3y - 3 = 2y +1
5x^2 +5 = 23 + 2x ^2 for x
P is the subject of the formula 1 = PRT how do i find the answers please show the working out
R is the subject of the formula A = 2 pirh
B is the subject of the formula 2S = A +3B + 5C
R is the subject of the formula A = piR^2H
write a formula in one vairable and then solve the equation
two numbers have the sum of 92, if the second number is the three times as large as the first, what are the two numbers.
Answers:
1.) solve a.]-2x+7=13
=> -2x=13-7 (transpose 7 to the other side)
=> -2x=6
=> -2x/-2=6/-2 (divide both sides by -2)
=> x=-3
b.] 3y-3=2y+1
=>3y-2y=3+1 (combine like terms)
=> y=4
c.] 5x^2+5=23+2x^2 (solve for x)
=> 5x^2-2x^2=23-5 (combine like terms)
=> 3x^2=18
=>3x^2/3=18/3 (divide both sides by 3)
=> x^2=6
=> square root of x^2= square root of 6
=> x= - square root of 6 and + square root of 6
2.)1=PRT (solve for P)
=> 1/RT=PRT/RT (divide both sides by RT so that, P will remain on the right side)
=>P=1/RT
3.) A=2pirh (solve for R)
=> A/2pih=2pirh/2pih (divide both sides by 2pih so that, r will remain on the right side)
=>R=a/2pih
4.)2S=A+3B+5C (solve for B)
=> 2S-A-5C=3B (transpose A and 5C)
=>2S-A-5C/3=3B/3 (divide both sides by 3)
B=2S-A-5C/3 (this is said all over 3)
5.) A=piR^2H (solve for R)
=> A/piH=piR^2H/piH (divide both sides by piH)
=> A/piH=R^2
=> square root of A/piH= square root of R^2
R= square root of A/piH
6.) 2 nos. have the sum of 92.
the second no. is 3 times as large as the first:
let x be the first no. and 3x the second no.
their sum is 92:
x+3x=92
4x=92
4x/4=92/4
x=23
the first no. is 23
substitute 23 in the second no, 3x
3(23)= 69
69 is the second no.
prove:
69+23=92
=D
write a formula in one variable and then solve the equation
two numbers have the sum of 92, if the second number is the three times as large as the first, what are the two numbers.
If n = one number, 92-n is the other because n + 92 - n = 92.
so 92 - n = 3(n)
add 1n to both sides getting 92 = 4n
divide both by 4 getting 23 = n and 3(23) = 69 for the other
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