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4 years ago

15

the temperature at 5:00 PM is 4 degrees. By midnight, the temperature had dropped 12 degrees. What was the temperature at midnig

ht?

2 answers:

77julia77 [94]4 years ago

8 0

Answer:

-8 degrees

Step-by-step explanation:

4-12 = -8

It dropped 12 degrees in 7 hours.

Scilla [17]4 years ago

3 0

Answer: -8°

Step-by-step explanation:

4 - 12

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If RS=8y+4 , ST=4y+8 , and RT=15y-9, find RS, ST, and RT<br> someone please help me

Talja [164]

$RS+ST=RT\\\\
RS=8y+4\\
RT=15y-9\\
ST=4y+8\\\\
8y+4+4y+8=15y-9\\
12y+12=15y-9\ \ \ |Subtract\\
12y=15y-9-12\\
12y=15y-21\ \ \ | Subtract\ 15y\\
12y-15y=-21\\
-3y=-21 \ \ |Divide\ by\ (-3)\\
y=7\\\\
RS=8y+4=8*7+4=56+4=60\\
RT=15y-9=15*7-9=105-9=96\\
ST=4y+8=4*7+8=28+8=36$

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4 years ago

Read 2 more answers

A man is in a boat 2 miles from the nearest point on the coast. He is to go to point Q, located 3 miles down the coast and 1 mil

mafiozo [28]

Answer:

The answer is "0.45385".

Step-by-step explanation:

The time is taken to reach the coast $= \frac{\sqrt{(2^2 + x^2)}}{4}$

The time is taken to reach Q after reaching the coast $= \frac{\sqrt{(1 + (3-x)^2)}}{4}$

Total time, $T= \frac{\sqrt{(1 + (3-x)^2)}}{4}+ \frac{\sqrt{(1 + (3-x)^2)}}{4}$

This has to be minimum,

$\to \frac{dt}{dx} = 0$

$\to \frac{\sqrt{(2^2 + x^2)}}{2} + \frac{\sqrt{1 + (3-x)^2}}{3}\\\\ \frac{x}{4}\sqrt{(4+x^2)} + \frac{(x-3)}{(4\sqrt{(x^2-6x+10)}} = 0\\\\\frac{x^2}{16} \times (4+x^2) = \frac{(x-3)^2}{16 \times (x^2-6x+10)}\\\\x^2(4+x^2) = \frac{(x-3)^2}{(x^2-6x+10)}\\\\4x^2+x^4 = \frac{ x^2+9-6x}{(x^2-6x+10)}\\\\ 4x^2+x^4(x^2-6x+10) = x^2+9-6x\\\\4x^4-24x^3+40x^2+x^6-6x^5+10x^4= x^2+9-6x\\\\x^6-6x^5+ 14x^4-24x^3+39x^2+6x-9=0\\\\$

x=0.45385

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