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x = -5 and y = -7 and z = 0

(98*4)+(7*4)+7

BODMAS/BIDMAS

98 is 2 off 100 

(100*4)-(2*4) = 392

7 is 3 off 10 

(10*4)-(3*4) = 28

392+28+7

=400+20+7

=427

The total FV of the three payments will be:

R 8,141.59.

You can find their FV by evaluating them individually, which is the easiest way, then add them all up . This is the formula you would use to do that: Remember, the deposits are made at the BEGINNING of the year, which is important.

FV=PV[1 + R]^N, where FV=Future Value, PV=Present value, R=Interest rate per period, N=Number of periods.

If the terminal side of 'theta' is in quadrant II and the reference angle is pi/3,

find the exact values of sin(theta) and cos(theta)

II. Quadrant : \(\cos(\theta) <0 \\ \sin(\theta) >0\)

So the answer is c.

\(\begin{array}{|rcll|} \hline \sin^2(\theta) + \cos^2(\theta) &=& 1 \\ \left( \frac{\sqrt{3}}{2} \right)^2 + \left( -\frac12 \right)^2 &=& 1 \\ \frac{3}{4} + \frac14 &=& 1 \\ 1 &=& 1 \\ \hline \end{array} \)

I should have added that x has been taken to be positive.

9+10

9=3*3; 10=2*5

3*2=6; 3*5=15

6+15=old a*s vine

In arithmetic, the rule is that you are not allowed to divide by zero. The fraction 1/0 is undefined in the sense that the operation of dividing 1 by zero simply can't be carried out. You are not allowed to do it.

Compare that with the situation of asking what happens to the fraction 1/x in the limit as x tends to zero. Here you can say that as x gets smaller and smaller, (though not actually reaching zero), the value of the fraction gets bigger and bigger. It gets bigger without limit. Think of a number as big as you want and a value of x can be found so that the value of the fraction is bigger than your number, and the value of the fraction will be bigger than your number for all (non zero) values of x smaller than that. In this case we say that the limit is infinity, and write (inaccurately) that 1/0 = infinity.

Find the M is the midpoint of the line segment PQ, what is the coordinates of point Q? P (4, 5), M (-2, -3)

Answers:

\(a. (2,2) \\ b. (6,8) \\ c. (-8,-11) \\ d. (-8,-15) \\ e. (-4, -7.5) \\\)

\(\begin{array}{|rcll|} \hline \frac{\vec{P} + \vec{Q}} {2} &=& \vec{M} \quad & | \quad \cdot 2\\\\ \vec{P} + \vec{Q} &=& 2\cdot \vec{M} \quad & | \quad - \vec{P} \\\\ \vec{Q} &=& 2\cdot \vec{M} - \vec{P} \quad & | \quad \vec{M} = \binom{-2}{-3} \quad \vec{P} = \binom{4}{5}\\\\ \vec{Q} &=& 2\cdot \binom{-2}{-3} - \binom{4}{5} \\\\ \vec{Q} &=& \binom{-2\cdot 2}{-3\cdot 2} - \binom{4}{5} \\\\ \vec{Q} &=& \binom{-4}{-6} - \binom{4}{5} \\\\ \vec{Q} &=& \binom{-4-4}{-6-5} \\\\ \mathbf{ \vec{Q} } &\mathbf{=} & \mathbf{ \binom{-8}{-11} } \\\\ \hline \end{array}\)

Point Q is  (-8, -11)

(c.)

25,000*(1.15^10)

Hi Max and ElectricPavlov       

Let's consider this problem for a moment....

Here is the graph of      \(y=\frac{12}{x}\)

You can see tha as x approaches 0 from the positive side that   y approaches  +infinity

BUT

As x approaches 0 from the negative side then                            y approaches  - infinty

This is why   12/ 0    is undefined .....    it cannot be negative infinity and positive infinity both at the same time ://

so

\(\displaystyle\lim_{x\rightarrow 0^+}\:\frac{12}{x}=+\infty\\~\\ \displaystyle\lim_{x\rightarrow 0^-}\:\frac{12}{x}=-\infty\\~\\\)

I must have forgotten some rule... trying to figure out how to simplify this... 

\(x=\frac{4\left(a+b\right)±\sqrt{16\left(a+b\right)^2-48ab}}{24}\)

Simplifies to: \(x=\frac{\left(a+b\right)±\sqrt{a^2-ab+b^2}}{6}\)

\(\begin{array}{|rcll|} \hline x &=& \frac{4\left(a+b\right) \pm \sqrt{16\cdot \left(a+b\right)^2-48ab}}{24} \\ &=& \frac{4\left(a+b\right) \pm \sqrt{16\cdot \left(a+b\right)^2-16\cdot 3 \cdot ab}}{24} \\ &=& \frac{4\left(a+b\right) \pm \sqrt{ 16 \cdot[~ \left(a+b\right)^2-3\cdot ab ~]}}{24} \quad & | \quad \sqrt{a\cdot b} = \sqrt{a}\cdot \sqrt{ b} \\ &=& \frac{4\left(a+b\right) \pm \sqrt{ 16}\cdot \sqrt{ \left(a+b\right)^2-3ab } }{24} \quad & | \quad \sqrt{ 16}=4\\ &=& \frac{4\left(a+b\right) \pm 4\cdot \sqrt{ \left(a+b\right)^2-3ab } }{24} \\ &=& \frac{4 \cdot[~ \left(a+b\right) \pm \sqrt{ \left(a+b\right)^2-3ab} ~] }{24} \\ &=& \frac{4 \cdot[~ \left(a+b\right) \pm \sqrt{ \left(a+b\right)^2-3ab} ~] }{4\cdot 6} \\ &=& \frac{ \left(a+b\right) \pm \sqrt{ \left(a+b\right)^2-3ab} } { 6} \quad & | \quad (a+b)^2 = a^2 + 2ab + b^2\\ &=& \frac{ \left(a+b\right) \pm \sqrt{ a^2 + 2ab + b^2-3ab} } { 6} \\ &=& \frac{ \left(a+b\right) \pm \sqrt{ a^2 + 2ab-3ab + b^2} } { 6} \quad & | \quad 2ab-3ab = ab(2-3)= ab(-1) = -ab\\ &=& \frac{ \left(a+b\right) \pm \sqrt{ a^2 -ab + b^2} } { 6} \\ \hline \end{array}\)

no                                    

Oké                                                                         

Purple because aliens don't wear hats.

Say that the number of loaves of bread is N.

Morning sales are M = 33

Afternoon sales are A = 0.6 * (N - M)

And the rest is R = 1/8 * N

M + A + R = N

\(33 + 0.6*(N-33)+{1\over{8}}*N=N\)

\(33 + 0.6N-19.8+0.125N=N\)

\(13.2+0.725N=N\)

\(13.2=0.275N\)

\(N=13.2/0.275=48\)

Let's check: 33 sales leaves 15 loaves.

60% of 15 is 9 which leaves 6 loaves.

6 times 8 equals 48.

Original number = 'x'

(x-33)    #left after morning sales

.4(x-33) = # left after afternoon sales = 1/8x

.4x - 1/8x =13.2

x=48 loaves at start

x=position of ball  xo= original position of ball   vo = original velocity    t = time   a = acceleration of gravity 9.8 m/s^2

x=xo + vot + 1/2 a t^2    note the vo and a are DOWNWARD in this example and thus NEGATIVE

x = 60 + -3(1) + 1/2 (-9.8)(1^2)

x = 60 -3 -4.9 = 52.1 m above the ground  or 7.9 m down from its original position

if im not wrong, but I think its 3 meters. for distance = Speed* time, so for one second it would take the ball moving at 3meters per second, to move 3 meters. but if you were to account for the fact that the ball will be increasing in speed as it falls, then you would have to know the mass of the ball, then you would calculate the change in speed over time, or acceleration. 

The ratio of 'a' to 'b'  is expressed as   a/b   as a fraction.

Huh?      Did you mean convert degrees C to F?

F =  9/5 C + 32       9/5 (23) + 32 =   well, just use the calculator !

Possibly....if you give us the table of numbers.....

Just multiply  3/5   x  2500  

= 1500 allowable

(1/2) / 3   =   (1/2) / (3/1)  =  1/2  x  1/3  =  1/6

12 /x     as 'x'  APPROACHES  0   the value APPROACHES infinity.   Melody is correct....dividing by 0 is 'undefined'  or not allowed.

huh?