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2 years ago

What is the range of this relation?

Mathematics

1 answer:

gayaneshka [121]2 years ago

6 0

Answer:

{- 1, 0, 4 }

Step-by-step explanation:

the range is the y- coordinates of the plotted points

the coordinates of the points are

(- 4, - 1 ) , (1, 0 ) , (1, 4 ) with y- coordinates - 1, 0, 4

then range is { - 1, 0, 4 }

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What is the value of x ? round your answer to the nearest tenth .

djverab [1.8K]

Answer:

39 maybe i don't know

Step-by-step explanation:

4 0

3 years ago

Suppose you have a light bulb that emits 50 watts of visible light. (Note: This is not the case for a standard 50-watt light bul

natali 33 [55]

Answer:

0.049168726 light-years

Step-by-step explanation:

The apparent brightness of a star is

$\bf B=\displaystyle\frac{L}{4\pi d^2}$

where

L = luminosity of the star (related to the Sun)

d = distance in ly (light-years)

The luminosity of Alpha Centauri A is 1.519 and its distance is 4.37 ly.

Hence the apparent brightness of Alpha Centauri A is

$\bf B=\displaystyle\frac{1.519}{4\pi (4.37)^2}=0.006329728$

According to the inverse square law for light intensity

$\bf \displaystyle\frac{I_1}{I_2}=\displaystyle\frac{d_2^2}{d_1^2}$

where

$\bf I_1=$ light intensity at distance $\bf d_1$

$\bf I_2=$ light intensity at distance $\bf d_2$

Let $\bf d_2$ be the distance we would have to place the 50-watt bulb, then replacing in the formula

$\bf \displaystyle\frac{0.006329728}{50}=\displaystyle\frac{d_2^2}{(4.37)^2}\Rightarrow\\\\\Rightarrow d_2^2=\displaystyle\frac{0.006329728*(4.37)^2}{50}\Rightarrow d_2^2=0.002417564\Rightarrow\\\\\Rightarrow d_2=\sqrt{0.002417564}=\boxed{0.049168726\;ly}$

Remark: It is worth noticing that Alpha Centauri A, though is the nearest star to the Sun, is not visible to the naked eye.

7 0

3 years ago

Steven wrote this equation. 4×18=418 Explain the error in Steven's reasoning. Find the correct product. Enter your explanation a

Thepotemich [5.8K]

Given:

Steven wrote this equation:

$4\times 18=418$

To find:

The error in Steven's reasoning and then find the correct product.

Solution:

We have,

$4\times 18=418$

This statement is incorrect because we cannot write product directly as Steven wrote.

We need to multiply the place values of each digit of first number with the place values of second number and then we need to add the resulted values.

$4\times 18=4\times (10+8)$