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First you'll have to DEFINE what is accepted as HIGH BLOOD PRESSURE.  (I'm a doctor....I know what is accepted here in USA .)

do you know about a x and y axis?(x-coordinate,y-coordinate).which means you can create a 3 dot by 3 dot grid. use your mind.

OK....let's multiply it out

y-1/2 = 5x (x-1/2)

y-1/2=5x^2-2.5x

y=5x^2 -2.5x +1/2

Is THAT what you need?

what is cavemans favorite sandwich?

I expect that would be a club sandwich ://      

Thanks heaps, I get it now.

There are 4 squares that are 1 by 1

and there is 1 square that is 2 by 2

That is 5 squares.

Draw the dots on a peice of paper.  It is super easy. :)

Well, first you can express the depth of Superior as a fraction of a mile    

like this:

1,333ft/5280ft      and then just divide it out on a calculator     1,333 divided by 5280 equals  0.2524621212121212    rounded to the nearest HUNDREDTH is .25 miles (or a quarter of a mile deep!......WAY over your head)

Well, they just basically want you to do the math....

soooo the SECOND one will look like this (you should be able to do the first one after seeing this )

7x^2  + 4x - x^2 -4x

(7x^2  - x^2) + (4x - 4x)

6x^2 + 0  =  6x^2               Do you see how it works?   Can you do the firs one now? it is easier.....

I.m probally wrong. :/

IDK if I'm correct or not but the answer should be around 0.2524621212121212 miles...

Using one of the many trig identities:

\(cos(u)cos(v) = \frac{1}{2}[cos(u-v)+cos(u+v)]\\cos(2x)cos(4x) = \frac{1}{2}[cos(-2x)+cos(6x)]\)

2/5 University students have cars and the number of students is 850, how many of them have no car?

3/5 * 850   = 510     no tienen coche       [have no car]

cos (2x) times cos (4x) 

\(cos(2x)=cos^2x-sin^2x\\ cos(2x)=cos^2x-(1-cos^2x)\\ cos(2x)=2cos^2x-1\\ ...\\ cos(4x)=cos^2(2x)-sin^2(2x)\\ cos(4x)=[cos^2x-sin^2x]^2 -[2sinxcosx]^2\\ cos(4x) =[cos^2x-(1-cos^2x)]^2 -[4sin^2xcos^2x]\\ cos(4x) =[2cos^2x-1]^2 -[4(1-cos^2x)cos^2x]\\ cos(4x) =[4cos^4x-4cos^2x+1] -[4cos^2x-4cos^4x]\\ cos(4x) =4cos^4x-4cos^2x+1-4cos^2x+4cos^4x\\ cos(4x) =8cos^4x-8cos^2x+1\\ ...\\ cos(2x)(cos4x)=(2cos^2x-1)(8cos^4x-8cos^2x+1)\\ cos(2x)(cos4x)=2cos^2x(8cos^4x-8cos^2x+1)-1(8cos^4x-8cos^2x+1)\\ cos(2x)(cos4x)=6cos^6x-16cos^4x+2cos^2x-8cos^4x+8cos^2x-1\\ cos(2x)(cos4x)=6cos^6x-24cos^4x+10cos^2x-1\\ \)

That is if I did not make any careless mistakes.    

Hold on how did you know about this question? im the one who post it.

Still 3.....unless you've taken the dead one out......then, there are only 2

If you take E and R to be variables:

\(\ln{E}=1.2+1.81\ln{R} \\ E = e^{1.2+1.81\ln{R}}\\E = e^{1.2}e^{\ln{R^{1.81}}}\\E=R^{1.81}e^{1.2}\)

Which is an equivalent equation without the natural log

Cool!

(-64)^2\3

\(((-64)^{1/3})^2 = (-4)^2=16 \)

Just divide: (1.325 X 10^9) / (3.04 X 10^8)=4.4x more people in China than the US

Are E and R both just variables or is E really the number e ???

If there are 2 variables then you can't solve it with just one equation.

However ,if your question to find R given this then..

log(e) = 1.2 + 1.81 ln(R)

1= 1.2 + 1.81 ln(R)

-0.2=  1.81 ln(R)

(-0.2/1.81)=  ln(R)

R=e^(-0.2/1.81)

e^(-0.2/1.81) = 0.8953888036358034

80000*(1+0.014)^100=321,281.33

Si 3 kg de carne me cuestan 255 ¿cuánto pagaré por 8.5 kg?

 [If 3 kg of meat cost me 255 how much do I pay for 8.5 kg? ]

255/ 3   = x / 8.5      cross-multiplican

255 (8.5)  = 3x         dividir por 3

[ 255 (8.5)] / 3   = x  =  722.5

I can not understand your Question.