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leonid [27]
3 years ago
14

Which name does not apply to the figure? A. quadrilateral B. square C. parallelogram D. rhombus

Mathematics
2 answers:
aivan3 [116]3 years ago
8 0
The answer is B. 
-Corey

xz_007 [3.2K]3 years ago
7 0
B because you could make it into a parallelogram when you tilt it.
You can make a rhombus because it is a rhombus.
You can make a quadrilateral because it has 4 sides.
~Jz
Hope it helps.
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Destiny paid $6.60 for a dozen cupcakes for a birthday party. What is the unit cost for each cupcake?
balandron [24]

Answer:

The unit cost for each cupcake is .55 or 55 cents

Step-by-step explanation:

6.60 divided by 12 is .55  so therefore, your answer is .55 have a great day hopes this helps :)

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2 years ago
Help please, just number 3, thats all im asking, thanks.
Bezzdna [24]

The rectangle that has area 400 has width 10, so its length is 40.

A = LW; L = A/W = 400/10 = 40

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The multiplication is 10.2 * 12

40.2 * 12 = 40 * 10 + 0.2 * 10 + 40 * 2 + 0.2 * 2

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Ples help me find slant assemtotes
FrozenT [24]
A polynomial asymptote is a function p(x) such that

\displaystyle\lim_{x\to\pm\infty}(f(x)-p(x))=0

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Since this equation defines a hyperbola, we expect the asymptotes to be lines of the form p(x)=ax+b.

Ignore the negative root (we don't need it). If y=2x-1+2\sqrt{x^2-x}, then we want to find constants a,b such that

\displaystyle\lim_{x\to\infty}(2x-1+2\sqrt{x^2-x}-ax-b)=0

We have

\sqrt{x^2-x}=\sqrt{x^2}\sqrt{1-\dfrac1x}
\sqrt{x^2-x}=|x|\sqrt{1-\dfrac1x}
\sqrt{x^2-x}=x\sqrt{1-\dfrac1x}

since x\to\infty forces us to have x>0. And as x\to\infty, the \dfrac1x term is "negligible", so really \sqrt{x^2-x}\approx x. We can then treat the limand like

2x-1+2x-ax-b=(4-a)x-(b+1)

which tells us that we would choose a=4. You might be tempted to think b=-1, but that won't be right, and that has to do with how we wrote off the "negligible" term. To find the actual value of b, we have to solve for it in the following limit.

\displaystyle\lim_{x\to\infty}(2x-1+2\sqrt{x^2-x}-4x-b)=0

\displaystyle\lim_{x\to\infty}(\sqrt{x^2-x}-x)=\frac{b+1}2

We write

(\sqrt{x^2-x}-x)\cdot\dfrac{\sqrt{x^2-x}+x}{\sqrt{x^2-x}+x}=\dfrac{(x^2-x)-x^2}{\sqrt{x^2-x}+x}=-\dfrac x{x\sqrt{1-\frac1x}+x}=-\dfrac1{\sqrt{1-\frac1x}+1}

Now as x\to\infty, we see this expression approaching -\dfrac12, so that

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The other asymptote is obtained similarly by examining the limit as x\to-\infty.

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\displaystyle\lim_{x\to-\infty}(2x-1+2\sqrt{x^2-x}-b)=0

\displaystyle\lim_{x\to-\infty}(x+\sqrt{x^2-x})=\frac{b+1}2

(x+\sqrt{x^2-x})\cdot\dfrac{x-\sqrt{x^2-x}}{x-\sqrt{x^2-x}}=\dfrac{x^2-(x^2-x)}{x-\sqrt{x^2-x}}=\dfrac
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This time the limit is \dfrac12, so

\dfrac12=\dfrac{b+1}2\implies b=0

which means the other asymptote is the line y=0.
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