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Solve for y:

y^2 + 49 = 0

Subtract 49 from both sides:

y^2 = -49

Take the square root of both sides:

y = 7i        or        y = -7i

Solve for x:

x^4 + 10 x + 25 = 0

Subtract 10 x - 10 x^2 from both sides:

x^4 + 10 x^2 + 25 = 10 x^2 - 10 x

x^4 + 10 x^2 + 25 = (x^2 + 5)^2:

(x^2 + 5)^2 = 10 x^2 - 10 x

Add 2 (x^2 + 5) λ + λ^2 to both sides:

(x^2 + 5)^2 + 2 λ (x^2 + 5) + λ^2 = -10 x + 10 x^2 + 2 λ (x^2 + 5) + λ^2

(x^2 + 5)^2 + 2 λ (x^2 + 5) + λ^2 = (x^2 + 5 + λ)^2:

(x^2 + 5 + λ)^2 = -10 x + 10 x^2 + 2 λ (x^2 + 5) + λ^2

-10 x + 10 x^2 + 2 λ (x^2 + 5) + λ^2 = (2 λ + 10) x^2 - 10 x + 10 λ + λ^2:

(x^2 + 5 + λ)^2 = x^2 (2 λ + 10) - 10 x + 10 λ + λ^2

Complete the square on the right hand side:

(x^2 + 5 + λ)^2 = (x sqrt(2 λ + 10) - 5/sqrt(2 λ + 10))^2 + (4 (2 λ + 10) (λ^2 + 10 λ) - 100)/(4 (2 λ + 10))

Solve using the quadratic formula:

x = 1/2 (sqrt(2) sqrt(λ + 5) + sqrt(2) sqrt(-(25 + 10 λ + λ^2 + 5 sqrt(2) sqrt(λ + 5))/(λ + 5))) or x = 1/2 (sqrt(2) sqrt(λ + 5) - sqrt(2) sqrt(-(25 + 10 λ + λ^2 + 5 sqrt(2) sqrt(λ + 5))/(λ + 5))) or x = 1/2 (sqrt(2) sqrt((-25 - 10 λ - λ^2 + 5 sqrt(2) sqrt(λ + 5))/(λ + 5)) - sqrt(2) sqrt(λ + 5)) or x = 1/2 (-sqrt(2) sqrt(λ + 5) - sqrt(2) sqrt((-25 - 10 λ - λ^2 + 5 sqrt(2) sqrt(λ + 5))/(λ + 5))) where λ = -5 + (5 2^(2/3))/(3/5 (i sqrt(1119) + 9))^(1/3) + (5/6)^(2/3) (i sqrt(1119) + 9)^(1/3)

Substitute λ = -5 + (5 2^(2/3))/(3/5 (i sqrt(1119) + 9))^(1/3) + (5/6)^(2/3) (i sqrt(1119) + 9)^(1/3) and approximate:

x = -1.61762 - 1.035 i or x = -1.61762 + 1.035 i or x = 1.61762 - 2.04014 i or x = 1.61762 + 2.04014 i

2. Start with sin(x)        amplitude is 1      want 4   so multiply by 4

     4 sin(x)                    period is 2 pi     frequency is 1/2pi     want period = 4     frequency = 1/4

                                         1/2pi * m  = 1/4       m= modulator    

                                          m= 2pi/4 = pi/2

4 sin (x pi/2)

AND shift it up 1....just add 1

4 sin (x pi/2)  + 1   

Solve for x:

x^4 - 81 = 0

Add 81 to both sides:

x^4 = 81

Taking 4^th roots gives 3 times the 4^th roots of unity:

x = -3    or    x = -3 i    or    x = 3 i    or    x = 3

https://youtu.be/eHiH1NytwjM

Let the price of senior citizen ticket = S

Let the price of a child's ticket = C

3S + C = 38.....................................(1)

3S + 2C = 52, solve for C, S...........(2)   subtract (1) from (2)

C = $14 - price of 1 child's ticket

Sub this in (1)

3S + 14 = 38    subtract 14 from both sides

3S = 24            Divide both sides by 3

S = $8 - price of a senior citizen's ticket.

Here is the first one to get you started.....

sqrt (27/75) = sqrt((3*3*3) /(3*5*5) )  = sqrt (( 3*3)/(5*5) )  =  3/5

sqrt(x^4) = x^2

sqrt(1/y^2) = 1/y

result :  3/5  *  x^2/y

Solve for x:

x^2 + 6 x + 10 = 0

Using the quadratic formula, solve for x.

x = (-6 ± sqrt(6^2 - 4×10))/2 = (-6 ± sqrt(36 - 40))/2 = (-6 ± sqrt(-4))/2:

x = (-6 + sqrt(-4))/2 or x = (-6 - sqrt(-4))/2

Express sqrt(-4) in terms of i.

sqrt(-4) = sqrt(-1) sqrt(4) = i sqrt(4):

x = (-6 + i sqrt(4))/2 or x = (-6 - i sqrt(4))/2

Simplify radicals.

sqrt(4) = sqrt(2^2) = 2:

x = (-6 + i×2)/2 or x = (-6 - i×2)/2

Factor the greatest common divisor (gcd) of -6, 2 i and 2 from -6 + 2 i.

Factor 2 from -6 + 2 i giving -6 + 2 i:

x = (2 i - 6)/2 or x = (-6 - 2 i)/2

Cancel common terms in the numerator and denominator.

(2 i - 6)/2 = -3 + i:

x = i - 3 or x = (-6 - 2 i)/2

Factor the greatest common divisor (gcd) of -6, -2 i and 2 from -6 - 2 i.

Factor 2 from -6 - 2 i giving -6 - 2 i:

x = -3 + i or x = (-(2 i) - 6)/2

Cancel common terms in the numerator and denominator.

(-(2 i) - 6)/2 = -3 - i:

x = -3 + i        or        x = -i - 3

sqrt (36 g^6)  =   |g^3| sqrt (36)  =  6 |g ^3|      Because g^6 is POSITIVE result, you must make sure g^3 is positive

cbrt(125 x ^21y^24) = 5 x^7y^8

Larry has 4-cent stamps and 9-cent stamps, which he can combine to produce various amounts of postage.
For example, he can make 40 cents by using four 9-cent stamps and a 4-cent stamp, or by using ten 4-cent stamps.
However, there are some amounts of postage he can't make exactly, such as 10 cents.

What is the largest number of cents that Larry CANNOT make exactly from a combination of 4- and/or 9-cent stamps?

Explain how you know your answer is correct. (You should explain two things:
why Larry can't make the amount of your answer, and
why he CAN make any bigger amount.)

$\begin{array}{l} \cdots\{4\} \cdots \{8,9\}\cdots \{12,13\}\cdots \{16,17,18\} \cdots \{20,21,22\} \cdots {\color{red}23} \{\underbrace{24,25,26,27}_{\underset{27+4=31}{\underset{26+4=30,}{\underset{25+4=29,}{\underset{24+4=28,}{}}}} }\} \{\underbrace{28,29,30,31}_{\underset{31+4=35}{\underset{30+4=34,}{\underset{29+4=23,}{\underset{28+4=32,}{}}}} }\} \{\underbrace{32,33,34,35}_{\underset{35+4=39}{\underset{34+4=38,}{\underset{33+4=37,}{\underset{32+4=36,}{}}}} }\} \text{ and so on} \\\\ \{\text{reachable}\} \\ \cdots \text{not reachable}\\ \end{array}$

23 is the largest number of cents that Larry CANNOT make exactly from a combination of 4- and/or 9-cent stamps.

After four consecutive reachable numbers, 24, 25, 26, 27, he CAN make any bigger amount.

Here's the answer to part 1:

.

At the point in the rotation as it appears in the illustration, the horizontal force of the piston rods on the piston tend to want to push the piston to the right: i.e. towards the right side wall and away from the left side wall.  Hence there will be a non-zero reaction force from the right-side wall on the piston, but no reaction force from the left-side wall on the piston.  Later in the cycle, the opposite will hold.   

.

As follows:

.

If c is a constant such that 9x^2 + 10x + c is equal to the square of a binomial, then what is c?

$\begin{array}{|lrcll|} \hline &(9x^2+10x+c) &=& (3x+b)^2 \\ & &=& 9x^2+\underbrace{6b}_{=10}x+\underbrace{b^2}_{=c} \\\\ \hline (1) & 6b &=& 10 \\ & b &=& \frac53 \\\\ (2) & c &=& b^2 \\ & c &=& (\frac53)^2 \\ & c &=& \frac{25}{9} \\ \hline &\left(9x^2+10x+\dfrac{25}{9} \right) &=& \left(3x+\dfrac53 \right)^2 \\ \hline \end{array}$

If x^2 ends in 9 then the x must end in 3 or 7

If x^2 is four digits then x must be between 32 and 99

So the only possibilities for x are

33, 37, 43, 47, 53, 57, 63, 67, 73, 77, 83, 87, 93, 97

Square each of those and find the one/s that fit the other two requirements :)

Graph the image of the figure after a dilation with a scale factor of 2 centered at (−2, −2) .

$\begin{array}{lrcll} \text{Formula dilation:} & \boxed{\vec{A}' = (\vec{A}-\vec{C_{Center}})\cdot \lambda +\vec{C_{Center}}} \\ & \vec{A} \text{ before dilation } & \text{$\lambda$ scale factor } = 2 \\ & \vec{A}' \text{ after dilation } & \text{$\vec{C_{Center}}$ center at $\binom{-2}{-2} $ } \\ \end{array}$

$\begin{array}{lrcll} &\vec{A}' &=& (\vec{A}-\vec{C_{Center}})\cdot \lambda +\vec{C_{Center}}\\ & &=& \vec{A}\cdot \lambda +\vec{C_{Center}}\cdot (1-\lambda ) \\ & &=& \lambda \cdot \vec{A} +(1-\lambda )\cdot \vec{C_{Center}} \\ \text{Formula dilation:} & \boxed{\vec{A}' =\lambda \cdot \vec{A} +(1-\lambda )\cdot \vec{C_{Center}}} \\ & (1-\lambda )\cdot \vec{C_{Center}} &=& (1-2)\binom{-2}{-2} \\ & (1-\lambda )\cdot \vec{C_{Center}} &=& (-1)\binom{-2}{-2} \\ & (1-\lambda )\cdot \vec{C_{Center}} &=& \binom{2}{2} \\ \end{array}$

$\begin{array}{lrcll} \boxed{\vec{A}'=\lambda \cdot \vec{A} +\binom{2}{2}}& \vec{A} = \dbinom{-4}{-2}\\ \boxed{\vec{B}'=\lambda \cdot \vec{B} +\binom{2}{2}}& \vec{B} = \dbinom{-5}{2}\\ \boxed{\vec{C}'=\lambda \cdot \vec{C} +\binom{2}{2}}& \vec{C} = \dbinom{-1}{3}\\ \boxed{\vec{D}'=\lambda \cdot \vec{D} +\binom{2}{2}}& \vec{D}= \dbinom{1}{1}\\ \end{array}$

$\begin{array}{lrcll} \boxed{\vec{A}'=2 \cdot \binom{-4}{-2} +\binom{2}{2}}& \vec{A}' = \dbinom{-6}{-2}\\ \boxed{\vec{B}'=2 \cdot \binom{-5}{2} +\binom{2}{2}}& \vec{B}' = \dbinom{-8}{6}\\ \boxed{\vec{C}'=2 \cdot \binom{-1}{3} +\binom{2}{2}}& \vec{C}' = \dbinom{0}{8}\\ \boxed{\vec{D}'=2 \cdot \binom{1}{1} +\binom{2}{2}}& \vec{D}' = \dbinom{4}{4}\\ \end{array}$

Point (-4, -2 ) goes to (-6,-2)

Point (-5, 2) goes to (-8. 6)

Point ( -1, 3) goes to (0, 8)

Point ( 1, 1 ) goes to (4, 4 )

$\sum_{1}^{4}$ 16 (-1/2)^n         is one possibility

The displacement (in centimeters) of a particle moving back and forth along a straight line is given by the equation of motion 

s = 5 sin(πt) + 2 cos(πt),

 where t is measured in seconds. (Round your answers to two decimal places.)

$s = 5 sin(\pi t) + 2 cos(\pi t)\\ \frac{ds}{dt}=v(t)=5\pi cos(\pi t)-2\pi sin(\pi t)$

a) Find the average velocity during each time period.

(i)    [1, 2]  cm/s     

$s(t) = 5 sin(\pi t) + 2 cos(\pi t)\\ s(1) = 5 sin(\pi ) + 2 cos(\pi )=-2\\ s(2) = 5 sin(2\pi ) + 2 cos(2\pi )=0+2=2\\ Average\;\; velocity = \frac{2--2}{2-1} =4cm/sec$

(ii)    [1, 1.1]  cm/s 

(iii)    [1, 1.01]  cm/s 

(iv)    [1, 1.001]  cm/s 

(b) Estimate the instantaneous velocity of the particle when t = 1. 
 cm/s

$v(1)=5\pi cos(\pi )-2\pi sin(\pi )=-5\pi-0=-5\pi\;cm/sec$

You can do the others yourself :)

63^2 = 3,969  Meets ALL the conditions.

Solve for x:

x^2 - 8 x + 14 = 0

Solve the quadratic equation by completing the square.

Subtract 14 from both sides:

x^2 - 8 x = -14

Take one half of the coefficient of x and square it, then add it to both sides.

Add 16 to both sides:

x^2 - 8 x + 16 = 2 [THIS STEP IS THE SECOND ONE OF YOUR 4 CHOICES]

Factor the left hand side.

Write the left hand side as a square:

(x - 4)^2 = 2

Eliminate the exponent on the left hand side.

Take the square root of both sides:

x - 4 = sqrt(2) or x - 4 = -sqrt(2)

Look at the first equation: Solve for x.

Add 4 to both sides:

x = 4 + sqrt(2) or x - 4 = -sqrt(2)

Look at the second equation: Solve for x.

Add 4 to both sides:

x = 4 + sqrt(2)       or         x = 4 - sqrt(2)

Solve for x:

x^2 + 6 x + 10 = 0

Subtract 10 from both sides:

x^2 + 6 x = -10

Add 9 to both sides:

x^2 + 6 x + 9 = -1

Write the left hand side as a square:

(x + 3)^2 = -1

Take the square root of both sides:

x + 3 = i or x + 3 = -i

Subtract 3 from both sides:

x = -3 + i or x + 3 = -i

Subtract 3 from both sides:

x = -3 + i      or      x = -3 - i

Solve for x:

2 x^2 - 8 x + 5 = 0

Divide both sides by 2:

x^2 - 4 x + 5/2 = 0

Subtract 5/2 from both sides:

x^2 - 4 x = -5/2

Add 4 to both sides:

x^2 - 4 x + 4 = 3/2

Write the left hand side as a square:

(x - 2)^2 = 3/2

Take the square root of both sides:

x - 2 = sqrt(3/2) or x - 2 = -sqrt(3/2)

Add 2 to both sides:

x = 2 + sqrt(3/2) or x - 2 = -sqrt(3/2)

Add 2 to both sides:

x = 2 + sqrt(3/2)      or      x = 2 - sqrt(3/2)

Solve for x:

5 x^2 - 34 x + 24 = 0

The left hand side factors into a product with two terms:

(x - 6) (5 x - 4) = 0

Split into two equations:

x - 6 = 0 or 5 x - 4 = 0

Add 6 to both sides:

x = 6 or 5 x - 4 = 0

Add 4 to both sides:

x = 6 or 5 x = 4

Divide both sides by 5:

x = 6        or        x = 4/5