* <em>The value of A is 5</em>

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