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Anastasy [175]
3 years ago
12

line 1 passes through (2,0) and (4,5). line 2 passes through (0,-2) and (2,3). Are the lines parallel or perpendicular

Mathematics
1 answer:
ziro4ka [17]3 years ago
5 0

Answer:

Step-by-step explanation:

line 1:

(2,0),(4,5)

slope = (5 - 0) / (4 - 2) = 5/2

line 2:

(0,-2),(2,3)

slope = (3 - (-2) / (2 - 0) = 5/2

parallel lines will have the same slope....perpendicular lines will have negative reciprocal slopes.

these lines are parallel

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Write the quadratic function in standard form.<br><br> y=2(x - 3)² +9
xz_007 [3.2K]

Answer:

y = 2x^2 - 12x + 27

Step-by-step explanation:

<u>Step 1:  Distribute the power</u>

y = 2(x - 3)² + 9

y = 2(x^2 - 6x + 9) + 9

y = 2x^2 - 12x + 18 + 9

y = 2x^2 - 12x + 27

Answer: y = 2x^2 - 12x + 27

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Mr. Moran chose 10 different test scores out of 50, and they are shown below. For the following set of data, find the mean test
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3 years ago
Simplify the expression 6h+​(−8.7d​)−13+6d−3.3h.<br> 6h+​(−8.7d​)−13+6d−3.3h=
mezya [45]

Answer:

 -2.7d+2.7h-13

6 0
3 years ago
(Can somebody find the sum of this) <br><br><br> (5+2i)+(8-3i)
7nadin3 [17]
Hey there!!

Okay so here are the steps....

(5 + 2i) + (8 - 3i)       <---- when we have exponents we should worry about the                                            "i" because it is equal to √-1 but here it isn't the case.
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7 0
3 years ago
X = ? y = ? 16 45 degrees
mezya [45]

Let's put more details in the figure to better understand the problem:

Let's first recall the three main trigonometric functions:

\text{ Sine }\theta\text{ = }\frac{\text{ Opposite Side}}{\text{ Hypotenuse}}\text{ Cosine }\theta\text{ = }\frac{\text{ Adjacent Side}}{\text{ Hypotenuse}}\text{ Tangent }\theta\text{ = }\frac{\text{ Opposite Side}}{\text{ Adjacent Side}}

For x, we will be using the Cosine Function:

\text{ Cosine }\theta\text{ = }\frac{\text{ Adjacent Side}}{\text{ Hypotenuse}}Cosine(45^{\circ})\text{ = }\frac{\text{ x}}{\text{ 1}6}(16)Cosine(45^{\circ})\text{ =  x}(16)(\frac{1}{\sqrt[]{2}})\text{ = x}\text{ }\frac{16}{\sqrt[]{2}}\text{ x }\frac{\sqrt[]{2}}{\sqrt[]{2}}\text{ = }\frac{16\sqrt[]{2}}{2}\text{ 8}\sqrt[]{2}\text{ = x}

Therefore, x = 8√2.

For y, we will be using the Sine Function.

\text{  Sine }\theta\text{ = }\frac{\text{ Opposite Side}}{\text{ Hypotenuse}}\text{ Sine }(45^{\circ})\text{ = }\frac{\text{ y}}{\text{ 1}6}\text{ (16)Sine }(45^{\circ})\text{ =  y}\text{ (16)(}\frac{1}{\sqrt[]{2}})\text{ = y}\text{ }\frac{16}{\sqrt[]{2}}\text{ x }\frac{\sqrt[]{2}}{\sqrt[]{2}}\text{ = }\frac{16\sqrt[]{2}}{2}\text{ 8}\sqrt[]{2}\text{ = y}

Therefore, y = 8√2.

5 0
2 years ago
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