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Suppose that p  is prime and \(1007_p+306_p+113_p+125_p+6_p=142_p+271_p+360_p\).

How many possible values of  p  are there?

I don't think that there are any....

Since there are numbers up to 7 in there, the base, p,  would have to be 11 or more.  (because it is prime)

Looking at just the last digit.

7+6+3+5+6=27                               2+1+0=3

27=3+24

This means that p must be 24 or less (not that that matters)

Anyway,

prime factors, 11 or bigger, of 24 don't exist.

So I think that there are no possible values of p that will satisfy this.

Correct CPhill!

did you take the sin inverse on the third step?

Note that angle CEA = angle DCA + angle DAC and angle BDE = angle DCE + angle EDC, so angle DCA = angle CEA - angle DAC and angle DCE = angle EDC - angle CED.

Also, angle BDE = angle ABD + angle ADB = angle AEC + angle DAC, so angle DCE + angle DCA = (angle EDC - angle CED) + angle DCA.

Therefore, angle ACE = angle ACD + angle DCE = angle CEA, which implies that triangle ACE is isosceles.

Volume_cyl  = pi r^2 h     Plug in the values given

8.5             = pi (r^2) h              lateral surface area = pi d h  = 3.5     so h =  3.5 / (pi2r)       sub into the equation

8.5 =  pi r^2 ( 3.5/ pi 2 r) 

8.5 =    3.5 r/2

17/3.5 = r   = 4.857 cm     

G=(1/3)K
(1/3)K+10=K-10
(2/3)K=20
K=30, after multiplying both sides by the reciprocal 3/2
G=10
The answer is 40 calendars.

If you draw it out , you will find   AD = 13  cm              13 + 12 + 4 + 9 = 38 cm

[Ar] 4s²   3d¹⁰  4p⁵        for 35 electrons in bromine

Ca   is normally    1 s²    2s² 2p⁶   3s² 3p⁶   4s²       for 20 electrons

Argon   AR has 18 electrons  

            so   Ca could be    [Ar] 4s²          now remove one electron   to get Ca⁺  =     [ Ar] 4s¹

How will the surface area of cone of height h and radius r compare to the surface area of a square pyramid of height h and apothem r?

There are 17 boys and 15 girls in a Math 30 class. The teacher is selecting 7 students to represent the school on the student advisory council.

Determine the probability, to the nearest hundredth of a percentage, that at most 2 boys will be selected.

%=25%

fraction=1/4

Because Bob has a 30% chance and Anne has a 38% chance of getting the job, we can add these up and subtract them from the total 100% that a person could possibly have.

Therefore:

30 + 38 = 68

100 - 68 = 32

So, Liz has a 32% chance of getting the job.

im really not sure

im really not sure

Argon

Argon is the closest noble gas before bromine on the periodic table

• 10.8186°

i just finished this question imao 

asin(0.1877) =~ 10.8186 ^o

...thanks

200 minutes

$249.40 

i think the   constant  term  will  be  C(8,4)  ( 6x)^4   ( 1/(3x))^4   = 70  ✕  ( 1296 x^4)  ( 1/ (81x^4)  =1120

AB / CE = AF / FC = BF / FE

AB and CD are parallel

Let the midpoint of arc AB be G         

CG is a diameter