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Each term in the given arithmetic sequence can be made by subtracting 5

To find the next term, subtract 5 from the last given term:

\(-17-5=-22\), so \(-22\) is the next term!

First, know the formula of a sector;

Let r= radius

Let m= measure of central angle in degrees

\(A_{sector}=\pi r^2*\frac{m^{\circ}}{360^{\circ}}\)

Now, let's substitute into this formula. 60mm is the radius, and the central angle is also given, 30 degrees:

Therefore, your answer is correct!

Correct  !!!!

The area   =  (1/2)(radius)^2 * ( radian measure of the central angle)  =

(1/2) (60^2) *(pi / 6)   =

(1/2) (3600) * (pi/6)  =

3600pi / 12

300pi  mm^2

Note that   5/7   =  0.7142857     [ the  7142857 is a repeating string ]

Try this  :

 a₂ = 1, a₃  = 1, a₄  = 1,  a₅  = 0, a₆ = 4, a₇ = 2

1/2! + 1/3! + 1/4! + 0/5! + 4/6! + 2/7!   =     5 / 7

I didn''t really have a mathematical way to solve this.....but, note  that

1/2!  + 1/3! + 1/4!  + 1/5! = 43/60 > 5/7

And 

1/2!  + 1/3!   + 1/4!  = 17/24  <  5/7

So......we need that    5/7  - 17/24   =  1/168   =   a₆/6! + a₇/7!

So........it was just a case of  letting a₅ = 0 and then  manipulating a₆  and a₇  to find a solution

Perhaps some other mathematician can find a more elegant way to the solution....!!!!

4.69

The first thing we do is move the a₇/7! to the other side:

\(\frac57-\frac{a_7}{7!}=\frac{5\cdot6!}{7!}-\frac{a_7}{7!}=\frac{5\cdot6!-a_7}{7}\frac1{6!},\)

we require the first fraction to be an integer. This is the case for a₇=2, now we can bring the next fraction to the left:

\(\frac{514}{6!}-\frac{a_6}{6!}=\frac{514-a_6}{6}\frac{1}{5!},\)

for the first fraction to be an integer a₆ needs to equal 4. We bring the next fraction to the left:

\(\frac{85}{5!}-\frac{a_5}{5!}=\frac{85-a_5}{5}\frac1{4!},\)

which requires a₅ to be zero. We bring the next fraction to the left:

\(\frac{17}{4!}-\frac{a_4}{4!}=\frac{17-a_4}{4}\frac1{3!},\)

which requires a₄ to be 1. We continue on our course:

\(\frac{4}{3!}-\frac{a_3}{3!}=\frac{4-a_3}3\frac12,\)

obviously a₃ should be one. We are left with:

\(\frac12=\frac{a_2}{2!},\)

and we get a₂=1, case solved. Checking our answer with the calculator:

\(\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+\frac{4}{6!}+\frac{2}{7!}-\frac57=0.0000000000000000143,\)

which is purely a machine error, since 1/7! equals 0.0001984126984127 which is far larger than our error.

10%  of  $84    =    10/100  *  $84    =    $840/100    =    $8.40

Using  (a^m)^n   =   a^(m * n), we have

(3^8)^4   =   3^(8 * 4)   =  3^32

the answer is 19

I believe so......

Here's another one....

Prove that

Sin (x)  /  n       =    6

Thank you! I really appreaciate it!

AE * EB   =   CE * ED

12 * 22    =   8 * x

264   =   8x                   Divide both sides by 8 .

33     =   x

Don't see any choices, but I believe the answer is that 

CE*DE  = AE*BE

I believe that we need to know something about at least one of the individual's rate of work per minute to have a definite solution for this.

Note that  Mike and Joshua   must do   1/120 of the job in one minute [since it takes them 120 minutes for the whole job]

And Mike and Kelvin must do 1/180 of the job in one minute [ since it takes them 180 minutes for the whole job ]

Let M be the part of the job that Mike can do in one minute , J the part of the job that Joshua can do in one minute and K the part of the job that Kelvin can do in one minute

Then we have the following system

M +  J  =  1/120

M + K  =  1/180

Subtracting the second equation from the first, we have that J - K = 1/360....i.e., Joshua can do 1/360  more of the job per minute than Kelvin can

Suppose that Mike can do   1/200 of the job each minute

Then Joshua can do  1/300 of the job each minute and Kelvin can  do 1/1800 of the job each minute......and  1/300 - 1/1800  =  1/360

Or......suppose that Mike can do  1/300 of the job each minute

Then Joshua can do 1/200 of the job each minute and Kelvin can do 1/450 of the job each minute

And  1/200 - 1.450  =  1/360

Thus....there are infinite solutions to this......all we definitely know is that each person must do less than 1/180  of the job every minute

I don't see any shaded regions, so would the answer be 0.00 units?

If there is a picture, please add it again, just in case.

or sqrt(-7²)

Isn´t it the same?

Maybe I meant n :) but I don´t mind x 

It should have been the same letter, I´m sorry

-3(m - 2)   =   21

Divide both sides of the equation by   -3   .

m - 2   =   21 ÷  -3

m - 2   =   -7

Add 2 to both sides of the equation.

m   =   -7 + 2

m   =   -5

1.

area of old rectangle = L * W

area of new rectangle = (1.5) * L * (1.5) * W

area of new rectangle = 2.25 * L * W

area of new rectangle = 2.25 * area of old rectangle

area of new rectangle = 2.25 * 112 cm²

area of new rectangle = 252 cm²

2.

\(\angle{STU}=\frac12\,\cdot\,\overset{ \frown}{SU} \\~\\ 64^{\circ}=\frac12\,\cdot\,\overset{\frown}{SU} \qquad \text{ Multiply both sides of the equation by 2.} \\~\\ 128^{\circ}=\overset{\frown}{SU}\)

3.

\(m\overset{\frown}{HJ}=180^{\circ}-101^{\circ}=79^{\circ}\)

Also, I noticed the 281º in the bottom right of the circle.

\(m\overset{\frown}{HJ}=360^{\circ}-281^{\circ}=79^{\circ}\)

Cease.

Cease.

\(\frac{E}{e}=\frac{R+r}{R-r}\)

Multiply both sides of the equation by  (R - r)  .

\((R-r)\,*\,\frac{E}{e}=R+r\)

Distribute the   \( \frac{E}{e} \)  .

\(R*\frac{E}{e}\,-\,r*\frac{E}{e}=R+r\)

Add   \( r*\frac{E}{e} \)   to both sides of the equation.

\(R*\frac{E}{e}=R+r+r*\frac{E}{e}\)

Subtract R from both sides of the equation.

\(R*\frac{E}{e}-R=r+r*\frac{E}{e}\)

Factor out an   r   on the right side.

\(R*\frac{E}{e}-R=r(1+\frac{E}{e})\)

Divide both sides of the equation by   \( (1+\frac{E}{e}) \)  .

\(\frac{R*\frac{E}{e}-R}{1+\frac{E}{e}}=r\)

Cease.

He no love you no more.