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2 years ago

30 is what percent of 60%

Mathematics

2 answers:

liq [111]2 years ago

8 0

Answer:

Step-by-step explanation:

(✿◡‿◡)

masha68 [24]2 years ago

6 0

Answer:

it's 50 percent because 30 is half of 60 which is determined as a 100 % and 30 is 50%

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What is the exterior angle?

olganol [36]

Answer:

angle YUT = 120

Step-by-step explanation:

in Triangle SUT

angle 70 + angle 50 + angle x = 180

70 + 50 + x = 180

120 + x = 180

x = 180 - 120

x = 60

angle SUT = 60

angle SUT + angle YUT = 180 ( straight line)

60 + x = 180

x = 120

Therefore angle YUT ( exterior angle ) = 120

7 0

3 years ago

Solve for y. 9x+5y=7 <br>

Yuliya22 [10]

Answer: y=7/5-9x/5

Step-by-step explanation:

im sorry if this dont help

5 0

2 years ago

Read 2 more answers

What is the value of m in the equation ‡m - 3n=16, when n=8?

soldier1979 [14.2K]

Step-by-step explanation:

equation: m-3n=16

Solution,

or, m - 3n = 16

or, m - 3×8 = 16

or, m - 24 = 16

or, m = 16 + 24

or, m = 40

Therefore, the value of m is '40'

8 0

1 year ago

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.

5 0

3 years ago

5x+2y=10; (-2, 9) if -2 is x and 9 is y

Dovator [93]

What are you looking for. slope?

4 0

3 years ago