Hannah, the twist to number 7 is the x= something times y. x= # of dimes 0.10x=$ of dimes y= # of nickles 0.05y=$ of nickles "There are five times as many dimes as nickels in the machine." x = 5y To use the elimination method, make x and y equal some number. x - 5y = 0 "The face value of the coins is $4.40." 0.10x + 0.05y = 4.40 You might want to multiply this equation by 100 to get rid of the decimals and be able to eliminate. 100(.10x+.05y)=100(4.40) 10x + 5y =440 Now you combine the two equations. 10x + 5y = 440 x - 5y = 0 11x = 440 1/11(11x)= 1/11(440) x = 40 Use the equation x=5y to find out the amount of nickles. x = 5y 40= 5y 1/5(40)= 1/5(5y) 8 = y y = 8 Your answer should come out to be 40 dimes and 8 nickles. I hope this helps you Hannah. Saloni B. Period 1
I'm having trouble with 4 problems: 11,13,15,16. On 11, I have this eqution set up: 2(r-4)=3.5(r+4). I thought it was right, but I keep trying to solve it and something is wrong. On 13, I don't know where to begin. This problem is different than the ones we've been doing. I can find the answer, but I don't know how to write an eqution for it. On 15, the same thing...but I can't even find the answer. And 16, I have rate and time filled in the chart but WHAT IS THE DISTANCE? Blah..
Someone please help me. Saloni? I know you're out there. :)
I'm having trouble with number 11. On 13: x = 100's digit y = 10's digit We have always had two-digit numbers, but now we have three-digit numbers. "The sum of two numbers is 248." x + y = 248 "Their difference is 64." x - y = 64 It has it ready for the elimination method. x + y = 248 x - y = 64 2x = 312 1/2(2x)= 1/2(312) x = 156 Now plug that into the first equation. x + y = 248 156 + y = 248 -156 -156 y = 92 So I'm pretty sure that's the answer. For 15 and 16, I'm not really sure if it is correct or not, and I don't want to give the wrong answer. Saloni B. Period 1
for #15, x-10 digit# y-1 digit# y+4=x 10x+y=4(x+y) 10x-you need the 10 digit # because "the original # is 4 times the sum of its digits." The 10 digit # times the actual #10.
10x+y=4(x+y) 10x+y=4x+4y -y from both sides 10x=4x+3y -4x from both sides 6x=3y Substitute from the other equation 6(y+4)=3y 6y+24=3y -6y from both sides 24=-3y (-1/3) to both sides -8=y Go back to substitute to other equation. y+4=x (-8)+4=x -4+x
11 comments:
can somebody help with with #7 and 9?
Hannah Park
Per.5
#7 is what we did in class, and #9 is explained in Example 3 in the "Book Lesson".
austin C
Per.2
Hannah, the twist to number 7 is the
x= something times y.
x= # of dimes 0.10x=$ of dimes
y= # of nickles 0.05y=$ of nickles
"There are five times as many dimes as nickels in the machine."
x = 5y
To use the elimination method, make x and y equal some number.
x - 5y = 0
"The face value of the coins is $4.40."
0.10x + 0.05y = 4.40
You might want to multiply this equation by 100 to get rid of the decimals and be able to eliminate.
100(.10x+.05y)=100(4.40)
10x + 5y =440
Now you combine the two equations.
10x + 5y = 440
x - 5y = 0
11x = 440
1/11(11x)= 1/11(440)
x = 40
Use the equation x=5y to find out the amount of nickles.
x = 5y
40= 5y
1/5(40)= 1/5(5y)
8 = y
y = 8
Your answer should come out to be 40 dimes and 8 nickles. I hope this helps you Hannah.
Saloni B.
Period 1
When is the chapter 8 project due?
Lauren C.
per.2
Do we have homework tonight? Someone in 6th said it was only classwork.
Austin C.
Per. 2
BEDEC
What is the chapter 8 project?
Saloni B.
Period 1
I'm having trouble with 4 problems:
11,13,15,16.
On 11, I have this eqution set up: 2(r-4)=3.5(r+4). I thought it was right, but I keep trying to solve it and something is wrong. On 13, I don't know where to begin. This problem is different than the ones we've been doing. I can find the answer, but I don't know how to write an eqution for it. On 15, the same thing...but I can't even find the answer. And 16, I have rate and time filled in the chart but WHAT IS THE DISTANCE? Blah..
Someone please help me. Saloni? I know you're out there. :)
Shiloh S. Per. 3
WE HAD A CH 8 PROJECT?????????
Hannah p
per.5
Shilo- I didn't get 11 either, but I think in the equation it should be 5.5 not 3.5. And how did you get the 2??
Brett S.
per.4
I'm having trouble with number 11.
On 13:
x = 100's digit
y = 10's digit
We have always had two-digit numbers, but now we have three-digit numbers. "The sum of two numbers is 248."
x + y = 248
"Their difference is 64."
x - y = 64
It has it ready for the elimination method.
x + y = 248
x - y = 64
2x = 312
1/2(2x)= 1/2(312)
x = 156
Now plug that into the first equation.
x + y = 248
156 + y = 248
-156 -156
y = 92
So I'm pretty sure that's the answer.
For 15 and 16, I'm not really sure if it is correct or not, and I don't want to give the wrong answer.
Saloni B.
Period 1
Shiloh,
for #15,
x-10 digit#
y-1 digit#
y+4=x
10x+y=4(x+y)
10x-you need the 10 digit # because "the original # is 4 times the sum of its digits." The 10 digit # times the actual #10.
10x+y=4(x+y)
10x+y=4x+4y
-y from both sides
10x=4x+3y
-4x from both sides
6x=3y
Substitute from the other equation
6(y+4)=3y
6y+24=3y
-6y from both sides
24=-3y
(-1/3) to both sides
-8=y
Go back to substitute to other equation.
y+4=x
(-8)+4=x
-4+x
Your original # should come out to be 48.
kylie a.
per. 3
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