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3 years ago

Can someone help me please!!

Mathematics

2 answers:

zhuklara [117]3 years ago

8 0

Answer:

It is (-4,3)

Step-by-step explanation:

shutvik [7]3 years ago

4 0

The image has the points at: {(-3,-4), (−3,1), (-1, 1), (1, -1) (4,4)}

Domain is the X value and range is y the Y value.

Domain: {-3,-1,1,4}

Range {-4,1,-1,4}

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If T= 101+102+103+...+199 find T

umka2103 [35]

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3 years ago

What is the equation of the line that passes through (–2, 3) and is parallel to 2x + 3y = 6?

Zinaida [17]

2x + 3y = 6
3y = - 2x + 6
y = -2/3x + 2....slope here is -2/3
A parallel line will have the same slope

y = mx + b
slope(m) = - 2/3
(-2,3)...x = -2 and y = 3
now we sub, we r looking for b, the y intercept
3 = -2/3(-2) + b
3 = 4/3 + b
3 - 4/3 = b
9/3 - 4/3 = b
5/3 = b

ur equation is : y = -2/3x + 5/3 (slope intercept form)

y = -2/3x + 5/3
2/3x + y = 5/3
2x + 3y = 5 (standard form)

7 0

3 years ago

Read 2 more answers

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mel-nik [20]

Answer:

Down

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

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

<em>L = luminosity of the star (related to the Sun) </em>

<em>d = distance in ly (light-years) </em>

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

Is this a function or no?

IgorLugansk [536]

Answer:

Step-by-step explanation:

Positives cannot point at negatives I think.

5 0

2 years ago