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4 years ago

Two hundred five thousandths in decimal form

Mathematics

2 answers:

zalisa [80]4 years ago

8 0

Answer:

.02

Step-by-step explanation:

I am Lyosha [343]4 years ago

3 0

Answer:

0.250

Step-by-step explanation:

It's easy, we learned this junk like 4 years ago

Thanks

-Brianna Edwards

VoTe AnD mArK bRaInLiEsT

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Pls solve both and show ur answers :))

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lianna [129]

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Step-by-step explanation:

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3 years ago

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Please help, the question below

Licemer1 [7]

Answer:

$\sf R=\left(3, -\dfrac{5}{4}\right)$

Step-by-step explanation:

Given:

P = (1, 5)
Q = (-2, 3)
R = (a, b)
R lies on line x = 3
PR = QR

If point R lies on the line x = 3, then the x-value of point R is 3.

⇒ a = 3

Distance between two points

$\sf d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

(where (x₁, y₁) and (x₂, y₂) are the two points)

Use the distance formula to derive equations for PR and QR.

Let (x₁, y₁) = P = (1, 5)

Let (x₂, y₂) = R = (3, b)

$\begin{aligned} \sf PR & =\sf \sqrt{(3-1)^2+(b-5)^2}\\ & = \sf \sqrt{4+(b-5)^2} \end{aligned}$

Let (x₁, y₁) = Q = (-2, 3)

Let (x₂, y₂) = R = (3, b)

$\begin{aligned} \sf QR & =\sf \sqrt{(3-(-2))^2+(b-3)^2}\\ & = \sf \sqrt{25+(b-3)^2} \end{aligned}$

As PR = QR, equate the derived equations and solve for b:

$\begin{aligned} \sf PR & = \sf QR \\\sf \sqrt{4+(b-5)^2} & = \sf \sqrt{25+(b-3)^2}\\\sf 4+(b-5)^2 & = \sf 25+(b-3)^2\\\sf 4+b^2-10b+25 & = \sf 25 + b^2-6b+9\\\sf b^2-10b+29 & = \sf b^2 -6b +34\\\sf -10b+29 & = \sf -6b + 34\\\sf -4b & = \sf 5\\\sf b & = -\dfrac{5}{4}\end{aligned}$