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Answer:

The value of f(30) is equal to 2.

Step-by-step explanation:

The given expression is :

[tex]f(x)= 10 \sin(x) - 3[/tex]

We need to find the value of f(30)

Put x = 30 in above expression.

So,

[tex]f(x)= 10 \sin(30) - 3\\\\=10\times \dfrac{1}{2}-3\\\\=5-3\\\\=2[/tex]

Hence, the value of f(30) is equal to 2.

Answer:

Step-by-step explanation:

From the given information:

a)

Assuming the shape of the base is square,

suppose the base of each side = x

Then the perimeter of the base of the square = 4x

Suppose the length of the package from the base = y; &

the height is also = x

Now, the restriction formula can be computed as:

y + 4x ≤ 99

The objective function:

i.e maximize volume V = l × b × h

V = (y)*(x)*(x)

V = x²y

b) To write the volume as a function of x, V(x) by equating the derived formulas in (a):

y + 4x ≤ 99   --- (1)

V = x²y          --- (2)

From equation (1),

y ≤ 99 - 4x

replace the value of y into (2)

V ≤ x² (99-4x)

V ≤ 99x² - 4x³

Maximum value V = 99x² - 4x³

At maxima or minima, the differential of [tex]\dfrac{d }{dx}(V)=0[/tex]

[tex]\dfrac{d}{dx}(99x^2-4x^3) =0[/tex]

⇒ 198x - 12x² = 0

[tex]12x \Big({\dfrac{33}{2}-x}}\Big)=0[/tex]

By solving for x:

x = 0 or x = [tex]\dfrac{33}{2}[/tex]

Again:

V = 99x² - 4x³

[tex]\dfrac{dV}{dx}= 198x -12x^2 \\ \\ \dfrac{d^2V}{dx^2}=198 -24x[/tex]

At x = [tex]\dfrac{33}{2}[/tex]

[tex]\dfrac{d^2V}{dx^2}\Big|_{x= \frac{33}{2}}=198 -24(\dfrac{33}{2})[/tex]

[tex]\implies 198 - 12 \times 33[/tex]

= -198

Thus, at maximum value;

[tex]\dfrac{d^2V}{dx^2}\le 0[/tex]

Recall y = 99 - 4x

when at maximum x = [tex]\dfrac{33}{2}[/tex]

[tex]y = 99 - 4(\dfrac{33}{2})[/tex]

y = 33

Finally; the volume V = x² y is;

[tex]V = (\dfrac{33}{2})^2 \times 33[/tex]

[tex]V =272.25 \times 33[/tex]

V = 8984.25 inches³

Answer:

92 dB

Step-by-step explanation:

Use the mean formula, mean = sum of elements / number of elements.

Since it is a 8 hour work day, there are 8 elements.

mean = sum of elements / number of elements

mean = (98 + 98 + 98 + 98 + 98 + 82 + 82 + 82) / 8

mean = 736 / 8

mean = 92

So, the average noise level is 92 dB

Answer:

254.47 mm

Step-by-step explanation:

Answer:

Step-by-step explanation: It could be 8,7, or 2. Because these are all positive

:)

Answer: B

Step-by-step explanation:

4x+3x+2x=180

9x = 180

x = 20

20x3 = 60

Answer:

Step-by-step explanation:

7

Answer:

The best conclusion is that we are 90% that the true population proportion of Republicans that voted for Mitt Romney is between 0.7273 and 0.8106.

Step-by-step explanation:

x% confidence interval:

A confidence interval is built from a sample, has bounds a and b, and has a confidence level of x%. It means that we are x% confident that the population mean is between a and b.

In this question:

The 90% confidence interval for the proportion of Republican voters that had voted for Mitt Romney is (0.7273, 0.8106). The best conclusion is that we are 90% that the true population proportion of Republicans that voted for Mitt Romney is between 0.7273 and 0.8106.

Answer:

I think its 0.

Step-by-step explanation:

Double negatives make a positive, add the like terms. Hope this helps

[tex] \frac{35e {}^{9} }{5 {e}^{8} } \ \\ \\ \frac{7e {}^{9} }{e {}^{8} } \\ \\ \\ = 7e[/tex]

Step By Step Explanation:

Alternate Forms:

Answer:

0

Step-by-step explanation:

(5.4)^2 - 5.4^2

= 5.4^2 - 5.4^2

= 5,4^2(1 - 1)

= 5.4^2(0)

= 0

9514 1404 393

Answer:

  B, C, A, D

Step-by-step explanation:

The depths are easier to compare if they are all in the same form. Here, it is convenient to use decimal numbers rounded to hundredths. Your calculator can help with the fractions if you are not familiar with decimal equivalents.

  A: -1.6 m = -1.60 m

  B: -4/3 m ≈ -1.33 m

  C: -1.36m = -1.36 m

  D: -17/9 m ≈ -1.89 m

Then the least deep site is the one with the depth number closest to 0.

In order from least to greatest depth, the sites are ...

  B (-1.33) > C (-1.36) > A (-1.60) > D (-1.89)

Answer:

yeah

Step-by-step explanation:

Answer:

Let the retail cost be x and the wholesale cost be y

Step-by-step explanation:

x = y + 0.50y

x = 1.50y

Therefore the retail cost is 1.50 times the wholesale cost.

Answer:

f(x) = (x+5)^2 +12

The minimum value is 12

Step-by-step explanation:

f(x)=x^2+10x+37

The vertex will be the minimum value since this is an upwards opening parabola

Completing the square by taking the coefficient of x and squaring it adding it and subtracting it

f(x) = x^2+10x  + (10/2) ^2 - (10/2) ^2+37

f(x) = ( x^2 +10x +25) -25+37

    = ( x+5) ^2+12

Th is in vertex form y = ( x-h)^2 +k  where (h,k) is the vertex

The vertex is (-5,12)

The minimum is the y value or 12

Question 1

Question 2

Question 3

9514 1404 393

Answer:

  180 in³

Step-by-step explanation:

  V = Bh

  V = (6 in × 3 in)(10 in) = 180 in³

The volume is the product of the cross section area (6 in × 3 in) = 18 in² and the height perpendicular to that cross section, 10 in.

Answer:

Correct.

Step-by-step explanation:

It looks good to me.

Good job!

9514 1404 393

Answer:

  5y = 5

Step-by-step explanation:

  -2(x -2y) +(2x +y) = -2(-1) +(3) . . . . -2 times [eq2] + [eq1]

  -2x +4y +2x +y = 2 +3 . . . . eliminate parentheses

  5y = 5 . . . . . . . . collect terms

Answer:

A: -21

Step-by-step explanation:

Vertex of a quadratic function:

Suppose we have a quadratic function in the following format:

[tex]f(x) = ax^{2} + bx + c[/tex]

It's vertex is the point [tex](x_{v}, y_{v})[/tex]

In which

[tex]x_{v} = -\frac{b}{2a}[/tex]

[tex]y_{v} = -\frac{\Delta}{4a}[/tex]

Where

[tex]\Delta = b^2-4ac[/tex]

If a<0, the vertex is a maximum point, that is, the maximum value happens at [tex]x_{v}[/tex], and it's value is [tex]y_{v}[/tex].

In this question:

Quadratic function:

[tex]g(x) = x^2 - 6x - 12[/tex]

So [tex]a = 1, b = -6, c = -12[/tex].

Minimum value:

This is the y-value of the vertex. So

[tex]\Delta = b^2-4ac = (-6)^2 - 4(1)(-12) = 36+48 = 84[/tex]

[tex]y_{v} = -\frac{\Delta}{4a} = -\frac{84}{4} = -21[/tex]

The minimum value is -21, and the correct answer is given by option A.

Answer:

f(t) = -16t² + 36

Step-by-step explanation:

f(t) = a(t - h)² + k

This is vertex form where (h, k) is the (x, y) coordinate of the vertex

The vertex is give as (0, 36)

f(t) = a(t - 0)^2 + 36

f(t) =at² + 36

use point (1, 20) to find "a"

20 = a(1²) + 36

20 = a + 36

-16 = a

f(t) = -16t² + 36

Answer:2647.5

Step-by-step explanation:

Answer(s):

[tex]\displaystyle y = 3sin\: (1\frac{1}{2}x + \frac{\pi}{2}) - 2 \\ y = 3cos\: 1\frac{1}{2}x - 2[/tex]

Explanation:

[tex]\displaystyle y = Asin(Bx - C) + D \\ \\ Vertical\:Shift \hookrightarrow D \\ Horisontal\:[Phase]\:Shift \hookrightarrow \frac{C}{B} \\ Wavelength\:[Period] \hookrightarrow \frac{2}{B}\pi \\ Amplitude \hookrightarrow |A| \\ \\ Vertical\:Shift \hookrightarrow -2 \\ Horisontal\:[Phase]\:Shift \hookrightarrow \frac{C}{B} \hookrightarrow \boxed{-\frac{\pi}{3}} \hookrightarrow \frac{-\frac{\pi}{2}}{1\frac{1}{2}} \\ Wavelength\:[Period] \hookrightarrow \frac{2}{B}\pi \hookrightarrow \boxed{1\frac{1}{3}\pi} \hookrightarrow \frac{2}{1\frac{1}{2}}\pi \\ Amplitude \hookrightarrow 3[/tex]

OR

[tex]\displaystyle y = Acos(Bx - C) + D \\ \\ Vertical\:Shift \hookrightarrow D \\ Horisontal\:[Phase]\:Shift \hookrightarrow \frac{C}{B} \\ Wavelength\:[Period] \hookrightarrow \frac{2}{B}\pi \\ Amplitude \hookrightarrow |A| \\ \\ Vertical\:Shift \hookrightarrow -2 \\ Horisontal\:[Phase]\:Shift \hookrightarrow 0 \\ Wavelength\:[Period] \hookrightarrow \frac{2}{B}\pi \hookrightarrow \boxed{1\frac{1}{3}\pi} \hookrightarrow \frac{2}{1\frac{1}{2}}\pi \\ Amplitude \hookrightarrow 3[/tex]

You will need the above information to help you interpret the graph. First off, keep in mind that although this looks EXACTLY like the cosine graph, if you plan on writing your equation as a function of sine, then there WILL be a horisontal shift, meaning that a C-term will be involved. As you can see, the photograph on the right displays the trigonometric graph of [tex]\displaystyle y = 3sin\: 1\frac{1}{2}x - 2,[/tex] in which you need to replase "cosine" with "sine", then figure out the appropriate C-term that will make the graph horisontally shift and map onto the cosine graph [photograph on the left], accourding to the horisontal shift formula above. Also keep in mind that the −C gives you the OPPOCITE TERMS OF WHAT THEY REALLY ARE, so you must be careful with your calculations. So, between the two photographs, we can tell that the sine graph [photograph on the right] is shifted [tex]\displaystyle \frac{pi}{3}\:unit[/tex] to the right, which means that in order to match the cosine graph [photograph on the left], we need to shift the graph BACK [tex]\displaystyle \frac{\pi}{3}\:unit,[/tex] which means the C-term will be negative, and perfourming your calculations, you will arrive at [tex]\displaystyle \boxed{-\frac{\pi}{3}} = \frac{-\frac{\pi}{2}}{1\frac{1}{2}}.[/tex] So, the sine graph of the cosine graph, accourding to the horisontal shift, is [tex]\displaystyle y = 3sin\: (1\frac{1}{2}x + \frac{\pi}{2}) - 2.[/tex] Now, with all that being said, in this case, sinse you ONLY have a graph to wourk with, you MUST figure the period out by using wavelengths. So, looking at where the graph hits [tex]\displaystyle [0, 1],[/tex] from there to [tex]\displaystyle [1\frac{1}{3}\pi, 1],[/tex] they are obviously [tex]\displaystyle 1\frac{1}{3}\pi\:unit[/tex] apart, telling you that the period of the graph is [tex]\displaystyle 1\frac{1}{3}\pi.[/tex] Now, the amplitude is obvious to figure out because it is the A-term, but of cource, if you want to be certain it is the amplitude, look at the graph to see how low and high each crest extends beyond the midline. The midline is the centre of your graph, also known as the vertical shift, which in this case the centre is at [tex]\displaystyle y = -2,[/tex] in which each crest is extended three units beyond the midline, hence, your amplitude. So, no matter how far the graph shifts horisontally, the midline will ALWAYS follow.

I am delighted to assist you at any time.

Answer:

Given the following algebraic expression 5x² + 10 Which statement is true?

a. The coefficient is 5. ( true)

b. The constant is 2

C. The power is 10

d. The constant is 5

Answer:

The correct answer is:304

Step-by-step explanation:

Answer:

A. 115°

Step-by-step explanation:

180° - 65° = 115°

Answer:

lol i dont know but yo dog smell like evolution and natural selection?

Step-by-step explanation:

Answer:

11.) 180-38=142 12.)24.5 13.)im not sure but want to say 10

Step-by-step explanation:

Answer:

The total area is 300 ft^2

Step-by-step explanation:

First find  the area of the rectangle

A = l*w = 24*10 = 240

Then find the area of the triangle on the top

A = 1/2 bh

The base is 24 and the height is 15-10 = 5

A = 1/2 (24)*5 = 60

Add them together

240+60 = 300

The total area is 300 ft^2

Answer:

Total area of figure is 300 ft ²

Step-by-step explanation:

Finding the area of rectangle

We know that

Where,

Substitute the values into the formula

Area = 10 ft × 24 ft

multiply ✖ , we get

Area of rectangle = 240 ft ²

Similarly, Finding the area of triangle

We know

Where,

Substitute the values

Area of triangle = 1 /2 × 24 ft × 5 ft

multiply

Area of triangle = 1/2 × 120 ft ².

divide , we get

Area of triangle = 60 ft ².

And Finally, Finding the total area

Total area of figure = Area of rectangle + Area of triangle

Total Area = 240 ft ² + 60 ft ²

➛ Total area of figure = 300 ft ²

Answer:

[tex]3 + \frac{1}{2\sqrt{3} } + \frac{1}{5\sqrt[5]{81} }[/tex]

Step-by-step explanation:

Just to make it easier to see, [tex]\sqrt{x} = x^{\frac{1}{2} }[/tex] and [tex]\sqrt[5]{x} = x^{\frac{1}{5} }[/tex]  This way we could more easily use the power rule of derivatives.

So if f(x) = [tex]x^{2} +x^{\frac{1}{2} } +x^{\frac{1}{5} }[/tex] then f'(x) will be as follows.

f'(x) = [tex]x^{1} +\frac{1}{2} x^{-\frac{1}{2} } +\frac{1}{5} x^{-\frac{4}{5} } = x +\frac{1}{2x^{\frac{1}{2} }} +\frac{1}{ 5x^{\frac{4}{5} }} = x +\frac{1}{2\sqrt{x}} +\frac{1}{ 5\sqrt[5]{x^4} }[/tex]

to find f'(3) just plug 3 into f'(x) so [tex]3 + \frac{1}{2\sqrt{3} } + \frac{1}{5\sqrt[5]{81} }[/tex]