What is 9.2518 rounded to the nearest tenth?

Consider Z1= 4-4 radical 3i and Z2= -16 i. What is Z1-Z2?

goldfiish [28.3K]

Answer:

Option (3)

Step-by-step explanation:

$Z_1=4-4\sqrt{3}i$

$Z_2=-16i$

$Z_1-Z_2=(4-4\sqrt{3}i)-(-16i)$

$=4-4\sqrt{3}i+16i$

$=4-(4\sqrt{3}i-16i)$

$=4-4(\sqrt{3}-4)i$

Therefore, Option (3) will be the correct option.

8 0

2 years ago

Click an item in the list or group of pictures at the bottom of the problem and, holding the button down, drag it into the corre

Irina18 [472]

Well if the perimeter is 36in, then divide by 6 sides (the prefix hex- means six) to get 6. Don't forget your unit of measurement to be 6in of each side. It would be nice to have the picture though so if you could suply me with that it would help out.

7 0

3 years ago

Read 2 more answers

HELP ASAP PLEASE I BEG TOU

nevsk [136]

Answer:

(0,-2)

(2,-1)

(4,0)

Step-by-step explanation:

The points are the above ones.

8 0

2 years ago

Read 2 more answers

I tell you these facts about a mystery number, $c$: $\bullet$ $1.5 < c < 2$ $\bullet$ $c$ can be written as a fraction wit

makkiz [27]

Answer:

Possible answer: $\displaystyle c = \frac{16}{10} = \frac{8}{5} = 1.6$ .

Step-by-step explanation:

Rewrite the bounds of $c$ as fractions:

The simplest fraction for $1.5$ is $\displaystyle \frac{3}{2}$ . Write the upper bound $2$ as a fraction with the same denominator:

$\displaystyle 2 = 2 \times 1 = 2 \times \frac{2}{2} = \frac{4}{2}$ .

Hence the range for $c$ would be:

$\displaystyle \frac{3}{2} < c < \frac{4}{2}$ .

If the denominator of $c$ is also $2$ , then the range for its numerator (call it $p$ ) would be $3 < p < 4$ . Apparently, no whole number could fit into this interval. The reason is that the interval is open, and the difference between the bounds is less than $2$ .

To solve this problem, consider scaling up the denominator. To make sure that the numerator of the bounds are still whole numbers, multiply both the numerator and the denominator by a whole number (for example, 2.)

$\displaystyle \frac{3}{2} = \frac{2 \times 3}{2 \times 2} = \frac{6}{4}$ .

$\displaystyle \frac{4}{2} = \frac{2\times 4}{2 \times 2} = \frac{8}{4}$ .

At this point, the difference between the numerators is now $2$ . That allows a number ( $7$ in this case) to fit between the bounds. However, $\displaystyle \frac{1}{c} = \frac{4}{7}$ can't be written as finite decimals.

Try multiplying the numerator and the denominator by a different number.

$\displaystyle \frac{3}{2} = \frac{3 \times 3}{3 \times 2} = \frac{9}{6}$ .

$\displaystyle \frac{4}{2} = \frac{3\times 4}{3 \times 2} = \frac{12}{6}$ .

$\displaystyle \frac{3}{2} = \frac{4 \times 3}{4 \times 2} = \frac{12}{8}$ .

$\displaystyle \frac{4}{2} = \frac{4\times 4}{4 \times 2} = \frac{16}{8}$ .

$\displaystyle \frac{3}{2} = \frac{5 \times 3}{5 \times 2} = \frac{15}{10}$ .

$\displaystyle \frac{4}{2} = \frac{5\times 4}{5 \times 2} = \frac{20}{10}$ .

It is important to note that some expressions for $c$ can be simplified. For example, $\displaystyle \frac{16}{10} = \frac{2 \times 8}{2 \times 5} = \frac{8}{5}$ because of the common factor $2$ .