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2 years ago

what are the coordinates of the point (-10, -3) after a 180 degree clockwise rotation about the origin?

2 answers:

6 0

The coordinates of the point (-10, -3) after a 180 degree clockwise rotation about the origin will be (10, 3).

As, we have a point which is in negative side after we turn it clockwise direction it will move to positive side so the coordinates will become positive.

lisabon 2012 [21]2 years ago

4 0

The coordinates would be (10, 3)

If you plot these points and rotated 180 degrees, you get the same axes only all positives become negatives and vice versa.

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Add .003, 265.8, 83.04 and 1972

liq [111]

Answer: 2.320.843

Explanation: Calculator

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3 years ago

16. What is the area of the shaded region in terms of Tr? [20 points]

tangare [24]

Area of the circle outside
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=25 π

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2 years ago

Need serious help!!!!!!!!!!!!!! Ryan changed 90 euros (€) into dollars ($) when the exchange rate was €1 = $1.10. He spent $63.

love history [14]

€1 = $1.10
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2 years ago

How do i work this out

maxonik [38]

so that graph is mostly showing you which ones are greater so the first one woulld me 0.9 then 5.2 and just keep on going so here is the answer.

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4 0

3 years ago

The table shows a proportional relationship between the mass, in kilograms

OlgaM077 [116]

Answer:

C) Dog: 40 kg, Medicine: 1.6 mL

D) Dog: 35 kg, Medicine: 1.4 mL

Step-by-step explanation:

we know that

A relationship between two variables, x, and y, represent a proportional variation if it can be expressed in the form $k=\frac{y}{x}$ or $y=kx$

Let

x ---> the mass, in kilograms of a dog

y ---> milliliters of flea medicine a veterinarian prescribes

Find the value of the constant of proportionality k for the given data

$k=\frac{y}{x}$

For x=50 kg, y=2 mL ----> $k=\frac{2}{50}=0.04\ mL/kg$

For x=10 kg, y=0.4 mL ----> $k=\frac{0.4}{10}=0.04\ mL/kg$

For x=5 kg, y=0.2 mL ----> $k=\frac{0.2}{5}=0.04\ mL/kg$

The linear direct equation is equal to

$y=0.04x$

Verify each choice

A) Dog: 15 kg, Medicine: 0.9 mL

Divide the milliliters of medicine by the kilograms of the dog and compare the result with the constant of proportionality

$\frac{0.9}{15}= 0.06\ mL/kg$

$0.06 \neq 0.04$

therefore

These numbers could not be used as the missing values in the table

B) Dog: 25 kg, Medicine: 0.8 mL

Divide the milliliters of medicine by the kilograms of the dog and compare the result with the constant of proportionality

$\frac{0.8}{25}= 0.032\ mL/kg$

$0.032 \neq 0.04$

therefore

These numbers could not be used as the missing values in the table

C) Dog: 40 kg, Medicine: 1.6 mL

Divide the milliliters of medicine by the kilograms of the dog and compare the result with the constant of proportionality

$\frac{1.6}{40}= 0.04\ mL/kg$

$0.04 = 0.04$

therefore

These numbers could be used as the missing values in the table

D) Dog: 35 kg, Medicine: 1.4 mL

Divide the milliliters of medicine by the kilograms of the dog and compare the result with the constant of proportionality

$\frac{1.4}{35}= 0.04\ mL/kg$

$0.04 \neq 0.04$

therefore

These numbers could be used as the missing values in the table

E) Dog: 20 kg, Medicine: 1.2 mL

Divide the milliliters of medicine by the kilograms of the dog and compare the result with the constant of proportionality

$\frac{1.2}{20}= 0.06\ mL/kg$

$0.06 \neq 0.04$

therefore

These numbers could not be used as the missing values in the table

5 0

2 years ago