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3 years ago

12

A football team won 75% of 120 games in a season. How many games is that?

2 answers:

Luba_88 [7]3 years ago

8 0

Answer:

90 games

Step-by-step explanation:

Of the 120 games the football team played, they won 75% of them.

Remember that in mathematics, "of" almost always means to multiply. Here, that means we must multiply 75% by 120:

75% * 120

Remember also that % just means divided by 100, so 75% = 75 / 100. So:

75% * 120 = (75/100) * 120 = 9000/100 = 90 games

Bess [88]3 years ago

7 0

Answer: 90 games
Step by step explanation:

50% of 120 is half of it so that would be 60
10% of 120 is just 12
5% is 6
12x7=70% which is 84
Add 5% which is 6
84+6=90 so they played 90 games

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Read 2 more answers

Among all pairs of numbers with a sum of 238238, find the pair whose product is maximum. Write your answers as fractions reduced

mylen [45]

Answer:

$x=119119,\;\;\;y=119119$

Step-by-step explanation:

Let the $x$ be the first number, $y$ be the second number and $p$ be the product of the first and second number.

i.e $x+y=238238$ $\Rightarrow x=238238-y$

$p=xy$

$\Rightarrow p=(238238-y)y\\\Rightarrow p=238238y-y^2$

$\Rightarrow \frac{dp}{dy} =238238-2y\\\Rightarrow 0=238238-2y\\\Rightarrow 2y=238238\\\Rightarrow y=119119$

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$x=238238-119119\\\\ \Rightarrow x=119119$

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2 years 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