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3 years ago

Find a, b and h so that f(x)= asin(b(x-h)). Round to two decimal places if necessary.

Mathematics

1 answer:

Natali5045456 [20]3 years ago

6 0

Answer:

The answer is B

Step-by-step explanation:

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Function 1 is defined by the equation p= --3/2r - 5. function 2 is defined by the following table. Which function has a greater

Rus_ich [418]

0-5. Here’s your answer

6 0

3 years ago

If <img src="https://tex.z-dn.net/?f=%5Cmathrm%20%7By%20%3D%20%28x%20%2B%20%5Csqrt%7B1%2Bx%5E%7B2%7D%7D%29%5E%7Bm%7D%7D" id="Tex

Harman [31]

Answer:

See below for proof.

Step-by-step explanation:

Given:

$y=\left(x+\sqrt{1+x^2}\right)^m$

First derivative

$\boxed{\begin{minipage}{5.4 cm}\underline{Chain Rule for Differentiation}\\\\If $f(g(x))$ then:\\\\$\dfrac{\text{d}y}{\text{d}x}=f'(g(x))\:g'(x)$\\\end{minipage}}$

$\boxed{\begin{minipage}{5 cm}\underline{Differentiating $x^n$}\\\\If $y=x^n$, then $\dfrac{\text{d}y}{\text{d}x}=xn^{n-1}$\\\end{minipage}}$

$\begin{aligned} y_1=\dfrac{\text{d}y}{\text{d}x} & =m\left(x+\sqrt{1+x^2}\right)^{m-1} \cdot \left(1+\dfrac{2x}{2\sqrt{1+x^2}} \right)\\\\ & =m\left(x+\sqrt{1+x^2}\right)^{m-1} \cdot \left(1+\dfrac{x}{\sqrt{1+x^2}} \right) \\\\ & =m\left(x+\sqrt{1+x^2}\right)^{m-1} \cdot \left(\dfrac{x+\sqrt{1+x^2}}{\sqrt{1+x^2}} \right)\\\\ & = \dfrac{m}{\sqrt{1+x^2}} \cdot \left(x+\sqrt{1+x^2}\right)^{m-1} \cdot \left(x+\sqrt{1+x^2}\right)\\\\ & = \dfrac{m}{\sqrt{1+x^2}}\left(x+\sqrt{1+x^2}\right)^m\end{aligned}$

Second derivative

$\boxed{\begin{minipage}{5.5 cm}\underline{Product Rule for Differentiation}\\\\If $y=uv$ then:\\\\$\dfrac{\text{d}y}{\text{d}x}=u\dfrac{\text{d}v}{\text{d}x}+v\dfrac{\text{d}u}{\text{d}x}$\\\end{minipage}}$

$\textsf{Let }u=\dfrac{m}{\sqrt{1+x^2}}$

$\implies \dfrac{\text{d}u}{\text{d}x}=-\dfrac{mx}{\left(1+x^2\right)^\frac{3}{2}}$

$\textsf{Let }v=\left(x+\sqrt{1+x^2}\right)^m$

$\implies \dfrac{\text{d}v}{\text{d}x}=\dfrac{m}{\sqrt{1+x^2}} \cdot \left(x+\sqrt{1+x^2}\right)^m$

$\begin{aligned}y_2=\dfrac{\text{d}^2y}{\text{d}x^2}&=\dfrac{m}{\sqrt{1+x^2}}\cdot\dfrac{m}{\sqrt{1+x^2}}\cdot\left(x+\sqrt{1+x^2}\right)^m+\left(x+\sqrt{1+x^2}\right)^m\cdot-\dfrac{mx}{\left(1+x^2\right)^\frac{3}{2}}\\\\&=\dfrac{m^2}{1+x^2}\cdot\left(x+\sqrt{1+x^2}\right)^m+\left(x+\sqrt{1+x^2}\right)^m\cdot-\dfrac{mx}{\left(1+x^2\right)\sqrt{1+x^2}}\\\\ &=\left(x+\sqrt{1+x^2}\right)^m\left(\dfrac{m^2}{1+x^2}-\dfrac{mx}{\left(1+x^2\right)\sqrt{1+x^2}}\right)\\\\\end{aligned}$

$= \dfrac{\left(x+\sqrt{1+x^2}\right)^m}{1+x^2}\right)\left(m^2-\dfrac{mx}{\sqrt{1+x^2}}\right)$

Proof

$(x^2+1)y_2+xy_1-m^2y$

$= (x^2+1) \dfrac{\left(x+\sqrt{1+x^2}\right)^m}{1+x^2}\left(m^2-\dfrac{mx}{\sqrt{1+x^2}}\right)+\dfrac{mx}{\sqrt{1+x^2}}\left(x+\sqrt{1+x^2}\right)^m-m^2\left(x+\sqrt{1+x^2\right)^m$

$= \left(x+\sqrt{1+x^2}\right)^m\left(m^2-\dfrac{mx}{\sqrt{1+x^2}}\right)+\dfrac{mx}{\sqrt{1+x^2}}\left(x+\sqrt{1+x^2}\right)^m-m^2\left(x+\sqrt{1+x^2\right)^m$

$= \left(x+\sqrt{1+x^2}\right)^m\left[m^2-\dfrac{mx}{\sqrt{1+x^2}}+\dfrac{mx}{\sqrt{1+x^2}}-m^2\right]$

$= \left(x+\sqrt{1+x^2}\right)^m\left[0]$

$= 0$

8 0

2 years ago

What matrix results from −8·A?

djverab [1.8K]

Is that all the question says can you give it into more details

So I can help you out

3 0

3 years ago

The figure above shows a square inscribed in a rectangle.

Debora [2.8K]

Answer:

31 units ²

Step-by-step explanation:

Area of a rectangle = length × width

Length = 8 units

Width = 5 units

Area of a rectangle = length × width

= 8 units × 5 units

= 40 units ²

Area of of the blue square = length ²

Length = 3 units

Area of of the blue square = length ²

= (3 units) ²

= 9 units ²

Area of the orange colored part of the figure = Area of a rectangle - Area of of the blue square

= 40 units ² - 9 units ²

= 31 units ²

Area of the orange colored part of the figure = 31 units ²

6 0

3 years ago

1. What is the value of the y variable in the solution to the following system of equations?

ollegr [7]

Part 1)

we have

$3x-6y=3$ ------> equation A

$7x-5y=-11$ ------> equation B

Multiply by $-7$ the equation A

$-7(3x-6y)=-7*3$

$-21x+42y=-21$ ------> equation C

Multiply by $3$ the equation B

$3*(7x-5y)=-11*3$

$21x-15y=-33$ -------> equation D

Adds equation C and equation D

$-21x+42y=-21 \\21x-15y=-33\\---------- \\ 42y-15y=-21-33\\27y=-54 \\ y=-2$

therefore

the answer Part 1) is the option A

$y=-2$

Part 2)

we have

$3x+y=6$

$y=-3x+6$ ------> equation A

$6x+2y=8$

Simplify Divide by $2$ both sides

$3x+y=4$

$y=-3x+4$ ------> equation B

the lines A and B are parallel lines, because the slope m is equal

The system has no solution

therefore

the answer Part 2) is the option D

There is no x value as there is no solution to the system.

Part 3)

we have

$4x+2y=6$ ------> equation A

$x-y=3$

$y=x-3$ ------> equation B

substitute equation B in equation A

$4x+2[x-3]=6$

$6x-6=6$

$6x=12$

$x=2$

therefore

the answer part 3) is the option D

$x=2$

Part 4)

Let

x---------> The number of one-step equations

y---------> The number of two-step equations

we know that

$x+y=1,120$

$x=1,120-y$ -------> equation A

$3x-2y=1,300$ ------> equation B

substitute equation A in equation B

$3[1,120-y]-2y=1,300$

$3,360-5y=1,300$

$5y=3,360-1,300$

$5y=2,060$

$y=412$

therefore

the answer part 4) is the option D

$y=412$

8 0

3 years ago

Read 2 more answers