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steposvetlana [31]
4 years ago
5

P-2+qp;use p=7, and q=4

Mathematics
1 answer:
sergeinik [125]4 years ago
7 0
P-2 + qp
Substitute for p = 7 & q = 4

7-2 + 7*4
5 + 28 = 33

So, your final answer is 33

Hope this helps!
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Find the area of a square whose diagonal length is 32 CM​
BigorU [14]

Answer:

512cm^2

Step-by-step explanation:

Let's imagine cutting the square down that diagonal. Now, we have two isosceles right triangles with a hypotenuse with length 32cm. Let's take one of those triangles for now and focus on it. Let's give the one of the legs of a triangle some length x. Since both legs are the same length, by the Pythagorean theorem:

x^2+x^2=32^2\\\\2x^2=1024\\\\x^2=512\\\\x=\sqrt{512}

Since this leg makes up one side of the square, we can use it to calculate the area of the square, which is just the side length squared:

A=(\sqrt{512})^2 = 512

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3 years ago
Which list shows these lengths in order from greatest to least?
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Answer:

  √3/11, π/24, 3/25, 9/100

Step-by-step explanation:

The 2-decimal value of each of the numbers is ...

  9/100 = 0.09

  π/24 = 0.13

  3/25 = 0.12

  √3/11 = 0.16

Then the order from greatest to least is 0.16, 0.13, 0.12, 0.09. In the original form, that is ...

  √3/11, π/24, 3/25, 9/100 . . . . . matches choice B

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Step-by-step explanation:

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Graph each function. Label x-axis.​
alukav5142 [94]

Answer:

Here's what I get.

Step-by-step explanation:

Question 4

The general equation for a sine function is

y = a sin[b(x - h)] + k

where a, b, h, and k are the parameters.

Your sine wave is

y = 3sin[4(x + π/4)] - 2

Let's examine each of these parameters.

Case 1. a = 1; b = 1; h = 0; k = 0

y = sin x

This is a normal sine curve (the red line in Fig. 1).

(Sorry. I forgot to label the x-axis, but it's always the horizontal axes)

Case 2. a = 3; b = 1; h = 0; k = 0

y = 3sin x

The amplitude changes from 1 to 3.

The parameter a controls the amplitude of the wave (the blue line in Fig. 1).

Case 3. a = 3; b = 1; h = 0; k = 2

y = 3sin x - 2

The graph shifts down two units.

The parameter k controls the vertical shift of the wave (the green line

in Fig. 1).

Case 4. a = 3; b = 4; h = 0; k = 2

y = 3sin(4x) - 2

The period decreases by a factor of four, from 2π to π/2.

The parameter b controls the period of the wave (the purple line in Fig. 2).

Case 5. a = 3; b = 4; h = -π/4; k = 2

y = 3sin[4(x + π/4)] - 2

The graph shifts π/4 units to the left.

The parameter h controls the horizontal shift of the wave (the black dotted line in Fig. 2).

\boxed{a = 3; b = 4; h = \frac{\pi}{2}; k = -2}}

\text{amplitude = 3; period = } \dfrac{\pi}{2}}

\textbf{Transformations:}\\\text{1. Dilate across x-axis by a scale factor of 3}\\\text{2. Translate down two units}\\\text{3. Dilate across y-axis by a scale factor of } \frac{1}{4}\\\text{4. Translate left by } \frac{\pi}{4}

Question 6

y = -1cos[1(x – π)] + 3

\boxed{a = -1, b = 1, h = \pi, k = 3}

\boxed{\text{amplitude = 1; period = } \pi}

Effect of parameters

Refer to Fig. 3.

Original cosine: Solid red line

m = -1: Dashed blue line (reflected across x-axis)

 k = 3: Dashed green line (shifted up three units)

 b = 1: No change

h = π: Orange line (shifted right by π units)

\textbf{Transformations:}\\\text{1. Reflect across x-axis}\\\text{2. Translate up three units}\\\text{3. Translate right by } \pi

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4 years ago
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