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6/2(2 + 1) → 6/2*3 → 3*3 → 9

6/(2(2 + 1)) → 6/(2*3) → 6/6 → 1

    6/2(2+1)           First evaluate the expression in parenthesees.

=  6/2(3)               This can be written like this...

=  \(\frac62(3) \qquad\leftarrow\) Only the thing immediately following the division sign goes in the denominator.

                            Now we can do multiplication/division.

=  3(3)

=  9

"Do you not need to know the derivative of e^x in order to write down its series expansion ?"

No.

e^x is often defined as limit_n→infinity(1+x/n)ⁿ → limit_n→infinity(1 + n(x/n) + (1/2!)n(n-1)(x/n)² + ...)

→ 1 + x + x²/2! + ...

Or sometimes just directly as 1 + x + x²/2! + ...

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what is 6^2/2(3)+4

Power calculation before point calculation before line calculation,
then from left to right.

\(\frac{6^2}{2\times 3}+4=\frac{36}{6}+4=6+4\color{blue}=10\)

Whoever wants can also shorten de Bruch by 6.    (\(\frac{36}{6}=\frac{6}{1}\) )

  !

Heureka is absolutely right on this one - the question is testing your knowledge of vectors and the vector dot product.

It also shows you how vectors greatly simplify problems like this ,   X squared's solution is also right ,but look at the amount of work involved as opposed to using  vectors and the dot product.

ps  If you have the co-ordinates of any triangle's vertices you can of course find all 3 angles using the dot product. Always try and use it for problems like these as it will save you a lot of time and effort.

Isn't that a circular argument ?

Do you not need to know the derivative of e^x in order to write down its series expansion ?

You can extend Guest#1 answer to include the following:

Since the LCM of: 6, 45, 20 =180, therefore:

x =180n + 49, where n=0, 1,2,3.....etc.

So that: x=49, 229, 409, 589.....etc.

what is 0.05% of $700 000

0.05% means: 0.05 / 100 =0.0005, then:

0.0005 x $700,000 =$350.00

Here is the rule and its derivation:

You should now be able to do your questions by choosing an appropriate value for k.

.

two vertices of right triangle ABC are A(-2,6) and C(7,3).

If the right angle is at vertex A and vertex B is on the x-axis,

identify the coordinates of point B

Let \(\vec{A} = \binom{-2}{6}\)
Let \(\vec{B} = \binom{x}{0}\)
Let \(\vec{C} = \binom{7}{3}\)

\(\begin{array}{rcll} \vec{AC} = \ ? \\ \vec{AC} &=& \vec{C} - \vec{A} \\ \vec{AC} &=& \binom{7}{3} - \binom{-2}{6} \\ \vec{AC} &=& \binom{7+2}{3-6} \\ \vec{AC} &=& \binom{9}{-3} \\ \end{array}\)

\(\begin{array}{rcll} \vec{AB} = \ ? \\ \vec{AB} &=& \vec{B} - \vec{A} \\ \vec{AB} &=& \binom{x}{0} - \binom{-2}{6} \\ \vec{AB} &=& \binom{x+2}{-6} \\ \end{array}\)

\(\triangle ABC\) is a right triangle then \(\vec{AC}\cdot \vec{AB} = 0\)

\(\begin{array}{rcll} \mathbf{ \vec{AC}\cdot \vec{AB} } & \mathbf{=} & \mathbf{0} \\ \binom{9}{-3} \cdot \binom{x+2}{-6} &=& 0 \\ 9\cdot(x+2) + (-3)\cdot (-6) &=& 0 \\ 9x+18+18 &=& 0 \\ 9x+ 36 &=& 0 \quad & \quad :9 \\ x+ 4 &=& 0 \\ \mathbf{x} & \mathbf{=} & \mathbf{-4} \\ \end{array} \)

\(\mathbf{\vec{B} = \binom{-4}{0} } \)

Maybe you want to factor this?

    x² - 16                    This is a "difference of squares," so we can factor it like this...

=  (x + 4)(x - 4)

The prefix "iso" indicates "the same as", so  isoneutronic means the same number of neutrons.

The mass number tells you the number of nuclides that comprise the nucleus of an atom. These are protons and neutrons.  The atomic number tells you the number of protons on the nucleus, so:

mass number = number of protons + number of neutrons.

A:   76 = 32 + number of neutrons, so number of neutrons = 76 - 32 → 44

B: Has same number of neutrons as A, so 77 = number of protons + 44

    Hence number of protons = 77 - 44 → 33. This is the atomic number of B.

how do i prove triangleABC is a right triangle?

vertices: A(-3, 5), B(4,7), and C(-1,-2)

Let \(\vec{A} = \binom{-3}{5}\)
Let \(\vec{B} = \binom{4}{7}\)
Let \(\vec{C} = \binom{-1}{-2}\)

\(\begin{array}{rcll} \vec{CA} = \ ? \\ \vec{CA} &=& \vec{A} - \vec{C} \\ \vec{CA} &=& \binom{-3}{5} - \binom{-1}{-2} \\ \vec{CA} &=& \binom{-3+1}{5+2} \\ \vec{CA} &=& \binom{-2}{7} \\ \end{array}\)

\(\begin{array}{rcll} \vec{AB} = \ ? \\ \vec{AB} &=& \vec{B} - \vec{A} \\ \vec{AB} &=& \binom{4}{7} - \binom{-3}{5} \\ \vec{AB} &=& \binom{4+3}{7-5} \\ \vec{AB} &=& \binom{7}{2} \\ \end{array}\)

\(\triangle ABC\) is a right triangle if \(\vec{CA}\cdot \vec{AB} = 0\)

\(\begin{array}{rcll} \vec{CA}\cdot \vec{AB} = \ ? \\ \vec{CA}\cdot \vec{AB} &=& \binom{-2}{7} \cdot \binom{7}{2} \\ \vec{CA}\cdot \vec{AB} &=& (-2)\cdot 7 + 7 \cdot 2 \\ \vec{CA}\cdot \vec{AB} &=& -14+14 \\ \vec{CA}\cdot \vec{AB} &=& 0 \\ \end{array}\)

\(\mathbf{\triangle ABC}\) is a right triangle

x mod 20=9, x mod 45=4, x mod 6 =r, solve for x, r

By simple iteration, the smallest positive integer that satisfies all the congruences is=49

Check:

49 / 20 = 2 with a remainder of 9, and:

49 /45 =1  with a remainder of 4, and:

49 / 6 =8   with a remainder of 1. Therefore:

r=1 and x=49

    3.5e-9

=  3.5 * 10^-9            Move the decimal point left (we want to make the number smaller) 9 places.

                               So now there will be 8 zeros and a 3 after the decimal point.

=  0.0000000035

The reciprocal of tangent is cotangent.

\(\tan=\frac{\sin}{\cos} \\~\\ \cot=\frac{\cos}{\sin}\)

Welcome to the forum lovepie!!! 

As far as I know, there is no way to delete your own question....only the mods can delete something.

Here is one way...

These angles form a straight line, so they sum to 180°  .

b + c + d  =  180                   Since  c  and  d  are vertical angles,  c = d .

b + d + d  =  180

b + 2d  =  180

2d  =  180 - b                        Divide through by  2  .

d  =  90 - b/2

These angles form a straight line, so they sum to 180° .

a + d + (a - b)  =  180            Since  a  and  d  are vertical angles,  a = d .

d + d + (d - b)  =  180

d + d + d - b  =  180

3d - b  =  180                       Substitute  90 - b/2  in for  d .

3(90 - b/2) - b  =  180           Distribute the  3  .

270 - 3b/2 - b  =  180           Subtract  270  from both sides.

-3b/2 - b  =  -90                    Multiply through by  2  .

-3b - 2b  =  -180

-5b  =  -180                          Divide both sides by  -5  .

b  =  36

n mod 17 = 3, n mod 16 = 10, n mod 11 = 4, n mod 7 = 0, solve for n.

Since n mod 7=0, then the smallest number that satisfies all the conditions must be a whole multiple of 7 with no remainder. By simple iteration, it turns out that the multiple of 7 that satisfies all the conditions is =406. Therefore, the smallest number of cards in her collection would have to be:

7 x 406 =2,842 cards.

Check: 

2,842 / 17 =167 with a remainder of 3, and:

2,842 / 16 =177 with a remainder of 10, and:

2,842 / 11 =258 with a remainder of 4, and:

2,842 / 7  =406 with a remainder of 0

If the compounding period is not specified, then you take the interest rate to mean that it is compounded to match the periodic payments, which in this case is compounded monthly.

This is the formula you would use to calculate the Future Value of the monthly payments:

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

FV = 2,500 x {[1 + 0.07/12]^(10*12) - 1 / (0.07/12)}

FV = 2,500 x {[ 1.0058333]^120 - 1 / (0.0058333)}

FV = 2,500 x                173.08481.........

FV =$432,712.02 - This is the Future Value of the monthly deposits for 10 years @ 7% comp. monthly.

There are many ways to prove that a triangle is right, but this method, I believe, is the best and most efficient. Our first task is to find the actual length of the sides. To do this, I will utilize a fomula called the distance formula. The formula is the following:

\(d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\)

Let's use this to find the distances:

Do you remember that famous formula called the Pythagorean theorem? It describes the relationship of the sides of a right triangle, stating that the sum of the squares of the legs is equal to the hypotenuse squared:

\(a^2+b^2=c^2\)

The converse happens to also be true; in other words, if \(a^2+b^2=c^2\), then the triangle is right. Let's check to see if this condition is true.

Before we do, however, we need to understand which sides are the legs and which is the hypotenuse. The hypotenuse is always the longest side of the triangle. Although we do not know the exact length of the segments AB, BC, and CA, we can determine which side is the longest since we know that \(AB=\sqrt{53}\), \(BC=\sqrt{106}\), and \(CA=\sqrt{53}\). Logically speaking,  \(\sqrt{106}>\sqrt{53}\) because the higher the radicand, the larger the actual value is. Now that we understand the hypotenuse is \(BC\), the legs do not matter. One is assigned the value a and one is assigned the value b. To confirm that this triangle is indeed right, plug in the values into the equation:
 

Because \(a^2+b^2=c^2\), we can conclude that this triangle is indeed right.

Hi Rosala,

I’m a genetically enhanced chimp, with advanced mind reading ability.

The (e/m) ratio ascends in this order:  n < α < p < e (neutron, alpha particle, proton, and electron)

The neutron does not have a charge, so the value is zero. An alpha particle is a helium nucleus (2 protons, 2 neutrons); it is very heavy compared to the other particles. The proton and electron are next. The proton has the same charge as an electron but it is 1837 times heaver.

I don’t know which of your choices is correct; my mind reading ability isn’t that advanced. 

GA

A convex, 11-sides polygon (known as a hendecagon) may only have, at most, 3 acute interior angles.

This is because each acute angle results in an exterior angle that is greater than 90 degrees. If you were to add a fourth exterior angle, the sum of all the exterior angles will be greater than 360 degrees. Of course, the sum of all the exterior angles in a convex polygon must be equal to 360, so if the sum of all are greater than this number, you can cannot create a convex polygon. 

Wow! That is such a mouthful of an expression, but I will attempt to evaluate it anyway. 

If I am not mistaken, the expression is the following:
 

\((\frac{62^{62}-62^{62-1}}{62^{62-1}})^{\sqrt{62^{62-1}+62^{62}-62^{62-2}}}\)

At this point, I cannot simplify the exponent any further. I have done an impressive job of simplifying 

\((\frac{62^{62}-62^{62-1}}{62^{62-1}})^{\sqrt{62^{62-1}+62^{62}-62^{62-2}}}\) into \(61^{({62^{30}*\sqrt{3905}})}\). I will now have to utilize an extremely accurate calculator for a calculation of this size. My only option is to provide you with an exponent tower; that's how huge it is!

\((\frac{62^{62}-62^{62-1}}{62^{62-1}})^{\sqrt{62^{62-1}+62^{62}-62^{62-2}}}=61^{({62^{30}\sqrt{3905}})}\approx 10^{10^{55.81927966668297}}\)