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stellarik [79]
3 years ago
5

If 3 3/ 4 lb. of candy cost $18.75, how much would 1 lb. of candy cost?

Mathematics
1 answer:
Vladimir [108]3 years ago
6 0

Answer:

5.

18.75/3.75 (3 3/4) = 5

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Please help me i'm being timed
Evgen [1.6K]

To Find :-

  • The algebraic equations to solve the above problem .

Solution :-

Let ,

  • Large number = L
  • Small number = S

— According to first statement ,

  • L - S = 4
  • L - S - 4 = 0

— According to second statement ,

  • 3L + 4S = 54
  • 3L + 4S - 54 = 0 .

These equations matches with the second option b .

Option b is the correct answer.

7 0
2 years ago
The sides of the square shown below have a length of
USPshnik [31]

Answer:2\sqrt{6}because it is a 45 45 90

Step-by-step explanation:

8 0
3 years ago
If (x^2 - 1) is a factor of ax^4 + bx^3 + cx^2 + dx + e, show that a + c + e = b + d = 0
alisha [4.7K]
x^2-1=x^2-1^2=(x-1)(x+1)

If x²-1 is a factor of the polynomial, both x-1 and x+1 are factors of it.

According to the remainder theorem, if a binomial x-a is a factor of a polynomial p(x), then p(a)=0.

If x-1 and x+1 are factors of the polynomial p(x)=ax⁴+bx³+cx²+dx+e, then p(1)=0 and p(-1)=0.

p(1)=a \times 1^4+b \times 1^3 + c \times 1^2+ d \times 1+e \\
p(-1)=a \times (-1)^4 + b \times (-1)^3 + c \times (-1)^2 + d \times (-1)+e \\ \\
p(1)=a+b+c+d+e \\
p(-1)=a-b+c-d+e \\ \\
p(1)=0 \\
p(-1)=0 \\ \\ \hbox{add both equations:} \\
a+b+c+d+e=0 \\
\underline{a-b+c-d+e=0} \\
2a+2c+2e=0 \\
2(a+c+e)=0 \\
a+c+e=0 \\ \\
\hbox{substitute 0 for a+c+e in the first equation:} \\
a+b+c+d+e=0 \\
(a+c+e)+b+d=0 \\
0+b+d=0 \\
b+d=0 \\ \\
\boxed{a+c+e=b+d=0} \\
\hbox{proved } \checkmark
8 0
3 years ago
TRUE OR FALSE: DEFG is definitely a parallelogram.
LiRa [457]

Answer:

true

Step-by-step explanation:

ed and fg are parallel

6 0
2 years ago
If 1/2 of 12 were 8 what would 1/3 of 36 be
alexandr1967 [171]

Answer:

1/3 of 36 is 12

Step-by-step explanation:

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2 years ago
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