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Vilka [71]
3 years ago
8

How many times greater is the value of 3 in the product of 34 × 10000 in the value of 3 in 34

Mathematics
1 answer:
nika2105 [10]3 years ago
6 0

Answer:

It is 4 times greater.

Step-by-step explanation:

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Step 1: –10 + 8x < 6x – 4
Oduvanchick [21]
Click the link up there that answer is that
4 0
3 years ago
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
There are 15children ten children are each wearing 1cap how many more caps would you need to have one cap on each child? Draw to
Alex_Xolod [135]

Answer:2

Step-by-step explanation:

6 0
3 years ago
Find the area of this parallelogram. be sure to include the correct unit in your answer
gulaghasi [49]

Answer:

84ft

Step-by-step explanation:

The formula to find the area of a parallelogram is b x h.

The base is 14ft and the height is 6ft. Multiply these together.

14 x 6 = 84ft

6 0
3 years ago
Read 2 more answers
What is the slant height, s, of the triangular prism?
Rufina [12.5K]

Answer:

5.4

Explanation:

a²+b²=c²

2²+5²= 4+25

4+25=29

Now we have to square root the 29 which would make the answer 5.4

4 0
4 years ago
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