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ser-zykov [4K]
4 years ago
5

Choose an equation for the line that is parallel to the given line and passes through the given point for questions 1 and 2.

Mathematics
1 answer:
lutik1710 [3]4 years ago
3 0
To solve each of the problems, I first looked for the equation that had the same slope value and then plugged in the 'x' coordinate to see if it gave me the correct 'y' coordinate.

1.)
     D.) y = 5x + 4
          14 = 5(2) + 4
          14 = 10 + 4
          14 = 13

2.)
     C.) y = 1/5x - 19
        -16 = 1/5 (15) - 19
        -16 = (3) - 19
        -16 = -16

3.) First, get 'y' by itself by subtracting 4x from both sides and dividing the whole equation by -12. To find perpendicular lines, the slope must be the opposite reciprocal of the first slope (in this case, flip the fraction to get the three on top). Add the opposite sign to the slope as your final step.

4x - 12y = 2
-4x            -4x
-12y = -4x + 2
-12y / -12 = -4x / -12+ 2 / -12
y = 1/3x - 1/6

C.) y = -3x + 29
     -1 = -3(10) + 29
     -1 = (-30) + 29 
     - 1 = -1

     
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The 3rd selection is appropriate.

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Find the area of the triangle.
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Answer:

A. 496 ft2

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2 years ago
The area of the rhombus is 540 cm2; the length of one of its diagonals is 4.5 dm. What is the distance between the point of inte
malfutka [58]

1. The area of the rhombus can be found by the formula

A=\dfrac{d_1\cdot d_2}{2}, where d_1,\ d_2 are rhombus's diagonals.

Note that d_1=4.5\ dm=45\ cm, then

540=\dfrac{45\cdot d_2}{2},\\ \\540\cdot 2=45d_2,\\ \\d_2=24\ cm.

2. The diagonals of rhombus are perpendicular and are bisectors of each other. Then the triangle formed with halfs of diagonals is right triangles with legs

\dfrac{d_1}{2}=22.5\ cm,\ \dfrac{d_2}{2}=12\ cm.

The hypotenuse of this triangle is the rhombus's side. By the Pythagorean theorem

\text{rhombus's side}^2=(22.5)^2+12^2=506.25+144=650.25,\\ \\\text{rhombus's side}=25.5\ cm.

3. The distance between the point of intersection of the diagonals and the side of the rhombus is the height of right triangle considered above.

Use twice the Pythagorean theorem to find this height:

\left\{\begin{array}{l}x^2+h^2=12^2\\(25.5-x)^2+h^2=22.5^2,\end{array}\right.

where x is projection of leg 12 cm and h is height.

Subtract the first equation from the second:

(25.5-x)^2+h^2-x^2-h^2=22.5^2-12^2,\\ \\650.25-51x=506.25-144,\\ \\51x=650.25-362.25=288,\\ \\x=\dfrac{96}{17}\ cm.

Then

h^2=144-\left(\dfrac{96}{17}\right)^2=144-\dfrac{9216}{289}=\dfrac{32400}{289},\\ \\h=\dfrac{180}{17}\ cm.

Answer: h=\dfrac{180}{17}\ cm.

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