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shusha [124]
3 years ago
6

The graph shows quadrilateral FGHI and the location of vertex F' after a dilation with respect to the origin. What are the coord

inates of G' ?
a. (2, 3)
b. (0, 2)
c. (3, 2)
d. (2, 0)

Mathematics
2 answers:
Deffense [45]3 years ago
8 0
G is at (0, 2). Good luck!
PolarNik [594]3 years ago
3 0

Answer:

b. (0,2)

Step-by-step explanation:

The distance between origin and F is double than between origin and F'. With that in mind, we can deduce that the distance between origin and G must be double than between origin and G'. If the distance between origin and G is 4, then the distance between origin and G' has to be 2. Given that point G is on the y-axis, G' must be on the y-axis too. So, point G' is (0,2)

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Answer:

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Step-by-step explanation:

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2 years ago
a)A fruitseller bought 50kg of the fruits. He sold 30kg of fruits for the cost price of 35kg of fruits and he sold the remaining
alex41 [277]
<h2><u>Question</u><u>:</u><u>-</u></h2>

A fruitseller bought 50kg of the fruits. He sold 30kg of fruits for the cost price of 35kg of fruits and he sold the remaining quantity for the cost Price of 18kg of fruits. calculate his profit or loss percent in the total transaction.

<h2><u>Answer</u><u>:</u><u>-</u></h2>

let the cost price be 50x

→he sells 30kg of fruits on it's CP of 35 kg

→CP of 30kg fruits = 30x

→SP of 35kg fruits = 35x

→remaing fruits are 20kg

→he sells 20kg of fruits on CP of 16kg

→CP of 20kg fruits = 20x

→SP of 20kg fruits = 16x

→total CP is = 50x

→total SP is = (35 + 16) = 51x

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→ 51-50

→ 1

<h2 /><h2><u>Now,</u></h2>

→ Profit% = gain/CP × 100

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7 0
3 years ago
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vampirchik [111]
\displaystyle\lim_{h\to0}\frac{\tan\sqrt{x+h}-\tan x}h

Employ a standard trick used in proving the chain rule:

\dfrac{\tan\sqrt{x+h}-\tan x}{\sqrt{x+h}-\sqrt x}\cdot\dfrac{\sqrt{x+h}-\sqrt x}h

The limit of a product is the product of limits, i.e. we can write

\displaystyle\left(\lim_{h\to0}\frac{\tan\sqrt{x+h}-\tan x}{\sqrt{x+h}-\sqrt x}\right)\cdot\left(\lim_{h\to0}\frac{\sqrt{x+h}-\sqrt x}h\right)

The rightmost limit is an exercise in differentiating \sqrt x using the definition, which you probably already know is \dfrac1{2\sqrt x}.

For the leftmost limit, we make a substitution y=\sqrt x. Now, if we make a slight change to x by adding a small number h, this propagates a similar small change in y that we'll call h', so that we can set y+h'=\sqrt{x+h}. Then as h\to0, we see that it's also the case that h'\to0 (since we fix y=\sqrt x). So we can write the remaining limit as

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which in turn is the derivative of \tan y, another limit you probably already know how to compute. We'd end up with \sec^2y, or \sec^2\sqrt x.

So we find that

\dfrac{\mathrm d\tan\sqrt x}{\mathrm dx}=\dfrac{\sec^2\sqrt x}{2\sqrt x}
7 0
3 years ago
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