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Sophie [7]
3 years ago
14

Solve for x please and thank you

Mathematics
2 answers:
mariarad [96]3 years ago
3 0

Answer:X=3

Step-by-step explanation:

Troyanec [42]3 years ago
3 0

Answer:

x=3

Step-by-step explanation:

Hi there!

We are given ΔRTS, m<T=25x, m<S=57+x (both inside triangle RTS), and m<TRQ (outside of ΔRTS)=45x

We want to solve for x.

We can use the exterior angle theorem, which states that a measure of an exterior angle (an angle outside of a triangle) is equal to the sum of two remote interior angles (adding the values of two different angles that are inside the triangle gives us the value of the angle outside of a triangle).

Based on the theorem, this is true:

m<T+ m<S= m<RTQ

We can plug in values we know into the equation to get:

25x+57+x=45x

Now combine like terms

26x+57=45x

Subtract 26 from both sides

57=19x

Divide both sides by 19

3=x

Hope this helps!

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What is the volume of the extra large tube
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If by base of 10, you mean that the diameter of the base is 10, then the radius of the base is 5.

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A pizza pan is removed at 9:00 PM from an oven whose temperature is fixed at 450°F into a room that is a constant 70°F. After 5​
gladu [14]

Answer:

A) It will get to a temperature of 125°F at 9:19 PM

B) It will get to a temperature of 150°F at 9:16 PM

C) as time passes temperature approaches the initial temperature of 450°F

Step-by-step explanation:

We are given;

Initial temperature; T_i = 450°F

Room temperature; T_r = 70°F

From Newton's law of cooling, temperature after time (t) is given as;

T(t) = T_r + (T_i - T_r)e^(-kt)

Where k is cooling rate and t is time after the initial temperature.

Now, we are told that After 5​ minutes, the temperature is 300°F.

Thus;

300 = 70 + (450 - 70)e^(-5k)

300 - 70 = 380e^(-5k)

230/380 = e^(-5k)

e^(-5k) = 0.6053

-5k = In 0.6053

-5k = -0.502

k = 0.502/5

k = 0.1004 /min

A) Thus, at temperature of 125°F, we can find the time from;

125 = 70 + (450 - 70)e^(-0.1004t)

125 - 70 = 380e^(-0.1004t)

55/380 = e^(-0.1004t)

In (55/380) = -0.1004t

-0.1004t = -1.9328

t = 1.9328/0.1018

t ≈ 19 minutes

Thus, it will get to a temperature of 125°F at 9:19 PM

B) Thus, at temperature of 150°F, we can find the time from;

150 = 70 + (450 - 70)e^(-0.1004t)

150 - 70 = 380e^(-0.1004t)

80/380 = e^(-0.1004t)

In (80/380) = -0.1004t

-0.1004t = -1.5581

t = 1.5581/0.1004

t ≈ 16 minutes.

Thus, it will get to a temperature of 150°F at 9:16 PM

C) As time passes which means as it approaches to infinity, it means that e^(-kt) gets to 1.

Thus,we have;

T(t) = T_r + (T_i - T_r)

T_r will cancel out to give;

T(t) = T_i

Thus, as time passes temperature approaches the initial temperature of 450°F

6 0
3 years ago
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