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777dan777 [17]
3 years ago
14

Find the temperature in Celsius if the temperature is-31°F,using the formula F=9/5C+32

Mathematics
2 answers:
Nady [450]3 years ago
8 0
F=9/5c+32
-31=9/5c+32
-63=9/5c
-35=c
Viefleur [7K]3 years ago
6 0
Plug the F temperature in for F in the eqUation and solve for C
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A new cellular phone tower services all phones within a 20 mile radius. Doreen lives 16 miles east and 12 miles south of the tow
Lunna [17]

Answer: Yes.

Step-by-step explanation:

Hi, since the situation forms a right triangle (see attachment) we have to apply the Pythagorean Theorem:

c^2 = a^2 + b^2

Where c is the hypotenuse of the triangle (in this case the distance between Doreen’s house and the tower) and a and b are the other sides.

Replacing with the values given:

c^2 = 16^2 + 12^2

c^2 = 256+144

c^2 = 400

c = √400

c = 20 miles

So, she is within the area serviced by the tower (20 miles)

Feel free to ask for more if needed or if you did not understand something.

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4 years ago
Help plz!!!!!!!! Which expression is equivalent to the expression below?
Goshia [24]

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8 0
3 years ago
You’ve got a garden that is 10 ft by 12 ft.
ch4aika [34]
That’s a pretty average sized garden
8 0
3 years ago
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ss7ja [257]

Answer:

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7 0
4 years ago
Suppose that the given line has a slope of -2/3 and a y-intercept of (0,5/3). which of the following points is also a solution t
otez555 [7]

Answer:

From five given points it is find that points (1 , 1) and (4 , -1) will apply

Step-by-step explanation:

According to question,

Slop of line (m) = \frac{-2}{3}

And y-intercept is ( 0, \frac{5}{3})

So from above slope and points, the equation of line can be written as

y = mx + c

i.e \frac{5}{3} = \frac{-2}{3} x + c

    \frac{5}{3} = \frac{-2}{3} (0) + c

    \frac{5}{3} = 0 + c

Or, c = \frac{5}{3}

A) With points ( 5, \frac{5}{3} )

   At x = 5, y =\frac{-2}{3} (5) + \frac{5}{3}

            or, y = \frac{-10}{3} + \frac{5}{3}

            so, y = \frac{-5}{3}

Hence this points do not apply

B) With points ( 1 , 1 )

   At x = 1, y =  \frac{-2}{3} (1) +  \frac{5}{3}

      or,      y =  \frac{-2+5}{3}

      So,     y =  \frac{3}{3}

                y = 1

Hence this points will apply

C) With points ( 4 , -1 )

    At x = 4 , y =  \frac{-2}{3} (4) +  \frac{5}{3}

                    y =  \frac{-8+5}{3}

                    y =  \frac{-3}{3}

             So,  y= -1

Hence this points will apply

D) With points (-3 ,7)

    At x = - 3, y = \frac{-2}{3} (-3) +  \frac{5}{3}

                     y = \frac{14}{3}

 Hence this point will not apply

E) with points (0 , 0)

    At x 0,     y =  \frac{-2}{3} (0) +  \frac{5}{3}

               Or, y = 0 +  \frac{5}{3}

                     y =  \frac{5}{3}

Hence this point will not apply

∴ From above five given points it is find that points (1 , 1) and (4 , -1) apply

Answer

7 0
3 years ago
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