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Find the vertex of the graph of the equation ....

Just use the minus sign :)

How do I solve

17-12+4/2.

=17-12+2

=5+2

=7

Two similar pyramids have surface areas 200 m2 and 500 m2. The smaller pyramid has volume 500 m3. Find the volume of the larger pyramid.

A truck with tyres 1 metre in diameter is travelling at 100 km per hour. What is the angular velocity of the tyres in radians per second.

Thanks Guest :)

Mmm, I will just muffle through logically and see if I get the same answer.   

circumference of tyre = \(2\pi r=2\pi*0.5=\pi\;metres\)

100km = 100000m

1hour = 60*60=3600seconds

I use the units to help me work out questions like this.  Units can cancel just like numbers can in fractions...so

I want to find out       \(\frac{radians }{sec}\)

These are the units that I know:

\(\frac{100000m}{hour}\qquad \frac{3600seconds}{hour} \qquad\frac{2\pi\;radians}{revolution}\qquad \frac{\pi \;metres}{revolution}\)

I want radians on the top and seconds on the bottom and I want all the other units to cancel out 

\(\frac{2\pi\;radians}{revolution}\times \frac{1\;revolution}{\pi\;metre}\times \frac{10,000m}{hour}\times \frac{1\;hour}{3600\;seconds}\\ =\frac{2\pi\times 100000\;radians}{3600\pi\;seconds}\\ =\frac{1000\;radian}{18\;seconds}\\ =55.\dot5\;radians/second\)

Find the sum of all possible positive integer values of b

\(\begin{array}{|rcll|} \hline b & < & \frac{25}{8} \\ b & < & 3.125 \\\\ &\text{positive integer:}& b = 1 \qquad b = 2 \qquad b = 3\\\\ &\text{rational roots:}& b = 2 \qquad b = 3\\\\ && 2+3 =5\\ \hline \end{array}\)

So the positive integers that satisfy the conditions are:  b = 2  [x = -1/2 and -2]  and b = 3  [x = -1 and -3/4]

Hence the requested sum is 2 + 3 → 5

Use *10^x instead

e.g.   5.4EE-7   would be entered as   5.4*10^-7 in the calculator on the home page here.

n is an unspecified positive integer.  If you want a single numerical result for the sum you must specify the value for n.

For example:

n = 5    means  n(n+1)(n+2)(3n+1) = 3360

n=9     means   n(n+1)(n+2)(3n+1) = 27720

etc.

The sum is a function of n.

I think you use the incorrect formula for angular velocity.

Angular velocity is indeed radians per second and it is symbol \(\omega\)

But if you multiply by the radius you get the circumferential velocity, which in this case is the same as the land speed, because the wheel grips the road.

This means \(v = \omega * r\)

\(\omega = v/r\)

\(\omega = {27.78\over{0.5}} = 27.78 * 2=55.56\space \space rad/s\)

Hope this clears it up for you.

K= 1/2(6.6*10^2kg)(2.11*10^4m/s)^2

Solve for K

K= 1/2(6.6*10 ^2 kg)(2.11*10^4m/s)^2

\(\begin{array}{|rcll|} \hline K &=& \frac12*(6.6*10 ^2 kg)*(2.11*10^4\frac{m}{s})^2 \\ K &=& \frac12*6.6*10 ^2 kg * (2.11*10^4\frac{m}{s})^2 \\ K &=& \frac12*6.6*10 ^2 kg * 2.11^2*10^8\frac{m^2}{s^2} \\ K &=& \frac12*6.6*10 ^2 * 2.11^2*10^8\frac{ kg * m^2}{s^2} \\ K &=& \frac12*6.6 * 2.11^2*10^{10} \frac{ kg * m^2}{s^2} \\ K &=& 14.69193 *10^{10} \frac{ kg * m^2}{s^2} \\ K &=& 1.469193 *10^{11} \frac{ kg * m}{s^2} \cdot m \qquad | \qquad 1~N = 1~\frac{kg*m}{s^2} \\ \mathbf{K} &\mathbf{=}& \mathbf{1.469193 *10^{11} N \cdot m} \\ \hline \end{array} \)

K = 1.469×10^11 kg m^2/s^2 (kilogram meters squared per second squared)

K= 1/2(6.6*10^2kg)(2.11*10^4m/s)^2

\(K=6963000 \frac{kgm}{s^{2} } \times \frac{kN}{1000\frac{kgm}{s^{2} } } \)

K = 6963 kN

Yes but I need it in Scientific Notation and i need the units

K = 146,919,300,000

24-4(5-2)^2-11=-23 How is it that?

24 - 4(3)^2 - 11 = -23

24 - 4*9 - 11 = -23

24 - 36 - 11 = -23

-12 - 11 = -23

Find the sum of all possible positive integer values of b such that the quadratic equation

\(2x^2 + 5x + b = 0\)

has rational roots.

\(\begin{array}{|rcll|} \hline && 2x^2 + 5x + b = 0 \\\\ x& =& \frac{-5\pm\sqrt{5^2-4\cdot 2\cdot b}}{2\cdot 2} \\\\ x &=& \frac{-5\pm\sqrt{25-8\cdot b}}{4} \\\\ 25-8\cdot b &\ge& 0 \\ 25 &\ge& 8\cdot b \\ 8\cdot b &\mathbf{<}& 25\\ \mathbf{b} &\mathbf{<}& \mathbf{\frac{25}{8}} \\ \hline \end{array}\)

Thank you so much!

Un total de 35000 nuevos soles se invierte a tres tasas de interés: 7%, 8% y 9%.El interés en el primer año fue de 2830 nuevos soles, que nose reinvirtio. El segundo año la cantidad invertida originalmente al 9% ganó 10%, y las otras tasas permanecieron iguales.Si interes total en el segundo año fue de 2960 nuevos soles, ¿cuanto se invirtio a cada tasa???

A total of 35,000 new soles was invested in three interest rates: 7%, 8% and 9% .Interest in the first year was 2830 new soles, which was not reinvested. The second year the amount originally invested at 9% gained 10%, and other rates remained unchanged.The total interest in the second year was 2960 new soles, how much was invested at each rate ???

P.S. Soles is the Peruvian currency.

Let the amount invested at 7%=A

Let the amount invested at 8%=B

The amount invested at 9% =35,000 - (A+B)

.07A+.08B+.09[35000-(A+B)]=2830

 .07A+.08B+.1[35000-(A+B)]=2960, solve for A,B

A=10,000 invested at 7%

B=12,000 invested at 8%

And 13,000 invested at 9% for the 1st year, which gained 10% in the 2nd year.      

Is it (17-12)+(4/2), ((17-12)+4)/2, (17-(12+4))/2, 17-(12+(4/2)), or 17-((12+4)/2). I'll solve all five.

(17-12)+(4/2)                               ((17-12)+4)/2                                                       17-(12+(4/2))

(17-12)       +(4/2)       17-12=5    ((17-12)+4)      /2        (17-12)+4=5+4=9             17-        (12+(4/2))    (12+(4/2))=12+2=14

(5)        +(4/2)       4/2=2              9/2                                                                        17-14              17-14=3

5+2                   5+2=7                 ((17-12)+4)/2=9/2                                                17-(12+4/2)=3

7

(17-12)+(4/2)=7

(17-(12+4))/2                                                                         17-((12+4)/2)

(17-   (12+4)   )    /2                 12+4=16                                17-   (   (12+4)   /2)                 12+4=16

(17-16)     /2                        17-16=1                                      17-     (16/2)                  16/2=8

1/2                                                                                          17-8                            17-8=9

(17-(12+4))/2=1/2                                                                   9

                                                                                               17-((12+4)/2)=9

(17-(12+4))/2=1/2, 17-((12+4)/2)=9, (17-12)+(4/2)=7, ((17-12)+4)/2=9/2, and 17-(12+4/2)=3.