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Dmitrij [34]
3 years ago
14

/ has vertices at (–5, 1) and (–3, 7). After a transformation, the image had endpoints at (3, –1) and (5, 5). Describe the trans

formation.
A.
translation right 2 units and up 6 units
B.
rotation 90° about the origin
C.
translation right 8 units and down 2 units
D.
reflection across the y-axis
Mathematics
1 answer:
Pie3 years ago
3 0
1st point: ( -5+8. 1-2 ) = ( 3, -1 )
2nd point: ( -3+8, 7-2 ) = ( 5, 5 )
Answer: C) translation right 8 units and down 2 units. 
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What is the value of x.<br> x/6=8 <br> show how you got the answer
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Answer:

x = 48

Step-by-step explanation:

since it's x/6, to get rid of the 6 you have to mutiply

so you times each side by 6, 8 multiplied by 6 is 48

so x = 48

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Find the dimensions of the square <br> Area=441cm over 2
MA_775_DIABLO [31]

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each side is 55.125cm

Step-by-step explanation:

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5 0
3 years ago
An island is 1 mi due north of its closest point along a straight shoreline. A visitor is staying at a cabin on the shore that i
Elanso [62]

Answer:

The visitor should run approximately 14.96 mile to minimize the time it takes to reach the island

Step-by-step explanation:

From the question, we have;

The distance of the island from the shoreline = 1 mile

The distance the person is staying from the point on the shoreline = 15 mile

The rate at which the visitor runs = 6 mph

The rate at which the visitor swims = 2.5 mph

Let 'x' represent the distance the person runs, we have;

The distance to swim = \sqrt{(15-x)^2+1^2}

The total time, 't', is given as follows;

t = \dfrac{x}{6} +\dfrac{\sqrt{(15-x)^2+1^2}}{2.5}

The minimum value of 't' is found by differentiating with an online tool, as follows;

\dfrac{dt}{dx}  = \dfrac{d\left(\dfrac{x}{6} +\dfrac{\sqrt{(15-x)^2+1^2}}{2.5}\right)}{dx} =  \dfrac{1}{6} -\dfrac{6 - 0.4\cdot x}{\sqrt{x^2-30\cdot x +226} }

At the maximum/minimum point, we have;

\dfrac{1}{6} -\dfrac{6 - 0.4\cdot x}{\sqrt{x^2-30\cdot x +226} } = 0

Simplifying, with a graphing calculator, we get;

-4.72·x² + 142·x - 1,070 = 0

From which we also get x ≈ 15.04 and x ≈ 0.64956

x ≈ 15.04 mile

Therefore, given that 15.04 mi is 0.04 mi after the point, the distance he should run = 15 mi - 0.04 mi ≈ 14.96 mi

t = \dfrac{14.96}{6} +\dfrac{\sqrt{(15-14.96)^2+1^2}}{2.5} \approx 2..89

Therefore, the distance to run, x ≈ 14.96 mile

6 0
2 years ago
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