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guapka [62]
2 years ago
9

A restaurant used 5 pounds of ground beef to make one dozen hamburger. At this rate,how many pounds of ground are needed to make

144 hamburger?
Mathematics
1 answer:
Firdavs [7]2 years ago
3 0
In order to make 144 hamburgers you’re need 60 pounds of ground beef. hope this helps!
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Find AM, if AM = 6y-4 and MC = 2y+12.
aleksley [76]

Answer: 20

Step-by-step explanation:

Since Line l is a segment bisector, we know that AM and MC are equal to each other.

6y-4=2y+12        [subtract both sides by 2y]

4y-4=12              [add both sides by 4]

4y=16                 [divide both sides by 4]

y=4

Now that we have y, we plug that into AM.

6(4)-4                 [multiply]

24-4                   [subtract]

20

Now, we know that AM is 20.

6 0
3 years ago
Evaluate 4(a^2 + 2b) - 2b when a = 2 and b = –2.
Rama09 [41]

Answer:

4

Step-by-step explanation:

4(4-4)+4=4

insert 2 in a and-2 in b

7 0
3 years ago
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You have read 4 of the books shown. Which choice shows 2 ways of writing the fraction of these books that you read?
Olegator [25]
  • Total books read: 4
  • Total books in list: 12

So it can be shown as:

\sf \dfrac{read \ books}{total \ books \ in \ list}

\hookrightarrow \sf \dfrac{4}{12}

on simplifying

\hookrightarrow \sf \dfrac{1}{3}

So, option 1 is correct.

8 0
1 year ago
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The ratio of 8 melons to 36 fruits​
Ivenika [448]

8/36

= 2/9 after dividing by 4

Please mark as brainliest.

4 0
2 years ago
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2. Write a quadratic function that has -intercepts (- 3, 0) and (4, 0) and passes through the point (5, 8)
Reptile [31]

Answer:

f(x)=x^2-x-12

Step-by-step explanation:

<u>Quadratic Function</u>

Standard Form of Quadratic Function

The standard representation of a quadratic function is:

f(x)=ax^2+bx+c

where a,b, and c are constants.

When the zeros of f (x1 and x2) are given, it can be written as:

f(x)=a(x-x1)(x-x2)

Where a is a constant called the leading coefficient.

We are given the two roots of f: x1=-3 and x2=4, thus:

f(x)=a(x+3)(x-4)

We also know that f(5)=8, thus:

f(5)=a(5+3)(5-4)=8

Operating:

a(8)(1)=8

Solving:

a=1

The function is:

f(x)=1(x+3)(x-4)

Operating:

\boxed{f(x)=x^2-x-12}

5 0
2 years ago
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