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3 years ago

Fill in the blank to make the statement true (-2+6)+__=-2+[6+(-8)

Mathematics

1 answer:

bixtya [17]3 years ago

4 0

Answer:

-8

Step-by-step explanation:

you have to make them equivalent so and the one thing missing from the other side of the equation

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The volume of a right circular cone that has a height of 13 m and a base with a diameter of 3.4 m. Round your answer to the near

Mamont248 [21]

Answer:

The volume of right circular cone having height = $13 \ m$ and Diameter = $3.4 \ m$ is $39.4\ cubic\ meter$ .

Step-by-step explanation:

Diagram of the given scenario is shown below.

Given that

Height of a right circular cone is $13 \ m$ .

Diameter of the right circular cone is $3.4 \ m$ .

To find: The volume of a right circular cone.

So, From the question,

$Height = \ 13 \ m\\diameter = 3.4\ m$

$Radius = \frac{diameter}{2}= \frac{3.4}{2}$

⇒ $Radius = 1.7\ m$

Now

Volume of right circular cone = $\frac{1}{3}\pi r^{2}h$

$=\frac{1}{3}\times\frac{22}{7} \times (1.7)^{2}\times 13$

= $\frac{826.54}{21}$

$= 39.4\ cubic\ meter$ .

Therefore,

The volume of right circular cone having height = $13 \ m$ and Diameter = $3.4 \ m$ is $39.4\ cubic\ meter$ .

8 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

Represent the following expression using an exponent

Andrew [12]

12^4 which is equal to 12 times itself 4 times

4 0

2 years ago

How to you solve -15=y-62

sdas [7]

First move the (Y) to the other side of the equal sign,
$- y - 15 = - 62$
Then subtract add (15) on both sides
$- y = - 47$
Since a variable can't be negative just divide by (-1) on both sides,
$\frac{ - y}{ - 1} = \frac{ - 47}{ - 1}$
The answer would be,
$y = 47$
Hope this helped
:D

7 0

2 years ago

The distance an object falls when dropped from a tower varies directly as the square of the time it falls. If the object falls 1

Tanzania [10]

Answer:

the object will fall 792 feet in 17 seconds

Step-by-step explanation:

first you can do 144 multiplied by 5 because that would be 15 of those seconds next you would do 144 divided by 2 because you only need 2 seconds to reach 17 your answer would be 72 then you would do 720+72=792 so it would fall 792ft in 17 seconds

3 0

2 years ago

Read 2 more answers

Fill in the blank to make the statement true (-2+6)+__=-2+[6+(-8)​

Fill in the blank to make the statement true (-2+6)+__=-2+[6+(-8)