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docker41 [41]
2 years ago
9

Can you help with this please.

Mathematics
2 answers:
Lubov Fominskaja [6]2 years ago
8 0

Answer:

<u>8.1 units</u>

Step-by-step explanation:

<u>Distance Formula</u>

  • d = √(x₂ - x₁)² + (y₂ - y₁)²

<u>Solving</u>

  • d = √(14 - 10)² + (12 - 5)²
  • d = √4² + 7²
  • d = √65
  • d = <u>8.1 units</u> (nearest tenth)
andrew-mc [135]2 years ago
5 0

\huge\underline{\red{A}\blue{n}\pink{s}\purple{w}\orange{e}\green{r} -}

  • Given - <u>two </u><u>points </u><u>P </u><u>(</u><u> </u><u>5</u><u> </u><u>,</u><u> </u><u>1</u><u>0</u><u> </u><u>)</u><u> </u><u>and </u><u>R </u><u>(</u><u> </u><u>1</u><u>2</u><u> </u><u>,</u><u> </u><u>1</u><u>4</u><u> </u><u>)</u><u> </u><u>on </u><u>the </u><u>c</u><u>artesian </u><u>plane</u>

  • To find - <u>distance </u><u>between </u><u>the </u><u>two </u><u>points</u>

<u>Using </u><u>the </u><u>distance </u><u>formula</u> ~

\qquad\underline{d(PR) =  \sqrt{( x_{2}  -  x_{1}) {}^{2}  + ( y_{2} -  y_{1}) {}^{2}   }}  \\  \\

we have ,

\longrightarrow x_{1} = 5 \\ \longrightarrow x_{2} = 12 \\  \\  \longrightarrow \: y_{1} = 10 \\  \longrightarrow \: y_{2} = 14

<u>substituting</u><u> </u><u>the </u><u>values </u><u>in </u><u>the </u><u>formula </u><u>,</u><u> </u><u>we </u><u>get</u>

d(PR) =  \sqrt{(12-5) {}^{2} + (14-10) {}^{2}  }  \\  \\ \implies \: d(PR) =  \sqrt{(7) {}^{2}  + ( 4) {}^{2} }  \\  \\ \implies \: d(PR) =  \sqrt{49 + 16}  \\  \\ \implies \: d(PR) =  \sqrt{65}  \\  \\ \implies \: d(PR) = 8.1 \: units \: (approx)

hope helpful :)

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wel

Hey there!

“4y + 3x”

If x = 5 and y = -3, then substitute it into the equation

4(-3) + 3(5)

4(-3) = -12

-12 + 3(5)

3(5) = 15

-12 + 15

= 3

Answer: 3

Good luck on your assignment and enjoy your day!

~Amphitrite140:)

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The domain is the set of possible x input values. Look at the left most point (-4,-1). The x coordinate here is x = -4. This is the smallest x value allowed. The largest x value allowed is x = 5 for similar reasons, but on the other side of the graph.

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Writing [-4,5] for interval notation tells us that we have an interval from -4 to 5 and we include both endpoints. The square brackets mean "include endpoint"

Writing \left\{x|x\in\mathbb{R}, \ -4\le x\le 5\right\} is the set-builder notation way of expressing the domain. The x\in\mathbb{R} portion means "x is a real number"

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