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3 years ago

Divide. -4 1/3 divided by 2 3/5

Mathematics

1 answer:

Alexus [3.1K]3 years ago

6 0

The answer is: -1.66666666667 to be exact

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Which quadrilateral’s diagonals do not create a 90° angle at its intersection?

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A Rhombus and a Kite are the same thing they are the answer

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3 years ago

Read 2 more answers

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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3 years ago

Solve for the variable "a" in this equation. Show work:<br><br> c=8πab

Contact [7]

C=8πab
Divide both sides by 8πb
c/(8πb) = a

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4 years ago

Factor the quadratic expression as the product of two binomials x^2-3x-10

vovikov84 [41]

X^2+2x-5x-10
x(x+2)-5(x+2)

(x-5)(x+2)

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3 years ago

The following relation is a function: {(-8, – 9), (-4, 5), (1, 5), (4, 7)}

Nookie1986 [14]

Answer:

Yes

Step-by-step explanation:

The following relation is a function. When graphed, it passes the vertical line test, and you don't go across any two points at the same time. So therefore, it is a function. Another easy way to determine if the relation is a function is by taking a look at the x-coordinates. If any x-coordinates are the same, then it's not a function. If all x-coordinates are the same, then it's a function.

Hope this helps. Please mark brainliest.

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3 years ago