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statuscvo [17]
2 years ago
12

Which point fits the description: the y- coordinate is half the x-coordinate?

Mathematics
1 answer:
Naily [24]2 years ago
8 0

Answer:

3,6

Step-by-step explanation:

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Help me find the answer to this area of a kite problem
julia-pushkina [17]

Answer:

Below in bold.

Step-by-step explanation:

Area of a kite = 1/2 * D1 * D2   were D1 and D2 are the 2 diagonals

D1  = 5 + 2.5 = 7.5 and D2 = 2*44444.5 = 9

So area = 1/2 * 7.5 * 9

= 33.75.

5 0
2 years ago
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Which best describes a transformation that preserves the size, shape, and angles of an object?
vovikov84 [41]
The transformation is ISOMETRY.

In mathematics, an isometry<span> (or congruence, or congruent transformation) is a distance-preserving transformation between metric spaces, usually assumed to be bijective. A composition of two opposite </span>isometries<span> is a direct </span>isometry<span>. A reflection in a line is an opposite </span>isometry<span>, like R </span>1<span> or R </span>2<span> on the image.</span>
3 0
3 years ago
What is the Difference Between a 16-Ounce Brick and a Carpenter?
Andrew [12]

One weighs a pound, and the other pounds away!

4 0
3 years ago
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5+2x15+1-6/76^3x2+73*0+0.0000273
NemiM [27]
  • 6+2(15)+1-6/(76^2×2)+73×0+0.0000273
  • 6+30+1-6/11552+0+0.0000273
  • 37-0.000519+0.0000273
  • 36.999

Done

6 0
2 years ago
Find gradient <br><br>xe^y + 4 ln y = x² at (1, 1)​
cricket20 [7]

xe^y+4\ln y=x^2

Differentiate both sides with respect to <em>x</em>, assuming <em>y</em> = <em>y</em>(<em>x</em>).

\dfrac{\mathrm d(xe^y+4\ln y)}{\mathrm dx}=\dfrac{\mathrm d(x^2)}{\mathrm dx}

\dfrac{\mathrm d(xe^y)}{\mathrm dx}+\dfrac{\mathrm d(4\ln y)}{\mathrm dx}=2x

\dfrac{\mathrm d(x)}{\mathrm dx}e^y+x\dfrac{\mathrm d(e^y)}{\mathrm dx}+\dfrac4y\dfrac{\mathrm dy}{\mathrm dx}=2x

e^y+xe^y\dfrac{\mathrm dy}{\mathrm dx}+\dfrac4y\dfrac{\mathrm dy}{\mathrm dx}=2x

Solve for d<em>y</em>/d<em>x</em> :

e^y+\left(xe^y+\dfrac4y\right)\dfrac{\mathrm dy}{\mathrm dx}=2x

\left(xe^y+\dfrac4y\right)\dfrac{\mathrm dy}{\mathrm dx}=2x-e^y

\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{2x-e^y}{xe^y+\frac4y}

If <em>y</em> ≠ 0, we can write

\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{2xy-ye^y}{xye^y+4}

At the point (1, 1), the derivative is

\dfrac{\mathrm dy}{\mathrm dx}\bigg|_{x=1,y=1}=\boxed{\dfrac{2-e}{e+4}}

4 0
3 years ago
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