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2 years ago

What is the complex formula for dilation with scale factor 2 and rotation by 90

Mathematics

1 answer:

zhuklara [117]2 years ago

7 0

Answer:

z' = 2iz +11 -2i

Step-by-step explanation:

Dilation multiplies each point by the scale factor. Rotation by 90° CCW is equivalent to multiplication by i.

When the center is not the origin, the transformation is applied to the difference from the center, then the result is added to the center.

z' = 2i(z -(3+4i)) + (3+4i)

Simplifying gives ...

z' = 2iz -6i +8 +3 +4i

z' = 2iz +11 -2i

_____

Check

For example, consider the point 1 unit east of the center of dilation/rotation. That point is z = 4+4i. Applying the transformation moves this point to ...

z' = 2i(4 +4i) +11 -2i = 8i -8 +11 -2i

z' = 3 +6i . . . . . 2 units north of the center of dilation/rotation

This is where it is expected to be after dilation by a scale factor of 2 and rotation 90° CCW.

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Help me again please

mash [69]

Answer:

answer C (0,-9)

Step-by-step explanation:

the line passes through (0,-9)

6 0

3 years ago

Suppose an identical ladder was put against a building that was 18 feet high and the ladder reached the top of the building. How

nikdorinn [45]

The distance between the bottom of the ladder be from the base of the building will be 17.32 ft.

<h3>What is the Pythagorean theorem?</h3>

It states that in the right-angle triangle the hypotenuse square is equal to the sum of the square of the other two sides.

As we can see in the figure the length of the ladder is 20 ft and the base of the ladder is 10 ft from the base of the building.

By using the Pythagorean theorem we will calculate the distance between the tip of the ladder and the base of the building.

H² = 20² - 10²

H²= 400 - 300

H² = 300

H = √300

H = 17.32 ft.

Therefore the distance between the bottom of the ladder is from the base of the building will be 17.32 ft.

To know more about the Pythagorean theorem follow

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7 0

1 year ago

A fence must be built to enclose a rectangular area of 20,000 ft^2. Fencing material costs $4 per foot for the two sides facing

Marysya12 [62]

The cost of the least expensive fence is $3200

Step-by-step explanation:

The given is:

A fence must be built to enclose a rectangular area of 20,000 ft²
Fencing material costs $4 per foot for the two sides facing north and south and $8 per foot for the other two sides

We need to find the cost of the least expensive fence

Assume that the length of each side opposite to North or South is x feet and the length of each other sides is y feet

∵ The length of the rectangle = x feet

∵ The width of the rectangle = y feet

∵ The rectangular area is 20,000 ft²

- Area of a rectangle = length × width

∴ x × y = 20,000

- Divide both sides by x to find y in terms of x

∴ $y=\frac{20,000}{x}$

The fence's length is equal to the perimeter of the rectangular area

∵ Perimeter of the rectangle = 2 length + 2 width

∴ Perimeter of the rectangle = 2x + 2y

∵ Fencing material costs $4 per foot for the two sides facing

North and South

∴ x costs $4 per foot

∵ The cost of the other two sides is $8 per foot

∴ y costs $8 per feet

The cost of the fence is the sum of the products of 4 , 2x and 8 , 2y

∵ The cost of the fence (C) = 4(2x) + 8(2y)

∴ C = 8x + 16y

- Substitute y by its value above

∴ $C=8x+16(\frac{20,000}{x})$

∴ $C=8x+\frac{320,000}{x}$

To find the least expensive differentiate C with respect to x and equate the answer by 0 to find the value of x

∵ $\frac{320,000}{x}$ can be written as $320,000x^{-1}$

∴ $C=8x+320,000x^{-1}$

∵ $\frac{dC}{dx}=8(1)x^{1-1}+320,000(-1)x^{-1-1}$

∴ $\frac{dC}{dx}=8-320,000x^{-2}$

- Equate $\frac{dC}{dx}$ by zero

∴ $8-320,000x^{-2}=0$

∵ $-320,000x^{-2}=-\frac{320,000}{x^{2}}$

∴ $8-\frac{320,000}{x^{2}}=0$

- Subtract 8 from both sides

∴ $-\frac{320,000}{x^{2}}=-8$

- Multiply both sides by x²

∴ - 320,000 = - 8x²

- Divide both sides by -8

∴ 40,000 = x²

- Take √ for both sides

∴ 200 = x

Substitute x in the equation of C to find the least cost of fence

∵ $C=8(200)+\frac{320,000}{(200)}$

∴ C = 1600 + 1600

∴ C = 3200

The cost of the least expensive fence is $3200

Learn more:

You can learn more about the differentiation in brainly.com/question/4279146

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5 0

2 years ago

The branch manager of an outlet (Store 1) of a nationwide

monitta

Answer:

1. ( 19.1416, 23.5384,)

2. (0.276348, 0.46651)

3. the sample size = 170.73 approximately 171

4. sample size = 334.07 approximately 334

5. sample of 334 should be taken by manager

Step-by-step explanation:

mean = bar x = 21.34 dollars

size of sample n = 70

standard deviation of sample = 9.22

we use t distribution as the population standard deviation is not known.

95% Confidence interval

1-α = 0.95

α = 0.05

degree of freedom = 70-1 = 69

α/2 = 0.025

using the t distribution tsble,

= 1.9949

confidence interval = $21.34+-1.9949*[\frac{9.22}{sqrt(70)}] \\$

= 21.34 +- (1.9949*1.10200)

= 21.34 + 2.1984, 21.34 - 2.1984

= (23.5384, 19.1416)

the confidence interval of the mean amount spent at the supply store can be written as 19.1416<u<23.5384

2. sixe of those who only have a cat

p = 26/70 = 0.371429

at 90 % confidence interval,

1-α = 0.90

α = 0.10

we use the z table here

z(0.10/2) = Z(0.05)

= 1.645

$0.371429+-1.645\sqrt} \frac{0.371429(1-0.371429)}{70}$

= 0.371429 +-( 1.645 x 0.0578)

= 0.371429 + 0.095081, 0.371429 - 0.095081

= (0.276348, 0.46651)

3. sd = 10$

margin of erro,r e = 1.50$

α = 0.05

using z table

α/2 = Z0.025

= 1.96

sample size = 1.96² * 10² / 1.50²

= 3.8416 * 100/ 2.25

= 170.73

the sample size is approximately 171

d. we have 0.5 as sample proportion now

margin of error = 0.045

α = 0.10

Zα/2 = 0.05

= 1.645

sample size = 1.645²x0.5(1-0.5) / 0.045²

= 0.676506/0.002025

= 334. 07

sample size = 334

5. sample of 334 should be taken by manager

3 0

3 years ago

Round to the nearest tenth.

laila [671]

Answer:

8.2 units

Step-by-step explanation:

Given that the only information for our triangle are 2 sides and 1 angle, we must use the Law of Cosines to find side BC

Recall the Law of Cosines

$a^2=b^2+c^2-2bc\cdot cos(A)$

Identify angles and sides

$m\angle A=23^\circ\\a=BC=?\\b=AC=14\\c=AB=6$

Solve for side "a"

$a^2=b^2+c^2-2bc\cdot cos(A)\\\\a^2=14^2+6^2-2(14)(6)\cdot cos(23^\circ)\\\\a^2=196+36-168cos(23^\circ)\\\\a^2=222-168cos(23^\circ)\\\\a=\sqrt{222-168cos(23^\circ)}\\ \\a\approx8.2$

Therefore, the length of line segment BC is about 8.2 units

8 0

2 years ago