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3 years ago

Question 7

Mathematics

1 answer:

umka21 [38]3 years ago

5 0

Answer:

A. √16

Step-by-step explanation:

A perfect square is any number that can be expressed as a product of an integer and itself. i.e. 4 can be represented as 2 × 2. Therefore, 4 is a perfect square.

From the options given, the only radical expression that has two perfect square is √16.

√16 consists of two perfect squares, which are 4 and 4.

4 can be expressed as 2 × 2.

Thus: √16 = √(2 × 2)(2 × 2).

Therefore, √16 consists of two perfect squares.

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Given the function:

anyanavicka [17]

You are correct. The answer is choice D

The only way for g(x) to be differentiable at x = 0 is for two things to happen
(1) g(x) is continuous at x = 0
(2) g ' (x) is continuous at x = 0

To satisfy property (1) above, the value of b must be 1. This can be found by plugging x = 0 into each piece of the piecewise function and solving for b.

So the piecewise function becomes
$g(x) = \begin{cases}x+1 \ \text{ if } \ x \ < \ 0\\ \cos(x) \text{ if } \ x \ge 0\end{cases}$
after plugging in b = 1

--------------------------------

Now differentiate each piece with respect to x to get
$g'(x) = \begin{cases}1 \ \text{ if } \ x \ < \ 0\\ -\sin(x) \text{ if } \ x \ge 0\end{cases}$
The first piece of g ' (x) is always going to be equal to 1. The second piece is equal to zero when x = 0
Because -sin(x) = -sin(0) = 0

So there's this disconnect on g ' (x) meaning its not continuous

Therefore, the value b = 1 will not work.

So there are no values of b that work to satisfy property (1) and property (2) mentioned at the top.

5 0

3 years ago

Please help answer this question will give brainlst if your correct :)

charle [14.2K]

A line passes throught the origin (-1, 1) and (4, n). Find the value of n. enter the correct answer in the box. I'd say (4, -2)

If I'm wrong sorry. If I'm correct May I Have Brainliest? <3

6 0

3 years ago

One factor of 7x2 +33x–10 is

PIT_PIT [208]

Answer:

OptionB

Step-by-step explanation:

6 0

4 years ago

Use vertical multiplication to find the product of:

Rom4ik [11]

Answer:

$x^6+x^4+4x^3-2x^2-x+3$

Step-by-step explanation:

$x^3+2x+3$

$\times(x^3-x+1)$

---------------------------------

First step multiply your terms in your first expression just to the 1 in the second expression like so:

$x^3+2x+3$

$\times(x^3-x+1)$

---------------------------------

$x^3+2x+3$ Anything times 1 is that anything.

That is, $(x^3+2x+3) \cdot 1=x^3+2x+3$ .

Now we are going to take the top expression and multiply it to the -x in the second expression. $-x(x^3+2x+3)=-x^4-2x^2-3x$ . We are going to put this product right under our previous product.

$x^3+2x+3$

$\times(x^3-x+1)$

---------------------------------

$x^3+2x+3$

$-x^4-2x^2-3x$

We still have one more multiplication but before we do that I'm going to put some 0 place holders in and get my like terms lined up for the later addition:

$x^3+2x+3$

$\times(x^3-x+1)$

---------------------------------

$0x^4+x^3+0x^2+2x+3$

$-x^4+0x^3-2x^2-3x+0$

Now for the last multiplication, we are going to take the top expression and multiply it to x^3 giving us $x^3(x^3+2x+3)=x^6+2x^4+3x^3$ . (I'm going to put this product underneath our other 2 products):