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3 years ago

PLEASE HELP I’M REALLY BAD WITH ANGLES AND ITS A MINI-QUIZ.

Mathematics

1 answer:

vampirchik [111]3 years ago

4 0

Answer:

1. 16°

2. 16°

3. 46°

4. 59°

Step-by-step explanation:

1) angles in a right angle add up to 90°

in that case x would be 90°-(44+30)=

x=16°

2)same explanation as in one

hence x=90-74

x=16°

3) angles on a straight line add up to 180°

Therefore x would be 180°-134°

x=46°

4)59° and x are vertical angles and vertical angles are equal.

Therefore x= 59°

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At 2 PM, a thermometer reading 80°F is taken outside where the air temperature is 20°F. At 2:03 PM, the temperature reading yiel

Alexeev081 [22]

Use Law of Cooling:
$T(t) = (T_0 - T_A)e^{-kt} +T_A$
T0 = initial temperature, TA = ambient or final temperature

First solve for k using given info, T(3) = 42
$42 = (80-20)e^{-3k} +20 \\ \\ e^{-3k} = \frac{22}{60}=\frac{11}{30} \\ \\ k = -\frac{1}{3} ln (\frac{11}{30})$

Substituting k back into cooling equation gives:
$T(t) = 60(\frac{11}{30})^{t/3} + 20$

At some time "t", it is brought back inside at temperature "x".
We know that temperature goes back up to 71 at 2:10 so the time it is inside is 10-t, where t is time that it had been outside.
The new cooling equation for when its back inside is:
$T(t) = (x-80)(\frac{11}{30})^{t/3} + 80 \\ \\ 71 = (x-80)(\frac{11}{30})^{\frac{10-t}{3}} + 80$
Solve for x:
$x = -9(\frac{11}{30})^{\frac{t-10}{3}} + 80$
Sub back into original cooling equation, x = T(t)
$-9(\frac{11}{30})^{\frac{t-10}{3}} + 80 = 60 (\frac{11}{30})^{t/3} +20$
Solve for t:
$60 (\frac{11}{30})^{t/3} +9(\frac{11}{30})^{\frac{-10}{3}}(\frac{11}{30})^{t/3} = 60 \\ \\ (\frac{11}{30})^{t/3} = \frac{60}{60+9(\frac{11}{30})^{\frac{-10}{3}}} \\ \\ \\ t = 3(\frac{ln(\frac{60}{60+9(\frac{11}{30})^{\frac{-10}{3}}})}{ln (\frac{11}{30})}) \\ \\ \\ t = 4.959$

This means the exact time it was brought indoors was about 2.5 seconds before 2:05 PM

8 0

4 years ago

70 POINTS!! AND BRAINLIEST HELPPPPP!!!

DerKrebs [107]

Answer:

3000 girls
38 fishes
37 points
115 marbles
19 gray coloured cars

Step-by-step explanation:

Let the unknown values be x

30 class mates = 1 class room

18 girls in 1 classroom

18 girls in 30 classmates

x girls in 5000 class mates

$18\:girls= 30\:students\\x\:girls = 5000\:students\\\\Cross\:Multiply\\\\30x = 90000\\\\Divide\:both\:sides\:by\:30\\\frac{30x}{30} = \frac{90000}{30}\\ \\x = 3000\\= 3000 girls$

$2.5\:miles = 12 \:fishes\\8\:miles = x\:fishes\\\\Cross\:Multiply\\2.5x = 96\\\\Divide\:both\:sides\:of\:the\:equation\:by\:2.5\\\\\frac{2.5x}{2.5}=\frac{96}{2.5}\\ \\x = 38.4\\Approximately \: =38\:fishes$

$8\:minutes = 14\: points\\21 \:minutes = x \:points\\\\Cross\:Multiply\\8x = 294\\\\\mathrm{Divide\:both\:sides\:by\:}8\\\frac{8x}{8}=\frac{294}{8}\\\\Simplify\\x=\frac{147}{4}\\\\x = 36.75\\\\Approximately\:= 37\:points$

$32\:marbles = 5\:inches\\x\:marbles = 18\:inches\\\\Cross\:Multiply\\5x = 576\\\\\mathrm{Divide\:both\:sides\:by\:}5\\\frac{5x}{5}=\frac{576}{5}\\\\x=\frac{576}{5}\\\\x =115.2\\= 115=\:marbles$

$2 \: gray\:colour= 15\:cars\\x\:gray= 140\:cars\\\\Cross\:Multiply\\\\15x = 280\\\\\mathrm{Divide\:both\:sides\:by\:}15\\\frac{15x}{15}=\frac{280}{15}\\\\Simplify\\x=\frac{56}{3}\\\\x =18.666\\Approximately ;\\19$