Solve for x: 11/8 = 8/x+3
a) 31/11
b) 11/31
c) 97/11
d) 31
and...
Solve the proportion: 2/k = 16/4
a) k=1/2
b) k=1/4
c) k=8
d) k=2
2007-11-26 15:47:51 · 6 answers · asked by Jessica 1 in Science & Mathematics ➔ Mathematics
1)
Ok, for this one you start by cross multiplying:
11(x+3) = 8*8
11x+33 = 64
Then you solve for x with subtraction then division
11x+33-33=64-33
11x=31
11x/11=31/11
x=31/11
2) This one is the same concept, first cross multiply
2*4=16*k
8=16k
8/16=16k/16
8/16=k
k=8/16=1/2
2007-11-26 15:56:08 · answer #1 · answered by someone2841 3 · 0⤊ 0⤋
Solve for x: 11/8 = 8/x+3
cross multiply
11x+33 = 64
subtract 33 from both sides
11x=31
isolate x by dividing by 11
x= 31/11
Answer: a) 31/11
Solve the proportion: 2/k = 16/4
first off we can simply 16/4 = 4
now we have
2/k = 4
isolate k by dividing both sides by 2
k= 2/4
k=1/2
Answer a) k=1/2
2007-11-26 16:01:22 · answer #2 · answered by Matty B 3 · 0⤊ 0⤋
Multiply 11/8 by x+3 to get (11x+33)/8. Multiply both sides by 8 to get 11x+33=64. subtract 33 from both sides, get 11x=31. divide by 11, get x=31/11
Cross multiply to get (2*4)=16k. Divide both sides by 16 to get k=1/2.
2007-11-26 15:53:28 · answer #3 · answered by Anonymous · 0⤊ 0⤋
for #1
11/8=8/(x+3)
11(x+3)=64
11x+33=64
11x=31
A-- x=31/11
#2
2/k=16/4
8=16k
A --1/2=k
2007-11-26 15:52:27 · answer #4 · answered by Anonymous · 0⤊ 0⤋
upload 2 to each area: 3x/5 = x + 7/3 locate a large determination that each of the denominators go into (3 and 5 both go into 15) so multiply each little thing by 15: 15(3x)/5 = 15(x) + 15(7)/3 Now simplify the fractions: 3(3x) = 15(x) + 5(7) strengthen each of the brackets: 9x = 15x + 35 Subtract 15x from each area: -6x = 35 Divide each area by -6: x = -35/6
2016-10-25 02:58:06 · answer #5 · answered by ? 4 · 0⤊ 0⤋
2/k=16/4
2/k=4
2=4k
1/2=k
2007-11-26 15:55:02 · answer #6 · answered by Anonymous · 0⤊ 0⤋