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3 years ago

5

30 multiplied by what number would give me the letter (y), 30 multiplicado por que numero me daria la letra (y),

1 answer:

LiRa [457]3 years ago

5 0

Thats like geometry
and are you good in this suject?

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Temperature in degrees Fahrenheit is equal to 32 more than 9/5 times the temperature in degrees Celsius . One day the high tempe

bazaltina [42]

The high temperature is 35 degree celsius

<em><u>Solution:</u></em>

Given that, Temperature in degrees Fahrenheit is equal to 32 more than 9/5 times the temperature in degrees Celsius

Let "f" be the temperature in degree Fahrenheit

Let "c" be the temperature in degree celsius

Therefore,

Temperature in degrees Fahrenheit = 32 + $\frac{9}{5}$ times the temperature in degrees Celsius

$f = 32 + \frac{9}{5} \times c$

One day the high temperature in Edison. Nj was 95 degrees Fahrenheit

f = 95 ; c = ?

Substitute f = 95 in above equation

$95 = 32 + \frac{9c}{5}\\\\95=\frac{32 \times 5 +9c}{5}\\\\95 \times 5 = 160+9c\\\\475 = 160 + 9c\\\\9c = 475 - 160\\\\9c = 315\\\\c = 35$

Thus high temperature in degrees Celsius is 35 degree celsius

5 0

3 years ago

Needhelpslovingthisproblem

Mnenie [13.5K]

Answer:

-4 to -1.5

Step-by-step explanation:

When the graph start to slope downwards or go down its decreasing. So when the hump starts to go down at (-4,0) it then starts to go up again after (-1.5,0). Hope this helps:)

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3 years ago

What does this mean? (Picture)

lianna [129]

9 times 43 is 387 split 9 and 43 and put It in the area model and then add them up

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3 years ago

Find the area of the circle... Round to the nearest tenth and include units in your answer

Fantom [35]

Answer:

452.4 cm

Step-by-step explanation:

the formula for the area of a circle is A = pi r^2

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

<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