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3 years ago

Perform the operation and write the result in standard form 3i/3+i+8/3-i

Mathematics

1 answer:

Vera_Pavlovna [14]3 years ago

8 0

2.7 + 1.7i

We require the fractions to have a common denominator.

multiply the numerator of the first fraction by (3 - i) and the numerator of the second fraction by (3 + 1)

= $\frac{3i(3-i)+8(3+i)}{(3+i)(3-i)}$

= $\frac{9i-3i^2+24+8i}{9-i^2}$

[ note that i² = ( $\sqrt{(-1)^2}$ = - 1]

= $\frac{17i+27}{10}$ = 2.7 + 1.7i ← in standard form

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Gre4nikov [31]

Answer:

1. (4,6)

2. (-1, 8)

3. Not enough information given.

Step-by-step explanation:

1. Reflecting the point over the y-axis will place the point (-4,6) 4 units on the other side of the axis. This would place it at (4,6) without changing the y-coordinate.

2. Reflections across the line y=x means y and x trade places. This will result in the point (8,-1) becoming (-1,8).

3. Not enough information about the shape is given. Need to know where it is now to know where it will go.

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3 years ago

HELP PLEASE

BlackZzzverrR [31]

I chose a box of cereal-------> rectangular prism

Part 1)List the dimensions of your box. Be sure to include the units (in, cm, ft, etc.).

Length: 20 cm

Width: 8 cm

Height: 32 cm

Part 2).

Describe the shape of the cross-section when the box is cut parallel to the base

the shape of the cross-section would be a rectangle of dimensions

20 cm x 8 cm

Part 3)

What is the surface area of the box?

surface area=2*area of the base+perimeter of the base*height

area of the base=20*8=160 cm²

perimeter of the base=2*[20+8]-----> 56 cm

height=32 cm

surface area=2*160+56*32------> 2,112 cm²

the answer part 3) is

2,112 cm²

Part 4)

What is the surface area of the box if it is scaled up by a factor of 10?

we know that

surface area of the larger box=[scale factor]²*surface area original box

scale factor=10

surface area original box=2,112 cm²

surface area of the larger box=10²*2,112-----> 211,200 cm²

the answer part 4) is

211,200 cm²

Part 5)

What is the volume of the box?

volume of the box=L*W*H-----> 20*8*32------> 5,120 cm³

the answer Part 5) is

5,120 cm³

Part 6)

What is the volume of the box if it is scaled down by a factor of 1/10

we know that

the volume of the smaller box=[scale factor]³*volume original box

scale factor=1/10

volume original box=5,120 cm³

volume of the smaller box=[1/10]³*5,120-----> 5.12 cm³

the answer part 6) is

5.12 cm³

The length is⇒ 6 in., the width is⇒ 2 in., and the height is⇒ 16 in.

please mark brainliest!

8 0

3 years ago

Lesha wants to prove that the measures of the 3 angles in a triangle always have a sum of 180°. She decided to prove

levacccp [35]

Answer:

Step-by-step explanation:

Reflecting it across the lines will bring the triangle on to the other triangles. This will show that when the triangle goes under a transformation such as reflection, the angles will stay the same.

5 0

3 years ago

Margaret is thinking of a number. She says, “My number is odd. It is a factor of 36 and a multiple of 3” There are two possible

Novay_Z [31]

Answer:

3,9

Step-by-step explanation:

Factors of 36 = 1,2,3,4,6,9,12,18,36

Those factors which are odd = 1,3,9

Odd Factors which are multiple of 3 = 3,9

So, these are the two numbers Margret might be thinking of.

4 0

3 years ago

The statistical difference between a process operating at a 5 sigma level and a process operating at a 6 sigma level is markedly

Svet_ta [14]

Answer:

True

Step-by-step explanation:

A six sigma level has a lower and upper specification limits between $\\ (\mu - 6\sigma)$ and $\\ (\mu + 6\sigma)$ . It means that the probability of finding no defects in a process is, considering 12 significant figures, for values symmetrically covered for standard deviations from the mean of a normal distribution:

$\\ p = F(\mu + 6\sigma) - F(\mu - 6\sigma) = 0.999999998027$

For those with defects operating at a 6 sigma level, the probability is:

$\\ 1 - p = 1 - 0.999999998027 = 0.000000001973$

Similarly, for finding no defects in a 5 sigma level, we have:

$\\ p = F(\mu + 5\sigma) - F(\mu - 5\sigma) = 0.999999426697$ .

The probability of defects is:

$\\ 1 - p = 1 - 0.999999426697 = 0.000000573303$

Well, the defects present in a six sigma level and a five sigma level are, respectively:

$\\ {6\sigma} = 0.000000001973 = 1.973 * 10^{-9} \approx \frac{2}{10^9} \approx \frac{2}{1000000000}$

$\\ {5\sigma} = 0.000000573303 = 5.73303 * 10^{-7} \approx \frac{6}{10^7} \approx \frac{6}{10000000}$

Then, comparing both fractions, we can confirm that a 6 sigma level is markedly different when it comes to the number of defects present:

$\\ {6\sigma} \approx \frac{2}{10^9}$ [1]

$\\ {5\sigma} \approx \frac{6}{10^7} = \frac{6}{10^7}*\frac{10^2}{10^2}=\frac{600}{10^9}$ [2]

Comparing [1] and [2], a six sigma process has 2 defects per billion opportunities, whereas a five sigma process has 600 defects per billion opportunities.

8 0

3 years ago