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Schach [20]
2 years ago
14

What are the coordinates of the point on the directed line segment from (1, -5)(1,−5) to (10, 1)(10,1) that partitions the segme

nt into a ratio of 1 to 5?
Mathematics
1 answer:
Hitman42 [59]2 years ago
5 0

Check the picture below.

\textit{internal division of a line segment using ratios} \\\\\\ A(1,-5)\qquad B(10,1)\qquad \qquad \stackrel{\textit{ratio from A to B}}{1:5} \\\\\\ \cfrac{A\underline{C}}{\underline{C} B} = \cfrac{1}{5}\implies \cfrac{A}{B} = \cfrac{1}{5}\implies 5A=1B\implies 5(1,-5)=1(10,1)

(\stackrel{x}{5}~~,~~ \stackrel{y}{-25})=(\stackrel{x}{10}~~,~~ \stackrel{y}{1})\implies C=\underset{\textit{sum of the ratios}}{\left( \cfrac{\stackrel{\textit{sum of x's}}{5+10}}{1+5}~~,~~\cfrac{\stackrel{\textit{sum of y's}}{-25+1}}{1+5} \right)} \\\\\\ C=\left( \cfrac{15}{6}~~,~~\cfrac{-24}{6} \right)\implies C=\left(\cfrac{5}{2}~~,~~-4 \right)

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Ivanshal [37]

Answer:

  • True
  • True

Step-by-step explanation:

This question asks you to compare the coordinates of the vertex of each function.

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The vertex of the function is its minimum, the point where the graph stops decreasing and starts increasing. It is the lowest point on the graph.

<h3>f(x)</h3>

The vertex is (-4, -1). The minimum is -1, located at x = -4.

<h3>g(x)</h3>

The vertex is (1, -25). The minimum is -25, located at x = 1. We know this is the minimum because there are no g(x) values that are lower (more negative).

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The minimum of f(x), -1, is greater than the minimum of g(x), -25. TRUE

The x-value of f(x) at its minimum, -4, is less than the x-value of g(x) at its minimum, 1. TRUE

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2 years ago
Can anyone solve these 3 equations?
Nesterboy [21]

Answer:

1. x = 1, -2

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Step-by-step explanation:


3 0
2 years ago
What is the range of y =5/2(x-2)² - 4
blagie [28]

Answer:

vertex:(2,-4)

Focus:(2,-\frac{39}{10} )

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Directrixt: y=-\frac{41}{10}

Step-by-step explanation:

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3 years ago
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Nina [5.8K]

Answer:

\boxed{5 \cdot \sqrt{2}  \cdot \sqrt[6]{5} }

Step-by-step explanation:

\sqrt[3]{250} \cdot \sqrt{\sqrt[3]{10} }

\sqrt{\sqrt[3]{10} } \implies (10^\frac{1}{3} )^\frac{1}{2} =10^\frac{1}{6} =\sqrt[6]{10}

\therefore \sqrt{\sqrt[3]{10} }=\sqrt[6]{10}

\text{Solving }\sqrt[3]{250} \cdot \sqrt{\sqrt[3]{10} }

250=2 \cdot 5^3

\sqrt[3]{250}=\sqrt[3]{2\cdot 5^3}=5  \sqrt[3]{2}

Once

\sqrt[6]{2}  \cdot \sqrt[6]{5} = \sqrt[6]{10}

We have

5  \sqrt[3]{2} \cdot \sqrt[6]{2}  \cdot \sqrt[6]{5}

We can proceed considering the common base of exponentials

\sqrt[3]{2}  \cdot \sqrt[6]{2}  =  2^{\frac{1}{3}} \cdot  2^{\frac{1}{6} }  = 2^{\frac{3}{6} } = 2^{\frac{1}{2} }=\sqrt{2}

Therefore,

5  \sqrt[3]{2} \cdot \sqrt[6]{2}  \cdot \sqrt[6]{5} = 5 \cdot \sqrt{2}  \cdot \sqrt[6]{5}

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