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Whitepunk [10]
3 years ago
8

Select the quadrant or axis where each ordered pair is located on a coordinate plane

Mathematics
1 answer:
Helen [10]3 years ago
5 0

(-5, 7) - Quadrant II

(8 1/2, -4) - Quadrant IV

(0,5) - y axis

Step-by-step explanation:

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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

(y+1)^2=4xy\implies y(x)=2x-1\pm2\sqrt{x^2-x}

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

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

So one asymptote of the hyperbola is the line y=4x-2.

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

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

\displaystyle\lim_{x\to-\infty}(2x-2x\sqrt{1-\frac1x}-ax-(b+1))=0

Reduce the "negligible" term to get

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

Now we take a=0, and again we're careful to not pick b=-1.

\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
 x{x-(-x)\sqrt{1-\frac1x}}=\dfrac1{1+\sqrt{1-\frac1x}}

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.
4 0
3 years ago
Find the volume of a triangle prism with the following dimensions triangle base: 5ft
Andrews [41]
Hey there :)

We know the volume formula of a triangular prism
Volume = base area × height of prism

We are given:
Base of triangle = 5 ft
Height of triangle = 3 ft
Height of prism = 7.5 ft

Apply the formula
V = \frac{1}{2} ( 5 )( 3 ) × 7.5
   = 7.5 × 7.5
   = 56.24 ft³

4 0
3 years ago
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Gina can buy a 5-pound bag of apples that cost $4.50, or she can buy them by the pound for $0.85 or pound. Which of these best e
sweet-ann [11.9K]
It's "D", because when you divide 4.5/5 you get .90 per pound so buying per pound it less.
5 0
3 years ago
Please help!!
timofeeve [1]

Answer:

y = 2/3x

Step-by-step explanation:

So we know, when lines are perpendicular that they share the same slope. The slope in the equation of y = 2/3x + 7 is 2/3x. So, we have our slope. Now, we just need to find our b. We plug in the point that you gave us into the equation y = mx + b.

3 = 2/3(2) + b

we multiply the denominator 3 by 2, then divide that by the numerator 2.

3 = 3 + b

subtract 3 from itself and from the opposite side of the equal side 3.

0 = b

in the end, there is no b for this equation. Our answer is

y = 2/3x

6 0
3 years ago
4×(34−12)+3×6<br><br> No links i am bad at math
Inga [223]

Answer:

4×(34-12)+3×6

4×22+3×6

88+18

=106

Step-by-step explanation:

Apply the BEDMAS rule

B=brackets

E=exponents

D=Division

M=multiplication

A=addition

S=subtraction

8 0
3 years ago
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