If CD = 10 and DE = 10, then CE = 20. CE = CB = AF = DF.
If DE = 10, then midpoint of BE (let's call this G) = 10. BG = AG = 10.
Remember that these are equilateral triangles, so BC = BE = CE.
It helps to use the diagram for this.
Perimeter = CB + BG + AG + AF + EF + DE + CD
= 20 + 10 + 10 + 20 + 10 + 10 + 10
= 90
take a look at this question, how can i approach it
let the function p be defined by p(x)=a(x-k)^2, where a and k are positive constants for what value of x will the function p have its minimum value
-a
-k
0
a
k
This is testing the 3rd-side rule for triangles (geometry). For ANY triangle, sum of the 2 shorter sides must be greater than the longest side. So x must be one of the shorter sides due to the restriction. 2+x>7 means that x must equal 6. Only 1 triangle is possible.
The sum of the digits of a three-digit number is 12. If the hundreds digit is 3 times the tens digit and the tens digit is 1 over 2 the units digit, what is the tens digit of the number?
A. 2
B. 3
C. 4
D. 6
E. 9
The CB gave an algebraic solution.
Here's an arithmetic one.
The ratio of the hundreds digit to the tens digit is 3:1, and the tens digit to the units digit is 1/2:1. To combine these two ratios into one let's change the second to 1:2, so the ratio of the hundreds digit to the tens digit to the units digit is 3:1:2.
For number 312 the sum of its digits is 6 - half of what's needed, so we should double each digit to keep the ratio: 624. The tens digit is 2.
Looks long, but literally takes seconds.
There is even a simpler solution based on plug'n'check method.
Only answers A and B (2 and 3) are possible (all others make "the hundreds digit" >9).
Answer A works:
624, 6=3x2, 2=(1/2)4.
At a certain hospital, 89 children were born in the month of June. If more children were born on the fifteenth of June than on any other day in June, what is the least number of children that could have been born on the fifteenth of June?
The answer is 4. This is another example of the pigeon-hole principle. You have 30 days in June to work with and 89 items to distribute. Deal them out one at a time -- the way to give each day the fewest is to have 3 births on all of them except one which will only have 2. But now, it you want the 15th to have the most, take away one from one of the others (doesn't matter which one) and give it to the 15th. In other words, if the children's birth's were distributed evenly (almost), 4 births on the 15th would be more than any other day.
Replies to: Math help center
If CD = 10 and DE = 10, then CE = 20. CE = CB = AF = DF.
If DE = 10, then midpoint of BE (let's call this G) = 10. BG = AG = 10.
Remember that these are equilateral triangles, so BC = BE = CE.
It helps to use the diagram for this.
Perimeter = CB + BG + AG + AF + EF + DE + CD
= 20 + 10 + 10 + 20 + 10 + 10 + 10
= 90
let the function p be defined by p(x)=a(x-k)^2, where a and k are positive constants for what value of x will the function p have its minimum value
-a
-k
0
a
k
the answer is k
If x = -a, then (x-k) would be negative. (x-k)^2 would thus be positive. a(x-k)^2 would therefore be positive also.
If x = -k, (x-k) = negative. (x-k)^2 = positive, so a(x-k)^2 = positive.
If x = 0, (x-k) = negative. (x-k)^2 = positive, so a(x-k)^2 = positive.
If x = a, (x-k)^2 would have to be positive, no matter whether x-k < 0 or > 0. a^2 would be positive.
If x = k, (x-k)^2 = 0. a(x-k)^2 = 0. Smallest value.
A)a/(a-1)
B)1/(a-1)
C)A^6-A^5
D) A^5
E)A^6
The book says E, and I guessed A. How do you do this one?
(A^6)(A-1) / (A-1)
The A-1's cancel and you're left with A^6
"If x is an integer and 2<x<7, how many diff triangles are there with sides of lengths 2, 7 and x?"
I read the online explanation and can't seem to figure out why the outer side of triangle is 20 Q_Q
http://www.collegeboard.com/prod_downloads/prof/counselors/tests/sat/2007-08_sat_preparation_booklet.pdf
Page 48 # 8
Page 51: 18
The Official SAT Question of the Day
I got 4.
A. 2
B. 3
C. 4
D. 6
E. 9
The CB gave an algebraic solution.
Here's an arithmetic one.
The ratio of the hundreds digit to the tens digit is 3:1, and the tens digit to the units digit is 1/2:1. To combine these two ratios into one let's change the second to 1:2, so the ratio of the hundreds digit to the tens digit to the units digit is 3:1:2.
For number 312 the sum of its digits is 6 - half of what's needed, so we should double each digit to keep the ratio: 624. The tens digit is 2.
Looks long, but literally takes seconds.
There is even a simpler solution based on plug'n'check method.
Only answers A and B (2 and 3) are possible (all others make "the hundreds digit" >9).
Answer A works:
624, 6=3x2, 2=(1/2)4.
At a certain hospital, 89 children were born in the month of June. If more children were born on the fifteenth of June than on any other day in June, what is the least number of children that could have been born on the fifteenth of June?
Thanks
Get a Ti-89!