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3 years ago

PLEASE HELP!!! Find the component form of the vector that translates P(4,5) to P'(-1,-6)

Mathematics

1 answer:

lisov135 [29]3 years ago

3 0

Answer:

The component form will be:

(x, y) = (-5, -11)

Step-by-step explanation:

We need to find the component form of the vector that translates P(4,5) to P'(-1,-6).

It can be written as:

P(4, 5) → P'(4+x, -6+y) = P'(-1,-6)

Calculating 'x'

4 + x = -1

x = -1 - 4

x = -5

Calculatng 'y'

5 + y = -6

y = -6 - 5

y = -11

Thus, the component form will be:

(x, y) = (-5, -11)

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(4x^(4))^(2) without exponents

vlabodo [156]

Answer:

256x * 16

Step-by-step explanation:

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16x * 16x = 256x

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256x * 16

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3 years ago

I need to k ow how to do this

Ainat [17]

You just need to trace it without moving your pencil off the paper

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3 years ago

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Write each rational number as a terminating decimal. 1. 19/20 2. -1/8 3. 17/5

Dima020 [189]

19/20= 0.95
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4 years ago

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The current population of a town is 10,000. If the population, P, increases by 20% each year, which equation could be used to fi

frozen [14]

Answer:

The equation would be:

p = 10,000 * (1 + 0.2)∧ t

Step-by-step explanation:

1. Let's review the information provided to us to answer the question correctly:

Current population of a town = 10,000 inhabitants

Growth rate of the population = 20% annually

2. Which equation could be used to find the population after T years?

p = Population of the town after t years

t = Number of years

We will use the following equation:

p = 10,000 * (1 + 0.2)∧ t

For example, let's calculate the population after 6 years, replacing with the real values, we have:

p = 10,000 * (1 + 0.2)⁶

p = 10,000 * 2.986

p = 29,860 inhabitants

7 0

3 years ago

Suppose that prices of recently sold homes in one neighborhood have a mean of $265,000 with a standard deviation of $9300. Using

Hatshy [7]

Answer:

Range = (237100, 292900)

Step-by-step explanation:

Using Chebyshevs Inequality:

$P(|X - \mu | \le k \sigma )\ge 1 -\dfrac{1}{k^2}= 0.889$

$1 -\dfrac{1}{k^2}= 0.889$

$\dfrac{1}{k^2}= 1- 0.889$

$\dfrac{1}{k^2}=0.111$

$k = \sqrt{\dfrac{1}{0.111}}$

$k \simeq 3$

Thus, 88.9% of the population is within 3 standard deviation of the mean with the Range = μ ± kσ

where;

μ = 265000

σ = 9300

Range = 265000 ± 3(9300)

Range = 265000 ± 27900

Range = (265000 - 27900, 265000 + 27900)

Range = (237100, 292900)

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3 years ago