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Second one......one possibility

Midline  = .6

Vertical shift is up  .6 units  

Amplitude  = .3

Period  = pi....so....2 periods in 2 pi

Phase shift....sine graph shifted to the right by 3pi/4 units

The equation is

y = .3sin2(x - 3pi/4) + .6

Here's the graph : https://www.desmos.com/calculator/vr3zoak70p

First one....one possibility

Midline  =  -1

Amplitude  =  2

Vertical shift down 1

Phase shift.....sine curve shifted to  the right by 3pi/4 units

1 period in 2pi  [ normal period ]

So  the equation  is :

y  =  2sin(x - 3pi/4)   - 1

Here's the graph : https://www.desmos.com/calculator/bgaimmpmmr

same

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Thanks, same to you.

The probability is given by

C (6,3) * (.7)^3 * (.3)^3   =  .1852

Here's the second one:

The red line passes through the points  (0, 1)  and  (1, 0) , so

its slope  =  [ 0 - 1 ] / [ 1 - 0 ]  =  -1

Using the point  (0, 1)  and the slope  -1 ,  the equation of the red line is…

y - 1  =  -1x          Add  1  to both sides of the equation.
y  =  -x + 1           Add  x  to both sides of the equation.
x + y  =  1

We want the solutions to include all points to one side of the line, so the inequality will be either

x + y  ≤  1          or          x + y  ≥  1

The point  (0, 0) is in the yellow region, so we want  (0, 0)  to make the inequality true.

0 + 0  ≤  1          or          0 + 0  ≥  1
0 ≤ 1  true          or          0 ≥ 1  false

So the inequality that has  (0, 0) as a solution is the one we want, which is…

x + y  ≤  1

For the first.....the points (1, -2) and (3,1) are on the graph

The slope between these is   [ 1 - - 2 ] / [ 3 - 1 ]  = 3/2

And the equation of the line connecting these points is

y  = (3/2)(x -1) - 2

y = (3/2)x -3/2 - 2

y  = (3/2)x - 7/2

Multiply through by 2

2y = 3x - 7      rearrange as

3x - 2y  = 7

Since we have a dashed line...we are going to have a "<" or ">" sign involved...put a point in the "yelllow"  into this equation to see which inequality sign we need.....(0,0) seems good

3(0) - 2(0)  =  7    ???

0  = 7       ???

And its clear that we need  the  "<"  sign

So.....the equation is

3x - 2y <  7

Here's the graph : https://www.desmos.com/calculator/vn7sb63doc

3x + y + 2z  = 6    (1)

x + y +  4z = 3      (2)

2x + 3y + 2z = 2    (3)

Manuel....we can use the first two equations....write them as

y = -3x - 2z + 6

y = - x - 4z + 3          subtract the first equation from the second

0  =  2x  - 2z - 3

2x  = 2z + 3        (4)

Multiply the second equation by 2

2x + 2y + 8z  = 6      (5)

Put   (4)   into  (3)  and (5)  for 2x

 (2z + 3) + 3y + 2z   = 2   ⇒     3y + 4z  = -1     (6)

(2z + 3) + 2y + 8z  =  6  ⇒       2y  + 10z =  3  (7)

Multiply   (6)  by 2  and (7)  by -3

6y + 8z =  -2

-6y - 30z  = -9        add these

-22z  =  - 11       divide both sides by -22

z  = 1/2

Using (4)     2x   =  2z + 3    

2x  = 2(1/2) + 3

2x  = 4

x  = 2

And using (7)     2y + 10z  = 3

2y + 10 (1/2)  = 3

2y  + 5  = 3

2y  = -2

y = -1 

So    .... { x, y, z }   =  { 1/2, -1, 2 }

If we exclude "0" as a beginning digit, we have 4 * 4!  = 96 possible numbers

The "6"  and "8"   in either the tens or thousands positions will each be larger than their neighboring digits.....

And they can be arranged in two ways.....

And given that the 6 and 8  occupy these positions....the first number can be chosen in two ways and the remaining digits can be arranged in 2 ways

So.......2 * 2 * 2  = 8 numbers are possible

So.....the probability is    8/96   =  1/12

If we subtract the "2" from both sides, we have that

√[ x + 1 ]  =  - 2

But this is impossible.....the radical will either always produce an answer  ≥ 0.....never a negative!!!!

(2/3)t - 1 < t+ 7 ≤ -2t + 15

First, we have

(2/3) t - 1 < t + 7     multiply through by 3

2t - 3 < 3t + 21    subtract  21, 2t from both sides

-24 < t

Then, we have

t + 7 ≤  -2t+ 15     add 2t to both sides, subtract 7 from both sides

3t ≤ 8    divide both sides by 3

t ≤  8/3

So......the answer is

(- 24, 8/3 ]

[F + L]/2 x 18 =n^5

[p + p + 34]/2 x 18 =n^5

18[p + 17] =n^5

18p + 306 =n^5

Iterate: [3^5 - 306] / 18 =No integer solution

              [4^5 - 306] / 18 =No integer solution

              [5^5 - 306] / 18 = No integer solution

              [6^5 -306] / 18 =415 as the first term. So, we have:

n = 6 and p =415. Therefore:

6 x 415 =2,490.

\(\frac{2}{3}t-1\,\text{<}\,t+7\le-2t+15 \\~\\ \text{Let's split it into two inequalities.} \\~\\ \frac{2}{3}t-1\,\text{<}\,t+7\\~\\ -1-7\,\text{<}\,t-\frac{2}{3}t\\~\\ -8\,\text{<}\,\frac{1}{3}t\\~\\ -24\,\text{<}\,t\\~\\ \text{and}\\~\\ t+7\le-2t+15\\~\\ t+2t\le15-7\\~\\ 3t\le8\\~\\ t\le\frac{8}{3}\\~\\ \text{So t is greater than -24 but less than or equal to 8/3.}\\~\\ \text{t is in the interval:}\quad (-24,\frac{8}{3}]\)

You made a small mistake in the second to the last expansion. It should be:

x^2 - 11 x + 28 + (3 - x) (x + 2)

-x^2 + x + 6 + x^2 - 11 x + 28

-10 x + (x^2 - x^2) + (28 + 6)

-10x + 34, or: 34 - 10x

Interval or integral?

Step by Step:

1. Write out expression 

(x-4)(x-7) + (3-x)(2+x)

2. Expand

x² - 7x - 4x - 28 + (-)x² - 2x + 3x + 6

3. Simplify for both sides around the '+'

x² -11x - 28 + (-)x² + x + 6

4. Simplify expression to get the answer.

-10x - 22

I C thx for the answer

Differences:

The frequency of Gamma rays is higher than the rest. The frequency of radio waves is lower than the rest. The wavelength of Gamma rays is shorter than the rest. The wavelength of radio waves is higher than the rest.

Frequency (in order of highest to lowest):

Gamma

X-Ray

UV

Visible

IR

microwaves

radio waves.

Wavelength (in order of highest to lowest):

radio waves

microwaves

IR

Visible

Uv

X-Ray

Gamma

Similarities:

They are all part of the electromagnetic spectrum.

This is the formula you would use to calculate this:

FV =PV x {[1 + R]^N - 1 / R}

FV =45 x {[1 + 0.08/12]^(30*12) - 1 / (0.08/12)}

FV =45 x {[1.00666667]^360 - 1 / 0.00666667}

FV =45 x                 1,490.35944866337.........

FV =$67,066.18 - Balance of the investment account after 30 years.

-2 1/3=7q

-7/3=7q

=> q=-3

during

Begin by moving F a little to the left so that the square has begun it's rotation.

Call the angle EFB theta, call the side of the square b, and BF and FC x and y respectively.

Fill in the angles of the two triangles BEF and FGC and apply the sine rule in both triangles.

Use the fact that x + y = a, the length of the side of the equilateral triangle to deduce that

\(\displaystyle b=\frac{a\sqrt{3}}{(1+\sqrt{3})(\sin\theta+\cos\theta)}\).

So the size of the square varies as it rotates.

Now look at your larger diagram, the one with the dotted line running from D.

Call the point where this line meets BC N, then  DN/DF = sin(angle NFD).

DF  is the diagonal of the square = b.sqrt(2) and angle NFD = theta + 45 deg, so you can now work out DN.

You should find that it's a constant, equal to  a.sqrt(3)/(1 + sqrt(3)).

Tiggsy