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3 years ago

Which rule yields the dilation of the figure JKLM centered at the origin?

Mathematics

2 answers:

Naddika [18.5K]3 years ago

7 0

Answer:

Step-by-step explanation:

Sphinxa [80]3 years ago

3 0

I’d say it’s between B and D

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Solve for − ∣ x − 2 ∣ + 4 ≤ 0

MakcuM [25]

Answer:

x ≥ 6 or x ≤ -2

Step-by-step explanation:

- I x - 2 I + 4 ≤ 0

- I x - 2 I ≤ -4

I x - 2 I ≥ 4

x - 2 ≥ 4 or x - 2 ≤ -4

x ≥ 6 or x ≤ -2

5 0

3 years ago

(2x³)⁴= ? What is the right answer, with full explanation please:

netineya [11]

So you multiply the 4th power

2 times 4 is 8
4 times 3 is 12
Then you plug it In 8x^12

So your answer is D

Hope this helps *smiles*

7 0

3 years ago

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American cars maker produce 5650000 cars each year in Europe that Americans make 6 million 550 cars the mistake did Ben make how

Vlad [161]

Answer:

Ben has mixed up the digit for millions and hundred thousands: he wrote 5 in the place for millions and 6 in the place for hundred thousands, but it should be the other way. I think that he should exchange those two digits in the faulty report.

3 0

3 years ago

thank you for hurrying here is the very last one Explain how you would find the perimeter of this triangle.

Lena [83]

Step-by-step explanation:

Use the method of TOA CAH SOH.

$tan(51.3) = \frac{opp}{adj} \\ \tan(51.3) = \frac{u}{4} \\ u = 4 \tan(51.3)$

$cos(51.3) = \frac{adj}{hypo} \\ \cos(51.3) = \frac{4}{v} \\ v \cos(51.3) = 4 \\ v = \frac{4}{ \cos(51.3) }$

Perimeter of triangle = Sum of all sides =

$u \: + \: v + 4$

5 0

3 years ago

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The slope of the line tangent to the curve y^2 + (xy+1)^3 = 0 at (2, -1) is ...?

Lina20 [59]

$\frac{d}{dx}(y^2 + (xy+1)^3 = 0)
\\
\\\frac{d}{dx}y^2 + \frac{d}{dx}(xy+1)^3 = \frac{d}{dx}0
\\
\\\frac{dy}{dx}2y + \frac{d}{dx}(xy+1)\ *3(xy+1)^2 = 0
\\
\\\frac{dy}{dx}2y + \left(\frac{d}{dx}(xy)+\frac{d}{dx}1\right)\ * 3(xy+1)^2 = 0
\\
\\\frac{dy}{dx}2y + \left(\frac{d}{dx}xy+x\frac{d}{dx}y+\frac{dx}{dx}\right)\ * 3(xy+1)^2 = 0
\\
\\\frac{dy}{dx}2y + \left(\frac{dx}{dx}y+x\frac{dy}{dx}+\frac{dx}{dx}0\right)\ * 3(xy+1)^2 = 0
\\
\\\frac{dy}{dx}2y + \left(y+x\frac{dy}{dx}\right)\ * 3(xy+1)^2 = 0$
$\frac{dy}{dx}2y + \left(y+x\frac{dy}{dx}\right)\ * 3(xy+1)^2 = 0
\\
\\\frac{dy}{dx}2y + \left(3y+3x\frac{dy}{dx}\right) (xy+1)^2 = 0
\\
\\2y\ y' + \left(3y+3x\ y'\right) (xy+1)^2 = 0
\\
\\2y\ y' + \left(3y+3x\ y'\right) ((xy)^2+2xy+1) = 0
\\
\\2y\ y' + \left(3y+3x\ y'\right) ((xy)^2+2xy+1) = 0
\\
\\2y\ y' + 3x^2y^3 +6xy^2+3y+(3x y' +6x^2y y'+3x^2y y') = 0
\\
\\3x^2y^3 +6xy^2+3y+(3x y' +6x^2y y'+3x^2y y'+2yy' ) = 0
\\
\\y'(3x +6x^2y+3x^2y+2y ) = -3x^2y^3 -6xy^2-3y

$
$y' = \frac{-3x^2y^3 -6xy^2-3y}{(3x +6x^2y+3x^2y+2y )};\ x=2,y=-1
\\
\\y' = \frac{-3(2)^2(-1)^3 -6(2)(-1)^2-3(-1)}{(3(2) +6(2)^2(-1)+3(2)^2(-1)+2(-1) )}
\\
\\y' = \frac{-3(4)(-1) -6(2)(1)+3}{(6 +6(4)(-1)+3(4)(-1)-2 )}
\\
\\y' = \frac{12 -12+3}{(6 -24-12-2 )}
\\
\\y' = \frac{3}{( -32 )}$

7 0

3 years ago