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3 years ago

Helpif x ≠ 3, how would I write x^2 - 9/ x^2 - 3x?

Mathematics

1 answer:

maria [59]3 years ago

3 0

Do you want me to write it for you in standard form?

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Match each expression in the left column with the correct product in the right column.

lesya692 [45]

Answer:

-6

-6i

Step-by-step explanation:

1) √4 . √-3 . √-3

$\sqrt{4} = 2$

$\sqrt{-3} . \sqrt{-3} = (\sqrt{-3})^2$

$\sqrt{4} . (\sqrt{-3})^2 = 2 \times -3 =$ -6

2) √-4 . √-3 . √-3

$\sqrt{-4} = 2i$ .

Therefore, $\sqrt{-4} . \sqrt{-3} . \sqrt{-3} = 2. \sqrt{-1} \times -3 = 2i \times (-3) =$ - 6i

3) √4 . √3 . √-3

$\sqrt{4} = 2$

$\sqrt{3} . \sqrt{-3} = (\sqrt{3})^2 . \sqrt{-1}$

$\implies 2 \times 3i =$ 6i

4) √4 . √3 . √3

$\sqrt{4} = 2$

$\sqrt{3} . \sqrt{3} = (\sqrt{3})^2 = 3$

Therefore, √4 . √3 . √3 = 2 . 3 = 6

5 0

3 years ago

1. For each event, write the sample space and tell how many outcomes there are.

kozerog [31]

Answer:

A. 14

Step-by-step explanation:

6 0

2 years ago

Help pls this is for my math test.

Stells [14]

Answer:

D is the answers for the question

Step-by-step explanation:

please mark me as brainlest

5 0

3 years ago

Read 2 more answers

Don put $4,000 in a savings account with an interest rate of 4% for three years. If the interest is compounded annually, how muc

miskamm [114]

Answer:

$4,499.46

Step-by-step explanation:

We can use the compound interest formula for this problem:

$A=P(1+\frac{r}{n} )^{nt}$

P = initial balance

r = interest rate (decimal)

n = number of times compounded annually

t = time

First, lets change 4% into a decimal:

4% -> $\frac{4}{100}$ -> 0.04

Now lets plug the values into the equation as shown below:

$A=4,000(1+\frac{0.04}{1})^{1(3)}$

$A=4,499.46$

Don will have $4,499.46 at the end of the three years.

4 0

3 years ago

The as x approaches 0 of tan^2x/x

uysha [10]

L=Lim tan(x)^2/x x->0

Since both numerator and denominator evaluate to zero, we could apply l'Hôpital rule by taking derivatives.

d(tan^2(x))/dx=2tan(x).d(tan(x))/dx = 2tan(x)sec^2(x)

d(x)/dx = 1

L=2tan(x)sec^2(x)/1 x->0

= (2(0)/1^2)/1

=0/1

Another way using series,

We know that tan(x) = x+x^3/3+2x^5/15+.....

then tan^2(x), using binomial expansion gives

x^2+2*x^4/3+.... (we only need two terms)

and again apply l'Hôpital's rule, we have

L=d(x^2+2x^4/3+...)/d(x) = (2x+8x^3/3+...)/1

=0 as x->0

3 0

3 years ago