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2 years ago

NEED HELP PLEASE SOMEONE!!!!

Mathematics

1 answer:

weqwewe [10]2 years ago

4 0

Let me work this out really quick and I'll get back to you

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The difference of twice a number and 6 is less than - 18.

oee [108]

Answer:

2x-6<-18

Step-by-step explanation:

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2 years ago

A recipe called for 5/9 cup of chopped lettuce and 5/6 cup of diced lettuce. In total, how many cups of lettuce did the recipe c

77julia77 [94]

Answer:

1 7/18

Step-by-step explanation:

5/9 + 5/6 write all numerators above the least common denominator which is 18.

10+15/18 then you add the numerators which will make it 25/18 which alternative form will be 1 7/18, or in decimal 1.38

3 0

2 years ago

1. Rudy paid for 5 trees to be delivered.

sertanlavr [38]

Here you go, hope it helps!

8 0

2 years ago

3. Using the net, calculate and label the area of each face of the prism.

AysviL [449]

Answer:

Finding the areas of each of the rectangles and squares of the net of a rectangular prism and adding up those areas gives the surface area or total surface area of the prism. For example, if the length of one side of the cube 4 units then the area of one its face is 4 × 4 = 16 square units.

Step-by-step explanation:

3 0

2 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

2 years ago