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3 years ago

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Black and white copies cost $0.05 each, and color copies cost $0.49 each. The equation 0.05x+ 0.49y=5 represents the number of b

lack and white copies x and color copies y that can be made with $5. If no color copies are made how many black and white copies can be made with $5

1 answer:

alekssr [168]3 years ago

7 0

if no color copies are made, you can make 100 black and white copies with $5.

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MY LAST QUESTION PLEASE HURRY, SORRY FOR RUSHING BY THE WAY

77julia77 [94]

Answer:40

Step-by-step explanation:

dividing 1200 by 600 is 2

so you would do the same for 80

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3 years ago

Answers for all questions

abruzzese [7]

Answer:

3 and dived 30 multy is 182 is your answer

Step-by-step explanation:

1+1=2

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3 years ago

Complete the assignment on a separate sheet of paper<br><br> Please attach pictures of your work.

Irina18 [472]

Answer:

<u>TO FIND :-</u>

Length of all missing sides.

<u>FORMULAES TO KNOW BEFORE SOLVING :-</u>

$\sin \theta = \frac{Side \: opposite \: to \: \theta}{Hypotenuse}$
$\cos \theta = \frac{Side \: adjacent \: to \: \theta}{Hypotenuse}$
$\tan \theta = \frac{Side \: opposite \: to \: \theta}{Side \: adjacent \: to \: \theta}$

<u>SOLUTION :-</u>

1) θ = 16°

Length of side opposite to θ = 7

Hypotenuse = x

$=> \sin 16 = \frac{7}{x}$

$=> \frac{7}{x} = 0.27563......$

$=> x = \frac{7}{0.27563....} = 25.39568.....$ ≈ 25.3

2) θ = 29°

Length of side opposite to θ = 6

Hypotenuse = x

$=> \sin 29 = \frac{6}{x}$

$=> \frac{6}{x} = 0.48480......$

$=> x = \frac{6}{0.48480....} = 12.37599.....$ ≈ 12.3

3) θ = 30°

Length of side opposite to θ = x

Hypotenuse = 11

$=> \sin 30 = \frac{x}{11}$

$=> \frac{x}{11} = 0.5$

$=> x = 0.5 \times 11 = 5.5$

4) θ = 43°

Length of side adjacent to θ = x

Hypotenuse = 12

$=> \cos 43 = \frac{x}{12}$

$=> \frac{x}{12} = 0.73135......$

$=> x = 12 \times 0.73135.... = 8.77624....$ ≈ 8.8

5) θ = 55°

Length of side adjacent to θ = x

Hypotenuse = 6

$=> \cos 55 = \frac{x}{6}$

$=> \frac{x}{6} = 0.57357......$

$=> x = 6 \times 0.57357.... = 3.44145....$ ≈ 3.4

6) θ = 73°

Length of side adjacent to θ = 8

Hypotenuse = x

$=> \cos 73 = \frac{8}{x}$

$=> \frac{8}{x} = 0.29237......$

$=> x = \frac{8}{0.29237.....} = 27.36242.....$ ≈ 27.3

7) θ = 69°

Length of side opposite to θ = 12

Length of side adjacent to θ = x

$=> \tan 69 = \frac{12}{x}$

$=> \frac{12}{x} = 2.60508......$

$=> x = \frac{12}{2.60508....} = 4.60636....$ ≈ 4.6

8) θ = 20°

Length of side opposite to θ = 11

Length of side adjacent to θ = x

$=> \tan 20 = \frac{11}{x}$

$=> \frac{11}{x} = 0.36397......$

$=> x = \frac{11}{0.36397....} =30.22225....$ ≈ 30.2

5 0

2 years ago

Help anyone I’ll appreciate it thank you

Tomtit [17]

Answer:

150

Step-by-step explanation:

the total angle of a flat line is 180

since one side is 30 180-30 is 150

4 0

2 years ago

If a and b are positive numbers, find the maximum value of f(x) = x^a(2 − x)^b on the interval 0 ≤ x ≤ 2.

Ad libitum [116K]

Answer:

The maximum value of f(x) occurs at:

$\displaystyle x = \frac{2a}{a+b}$

And is given by:

$\displaystyle f_{\text{max}}(x) = \left(\frac{2a}{a+b}\right)^a\left(\frac{2b}{a+b}\right)^b$

Step-by-step explanation:

Answer:

Step-by-step explanation:

We are given the function:

$\displaystyle f(x) = x^a (2-x)^b \text{ where } a, b >0$

And we want to find the maximum value of f(x) on the interval [0, 2].

First, let's evaluate the endpoints of the interval:

$\displaystyle f(0) = (0)^a(2-(0))^b = 0$

And:

$\displaystyle f(2) = (2)^a(2-(2))^b = 0$

Recall that extrema occurs at a function's critical points. The critical points of a function at the points where its derivative is either zero or undefined. Thus, find the derivative of the function:

$\displaystyle f'(x) = \frac{d}{dx} \left[ x^a\left(2-x\right)^b\right]$

By the Product Rule:

$\displaystyle \begin{aligned} f'(x) &= \frac{d}{dx}\left[x^a\right] (2-x)^b + x^a\frac{d}{dx}\left[(2-x)^b\right]\\ \\ &=\left(ax^{a-1}\right)\left(2-x\right)^b + x^a\left(b(2-x)^{b-1}\cdot -1\right) \\ \\ &= x^a\left(2-x\right)^b \left[\frac{a}{x} - \frac{b}{2-x}\right] \end{aligned}$

Set the derivative equal to zero and solve for <em>x: </em>

$\displaystyle 0= x^a\left(2-x\right)^b \left[\frac{a}{x} - \frac{b}{2-x}\right]$

By the Zero Product Property:

$\displaystyle x^a (2-x)^b = 0\text{ or } \frac{a}{x} - \frac{b}{2-x} = 0$

The solutions to the first equation are <em>x</em> = 0 and <em>x</em> = 2.

First, for the second equation, note that it is undefined when <em>x</em> = 0 and <em>x</em> = 2.

To solve for <em>x</em>, we can multiply both sides by the denominators.

$\displaystyle\left( \frac{a}{x} - \frac{b}{2-x} \right)\left((x(2-x)\right) = 0(x(2-x))$

Simplify:

$\displaystyle a(2-x) - b(x) = 0$

And solve for <em>x: </em>

$\displaystyle \begin{aligned} 2a-ax-bx &= 0 \\ 2a &= ax+bx \\ 2a&= x(a+b) \\ \frac{2a}{a+b} &= x \end{aligned}$

So, our critical points are:

$\displaystyle x = 0 , 2 , \text{ and } \frac{2a}{a+b}$

We already know that f(0) = f(2) = 0.

For the third point, we can see that:

$\displaystyle f\left(\frac{2a}{a+b}\right) = \left(\frac{2a}{a+b}\right)^a\left(2- \frac{2a}{a+b}\right)^b$

This can be simplified to:

$\displaystyle f\left(\frac{2a}{a+b}\right) = \left(\frac{2a}{a+b}\right)^a\left(\frac{2b}{a+b}\right)^b$

Since <em>a</em> and <em>b</em> > 0, both factors must be positive. Thus, f(2a / (a + b)) > 0. So, this must be the maximum value.

To confirm that this is indeed a maximum, we can select values to test. Let <em>a</em> = 2 and <em>b</em> = 3. Then:

$\displaystyle f'(x) = x^2(2-x)^3\left(\frac{2}{x} - \frac{3}{2-x}\right)$

The critical point will be at:

$\displaystyle x= \frac{2(2)}{(2)+(3)} = \frac{4}{5}=0.8$

Testing <em>x</em> = 0.5 and <em>x</em> = 1 yields that:

$\displaystyle f'(0.5) >0\text{ and } f'(1)$

Since the derivative is positive and then negative, we can conclude that the point is indeed a maximum.

Therefore, the maximum value of f(x) occurs at:

$\displaystyle x = \frac{2a}{a+b}$

And is given by:

$\displaystyle f_{\text{max}}(x) = \left(\frac{2a}{a+b}\right)^a\left(\frac{2b}{a+b}\right)^b$

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3 years ago