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v = u + at\

I'm assuming u is the initiatial velocity.

u = 3m/s

a = 1.5m/s^2

t = 7

v = 3 + 1.5*7 = 3 + 10.5 = 13.5

velcocity is 13.5 m/s

If you don't understand anything feel free to ask.

Given \(x+y=10\) it follows, \(x=10-y\) thus, \(x^2=(10-y)^2=(100-20y+y^2)\) (1)

\(x^2+y^2=56\)  (2)

substitute (1) into (2)

\(100-20y+y^2+y^2=56\) simplify and subtract 56 (Setting the quadratic equation equals zero)

\(2y^2-20y+44=0\) Divide by 2

\(y^2-10y+22=0\)

Use the Quadratic formula 

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)            (a=1, b=-10 , c=22)

\(x = {10 \pm \sqrt{(-10)^2-4*1*22} \over 2*1}=5+\sqrt{3}\)  and \(5-\sqrt{3}\)

Now using x+y=10 again, substitute the x value to find y

\(5+\sqrt{3}+y=10\)

\(y=10-(5+\sqrt{3})\)

\(y=5-\sqrt{3}\)

Substitute the other x value

\(5-\sqrt{3}+y=10\)

\(y=10-(5-\sqrt{3})\)

\(y=5+\sqrt{3}\)

Thus the pairs of real numbers that satisfy this:

\((5+\sqrt{3},5-\sqrt{3})\) & \((5-\sqrt{3} ,5+\sqrt{3})\)

There should be a button next to "Flag" and "reply", which says "edit". You should edit your work to prevent further confusion.

Yes sorry that is what I meant

The numbers for 1) can't be -1 and 0, since it would be 2^(-1) = 2^0. You mean 1 right?

anything to the 0th power is 1.

So 2^x * 2^y = 1.

RIght?

so if you have a number with a neg. exponent, it's basically the reciprocal of that.

Well, a number and it's reciprocal multiplied equals one. 

Just stick to 2^-1 * 2^1 = 2^0. We have (-1, 1)

Well, we know about reciprocals, so 2^3/ 2^(-3) = 8/ ( 1/8). That's 8^8 = 64. 64 = 2^6. We have -3 and 6.

so 1/8 times x^-3 = 10^y.

Huh.

Well then y must also be negative.

so, if 10^-3, it's 1/1000, so 1/8 * 1/125 = 1/1000.

Funny. 1/125 = 5^-3.

If you don't understand anything feel free to ask.

1. When we multiply exponents we add the powers. So if we have one negative number we can do -1. -1+1 will equal 0 so the numbers can be -1 and 0.

2. For this one, we are dividing, or subtracting the powers. If we make the power of the quotient negative, the other has to be higher than three. Thus we could do the first blank to be four. 3-4=-1 so the second would be -1.

3. If the base of the product is ten and one factor is two, the other factor must be 5. When we multiple 2^-3 and 5^-3 we get 10^-6

Let me know if you need additional help!

Well said. Only your instructors will know.

Besides, it's not like everyone in your course knows you.

Chances are that NO ONE knows you.

My friends can't even guess my username even if I told them what letter it began with.

Asking for help is perfectly normal and the only way you can learn.

It said it was wrong.

I can't figure out the height.

D/10 = T................(1)

D/15 = T - 1/3.........(2)
When he cycles @ 15mph, that means it is going to take him 10/15 =2/3 of the time of cycling @ 10 mph. 
Solve (1) and (2) above and you will get:
D = 10 miles
T = 1 hour

[2*10] / [1+2/3] =20 x 3/5 =12 mph - his average speed.

I only need the answer but the working out to help me understand it would be amazing as well <3

A fair decision would mean that all people get the same number of slips.  This could only happen with 8 people from the possibile numbers supplied.

"What is the radius of the smallest circle that contains both of the circles x^2 + y^2 = 1 and (x - 1)^2 + (y - 2)^2 = 4?"

Quadrilateral ABCD is a square. Let a, b, c be parallel lines passing through A, B, C, respectively. The distance between lines a and b is 12, and the distance between lines b and c is 17. Find the area of square ABCD.

Quadrilateral ABCD is a square. Let a, b, c be parallel lines passing through A, B, C, respectively.
The distance between lines a and b is 12, and the distance between lines b and c is 17.
Find the area of square ABCD.

\(\begin{array}{|lrcll|} \hline (1) & \mathbf{\sin(\varphi)} &=& \mathbf{\dfrac{12}{x}} \\\\ (2) & \mathbf{\cos(\varphi)} &=& \mathbf{\dfrac{17}{x}} \\ \hline & \mathbf{\sin^2(\varphi)+\cos^2(\varphi)} &=& \mathbf{1} \\\\ & \left(\dfrac{12}{x}\right)^2+\left(\dfrac{17}{x}\right)^2 &=& 1 \\\\ & \dfrac{144}{x^2}+\dfrac{289}{x^2} &=& 1 \quad | \quad \cdot x^2 \\\\ & 144+289 &=& x^2 \\ & 433 &=& x^2 \\ & \mathbf{x^2} &=& \mathbf{433} \\ \hline \end{array}\)

The area of square ABCD is 433

N=2; R=0.06;PV=2700; FV=PV*(1 + R)^N; print"FV =$",FV

FV =3,033.72 pounds

3024.

substitue the values into the equation

A table of values of a linear function is shown below. Find the output when the input is n?

\(\begin{array}{|rcll|} \hline \mathbf{a_n} &=& \mathbf{a_1+(n-1)d} \quad | \quad a_1 = -11,\ d=-3 \\\\ a_n &=&-11+(n-1)*(-3) \\ a_n &=&-11-3n+3 \\ \mathbf{a_n} &=& \mathbf{-8-3n} \\ \hline \end{array}\)

Here is an easier approach.

The lowest common multiple of 5, 7, 9 is  5*7*9 = 315

579000/315 = 1838 and a bit

579999/315 = 1841 and a bit

so

1839*315 = 579285

1840*315 = 579600

1841*315 = 579915

They are your only 3 answers

Hi Chrissy, you should give yourself a point for all that work.

(Note: I have not checked it)

Maybe I will try it for myself......

Ok problem ... 579425 is not divisible by 9    

I am still impressed by your effort    

As follows:

Hana writes the following sums:

\(1^3 = 1, \\ 2^3 = 3 + 5, \\ 3^3 = 7 + 9 + 11, \\ 4^3 = 13 + 15 + 17 + 19\)

If the sum of \(n^3\) contains the sum 2021, then what is n?

I assume:

\(\begin{array}{|rcll|} \hline n^2-(n-1) ~\le~ 2021 ~\le~ n^2+(n-1) \\ \Rightarrow \mathbf{n=45} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline 45^3 &=& 1981+1983+\ldots+{\color{red}2021}+\ldots+2025+\ldots+2068+2069 \\\\ && \boxed{2025=45^2 \\ 1981=45^2-(45-1) \\ 2069=45^2+(45-1) }\\ \hline \end{array} \)

Here's a proof by induction: