Please answer this math problem... I will mark you brainliest
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y+1=-7(x-4)
y=-7x+27
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Help? I'll mark brainliest
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Number of packs of chewing gum=x+1
Number of chocolate bars=y
The number of mints was three times the number of chocolate bars; then
x=3y
We can suggest this system of equations:
x=3y
6(x+1)+3x+10y=80
we can solve this system of equations by substitution method.
6(3y+1)+3(3y)+10y=80
18y+6+9y+10y=80
18y+9y+10y=80-6
37y=74
y=74/37
y=2
x=3y
x=3*(2)=6
Number of mints=x=6
number of packs of chewing gum=x+1=6+1=7
number of chocolate bars=y=2
Answer: the number of mints is 6, the number of packs of chewing gum is 7, and the number of chocolate bars is 2.
Answer:
He bought 13 pkgs of gum
He bought 12 mints
He bought 4 candy bars
Step-by-step explanation:
Let +a+ = number of packages of gum Terry bought
Let +b+ = number of mints
Let +c+ = number of candy bars
------------------------------
(1) +a+=+b+%2B+1+
(2) +b+=+3c+
(3) +24a+%2B+10b+%2B+35c+=+572+ ( in cents )
-------------------------------------
There are 3 equations and 3 unknowns, so it's solvable
(2) +c+=+b%2F3+
Substitute (2) into (3)
(3) +24a+%2B+10b+%2B+35%2A%28+b%2F3%29+=+572+
Substitute (1) into (3)
(3) +24%2A%28+b+%2B+1+%29+%2B+10b+%2B+35%2A%28+b%2F3+%29+=+572+
(3) +24b+%2B+24+%2B+10b+%2B+%2835%2F3%29%2Ab+=+572+
Multiply both sides by +3+
(3) +3%2A34b+%2B+3%2A24+%2B+35b+=+3%2A572+
(3) +102b+%2B+72+%2B+35b+=+1716+
(3) +137b+=+1644+
(3) +b+=+12+
and
(1) +a+=+b+%2B+1+
(1) +a+=+12+%2B+1+
(1) +a+=+13+
and
(2) +c+=+b%2F3+
(2) +c+=+12%2F3+
(2) +c+=+4+
---------------------
He bought 13 pkgs of gum
He bought 12 mints
He bought 4 candy bars
-----------------------
check:
(3) +24a+%2B+10b+%2B+35c+=+572+ ( in cents )
(3) +24%2A13+%2B+10%2A12+%2B+35%2A4+=+572+
(3) +312+%2B+120+%2B+140+=+572+
(3) +572+=+572+
OK
Answers
Answer:
Step-by-step explanation:
b. y = 3(5) x
c. y = 3 x
d. y = 5(3) x
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The weight is at the rest position of the spring and moving upward
when x=6.
During the next 1.5 seconds, it moves up to the upward limit (the
'amplitude' of the oscillation) and then back down to the rest position.
When x=7.5, the weight is back at the spring's rest position, and moving
downward.
14abc and 28a^2b^2c^3
Answers
so between 14abc and 28a^2b^2c^3 the great common factor will be
between 14 and 28 is 14
btw. a and a^2 is a
btw. b and b^2 is b
btw. c and c^3 is c
---------------------------------
so the gcf will be 14abc
hope helped