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-4  is the answer.

Step by step:

2 - x + 3 - 5  =  4                    Combine 3 - 5:

2 - x - 2  =  4                          Combine 2 - 2:

   -x        =  4                          Divide both sides by -1:

         x  =  -4

Setting the polynomial = 0 and factoring, we have

(a -7)(a-3) = 0  ..... so a = 7 or a = 3

And for all a > 7, the polynomial is postive.

So, the greatest "a" that makes the polynomial ≤ 0 is when a = 7.

If you ask for a cos, the calculator will give you a value in the range from -1 up through +1.

For example, cos(45°) = 0.7071.

If you ask for invCos, your will be entering a number in the range from -1 up through +1, and it will give you the size of the angle that has that value for a cosine.   invCos = cos-1

For example, invCos(0.7071) = 45°.

To use on a calculator, depending upon the calculator, press either the (2nd) key or the (inv) key, then cos, then enter the value.

If you change the problem to this:  a² - 10a + 21  =  0

Then, factoring:                            (a - 7)(a - 3)  =  0    --->  a = 3 or a = 7.

Going back to your original problem, an inequalty:  ≤ 0:

If you try any number (in the original problem) smaller than 3, your answer is positive; they don't work.

If you try any number from 3 through 7, your answer is negative; they all work, with the largest number = 7.

If you try any number larger than 7, your answer is positive; they don't work.

Do you mean radians per second?

If you do:  One rotation is equal to  2π  radians; so 2000 rotations is equal to  2000 x 2π  =  4000π radians.

The rate  =   4000π radians / 40 sec  =  100π radians / sec.

Note that...for any given perimeter, the area is at a max when L = W.

So let L=W and so that LW = max area

Now, if we decrease W by 1 and increase L by 1, we have

(W - 1) (L +1) = WL + (W-L) -1  = max area + 0 - 1  = max area - 1

And if  we decrease W by 2 and increase L by 2, we have

(W - 2)(L+2) = WL + 2(W-L) - 4  = max area + 0 - 4 = max area - 4

And with successive increases/decreases of n = 3, 4, 5.......etc., the max area is decreased by n^2. So, the smallest of these decreases occurs between n = 0 and n = 1. Or, looking at it from another perspective, inceasing the Width from (W -1) to W and decreasing the Length from (L+1) to L ( where W = L), results in x = 1.

To the nearest tenth:  t_m = 0.1 seconds,  h_m = 14.9 cm.

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23.5 x 10= 235

Whenever multiplying a decimal by a power of ten, move the decimal place to the right how many times there are terminal zeros. It is similar with dividing, but you would move the decimal place to the left , instead, according to how many terminal zeros are in the power of ten. In this  case, because you are dividing by ten, you would move the decimal place over once, giving you the answer 235. 

The unit rate would be 50 times per second.

a).

b).

Use the first line above to help here.

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V_AB = -20V    the minus sign indicating that B is a higher voltage than A.

V_AC = -22V

Cut out the two resistances, which aren't relevant if you are only considering the voltages here (they are relevant if you want to go on and find currents of course!), then redraw what's left as a single line with A at the bottom.

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Interesting, I also need help with these four questions counting this one. Below this one, and the two above.

n^2 - 9n + 18

We only need to find out where the "zeroes" of this polynomial are.

So setting it to 0 and factoring, we have

(n - 6) (n-3) = 0

So.......n=6 and n=3

Note that when n = 5, the polynomial's value = -2  and when n = 7, the polynomial's value is 4. And for all integers, n > 7, this polynomial is > 0.

So the largest integer that makes this polynomial negative is when n = 5. Note that n = 6 just makes the polynomial = 0, as we found  !!!!!

So, 2+3-5 is 0. x has to be a negative to get a positive outcome. X is equal to -4. I think this is the answer.

This one is interesting....it may be a little difficult to picture what is going on....here is a diagram.....

Note that we have a set of lines drawn through B. And, drawing a line from C to each line through B such that they are perpendicular at their intersection generates a set of right triangles. And, as Anonymous has pointed out, the locus of all possible points at the perpendiculars generates a circle with a radius of BC/2. And if we let BC = the distance from (20,14) to (18,0), then BC = √200. And the area of the circle is pi*(√200/2)²  = pi*(200/4) = 50pi = 157.07. And the greatest integer not exceeding this is 157.

Thank you for this great answer, Anonymous......I couldn't picture what was going on until I read your response and "sussed it out" for myself with a picture!!!

2/3 divided by 1/3. So first, you want to get the reciprocal of 1/3. So, that would be 3/1. You know that dividing would be multiplying but with that reciprocal. 2/3*3/1 has 3 to be canceled out, and the answer would be 2. Hope this helped!

I have answers that I know are incorrect. Please give the answers as they are asked. For example, if it asks for interval, write it as interval. If it is interval notation, write it in interval notation. Thank you!

A^-n = 1/Aⁿ

So

(-6)^-4 = 1/6⁴    = 1/1296

Another way to see this is to note that that the sum of the first n even positive integers is n(n+1)

And the sum of the first n positive odd integers is n^2.

Therefore

n(n+1) - n^2   = n^2 + n - n^2  =  n

And since n = 102 in both cases, then that's the final sum.

The limit will be pi/2

see the graph

4 because 2+2=4

I like you manners  

Collin gets 3/10 so there is 7/10 left

Dave gets 2 shares out of 5 shares of what is left.

this means that he gets 2 fifths of 7 tenths

$$\left({\frac{{\mathtt{2}}}{{\mathtt{5}}}}\right){\mathtt{\,\times\,}}\left({\frac{{\mathtt{7}}}{{\mathtt{10}}}}\right) = {\frac{{\mathtt{7}}}{{\mathtt{25}}}} = {\mathtt{0.28}}$$

Anyway Dave ends up it 7 twenty fifths  or  28% of the entire amount.  

1/2+200=200.5

401/2=200.5

because 401 divide by 2 is 200.5 and 200+0.5 =200.5

so they are both equivalent 

You question looks a bit fishy.    

How can you have

 $$\begin{pmatrix}
1 \\
k
\end{pmatrix}$$

The big one has to be on the top.  

Like so?

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