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valina [46]
3 years ago
10

Terry has nuts and bolts. The ratio of the number of nuts to the

Mathematics
1 answer:
Rasek [7]3 years ago
5 0

Answer:

Step-by-step explanation:

Study the steps used to solve the equation.

Given: StartFraction c Over 2 EndFraction minus 5 equals 7

Step 1: StartFraction c Over 2 EndFraction minus 5 plus 5 equals 7 plus 5

Step 2: StartFraction c Over 2 EndFraction plus 0 equals 12

Step 3: StartFraction c Over 2 EndFraction equals 12

Step 4: 2 (StartFraction c Over 2 EndFraction) equals 12 (2)

Step 5: c equals 24

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Answer:

4.9

Step-by-step explanation:

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HELP ME OUT HERE! PLS TELL ME HOW U GOT THE ANSWER TOO!
IgorC [24]

Answer: Choice C

\begin{array}{|c|c|} \cline{1-2}x & y\\\cline{1-2}0 & 1\\\cline{1-2}2 & 2\\\cline{1-2}4 & 3\\\cline{1-2}6 & 4\\\cline{1-2}\end{array}

=========================================================

Explanation:

There are four marked points on the line.

Each point is of the form (x,y)

  • The first or left most point is (0,1)
  • The second point is (2,2)
  • The third is (4,3)
  • The fourth is (6,4)

Each of these points is then listed in the table format as shown above.

There are infinitely many other points on the line; however, we only select a few of them to make the table (or else we'd be here all day).

Extra side notes:

  • The slope of this line is m = 1/2 = 0.5
  • The y intercept is 1 located at (0,1)
  • The equation of this line is y = 0.5x+1
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1 year ago
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Hiro bought a small carton of milk at lunch. If the approximate dimensions of the milk carton are shown what is the minimum amou
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Answer:

C.\ 104\ in^2

Step-by-step explanation:

At first, the question looks like an optimization problem, but since all the dimensions of the carton are given, we only have to compute the total area of the given figure.

Let's calculate the front (and back) areas, which are rectangles

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Now with the lateral rectangles which happen to have the very same dimensions

A_2=19.5\ in^2

Next, we compute the front and back triangles of base 3 in and height 1.5 in

A_3=\frac{1}{2}(3)(1.5)=2.25\ in^2

Now, the lateral inclined rectangles of base 3 in and height 2 in

A_4=(3)(2)=6\ in^2

Finally, the base rectangle who happens to be a square of side 3 in

A_5=(3)(3)= 9\ in^2

This last area, unlike all others, is not doubled because its counterpart is inside the carton and is not part of the lateral area

Our total area of cardboard is

A_t=2(19.5)+2(19.5)+2(2.25)+2(6)+9=103.5\ in^2

The closest option to this answer is

C.\ 104\ in^2

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