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3 years ago

14

How can I simplify these?

1 answer:

Schach [20]3 years ago

4 0

Answer:

$\large\boxed{343=7\sqrt7}\\\boxed{10\sqrt{96}=40\sqrt6}$

Step-by-step explanation:

$343=7^3=7^2\cdot7\\\\\sqrt{343}=\sqrt{7^2\cdot7}\qquad\text{use}\ \sqrt{ab}=\sqrt{a}\cdot\sqrt{b}\\\\=\sqrt{7^2}\cdot\sqrt7\qquad\text{use}\ \sqrt{a^2}=a\ \text{for}\ a\geq0\\\\=7\sqrt7\\=====================\\\\96=16\cdot6\\\\10\sqrt{96}=10\sqrt{16\cdot6}=10\sqrt{16}\cdot\sqrt6=(10)(4)\sqrt6=40\sqrt6$

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Explain how you know that y is not directly proportianal to x

SVEN [57.7K]

You know this when the two values (x and y) are inversely proportional to each other by their product being constant (always remaining the same). This means when x increases y will decrease.

Hope this helped :)

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2 years ago

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zavuch27 [327]

Answer:

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Step-by-step explanation:

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2 years ago

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-Dominant- [34]

Answer:15/1

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2 years ago

3. A cone has a heighth and a base with radius r. You want to change the cone so its volume is tripled. What is the new height i

Sphinxa [80]

Answer:

Hence the new height is 3 times the original height .(h1=3h)

Step-by-step explanation:

Given:

A cone with has height and base with radius r .

To Find:

What is new height

Solution:

Consider as cone with height h base radius r, and volume v

Here given that only height changes for the cone i.e. r remains the unchanged or same or constant

The volume for a regular cone is given by ,

$V=pie*r^2*h/3$

Here V is directly proportional to h i.ee pie ,3 and r being constant

$V=K*h$

i.e V/h=constant

V1 and h1 are new dimensions for new cone

V/h=V1/h1

Here V1=3V

So V/h=3V/h1

1/h=3/h1

i.e h1=3h

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2 years ago

For which pair of functions is the vertex of k(x)7 units below the vertex of f(x)?

kvv77 [185]

Answer: Option C

$f(x) = x^2;\ k (x) = x ^ 2 -7$

Step-by-step explanation:

Whenever we have a main function f(x) and we want to transform the graph of f(x) by moving it vertically then we apply the transformation:

$k (x) = f (x) + b$

If $b> 0$ then the graph of k(x) will be the graph of f(x) displaced vertically b units down.

If $b> 0$ then the graph of k(x) will be the graph of f(x) displaced vertically b units upwards.

In this case we have

$f (x) = x ^ 2$

We know that this function has its vertex in point (0,0).

Then, to move its vertex 7 units down we apply the transformation:

$k (x) = f (x) - 7\\\\k (x) = x ^ 2 -7$ .

Then the function k(x) that will have its vertex 7 units below f(x) is

$k (x) = x ^ 2 -7$

3 0

2 years ago