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2 years ago

The question is below:

Mathematics

2 answers:

Andru [333]2 years ago

8 0

Answer:

$x=19.5^o$

$\angle RQS=43^o$

Step-by-step explanation:

Notice that the addition of these two angles give you and angle of $180^o$ , therefore we can write the following equation to represent such addition:

$(2x+4)^o + (6x+20)^o=180^o\\2x+6x+4^o+20^o=180^o\\8\,x+24^o=180^o\\8\,x=180^o-24^o\\8\,x=156^o\\x=156^o/8\\x=19.5^o$

Therefore, the value of the angle RQS is:

$\angle RQS=(2\,x+4)^o=(2\,*\,19.5^o)+4^o=43^o$

Sonja [21]2 years ago

4 0

Answer:

x = 19.5, RQS=43

Step-by-step explanation:

It is important to note that RQS and TQS are supplementary, meaning their angles will add up to 180. Knowing this, we can create and solve the equation to find x..

(2x+4) + (6x+20) = 180

8x + 24 = 180

8x = 156

x = 19.5

Now that we know the value of x, we can substitute it into the equation for RQS, 2x+4.

2(19.5)+4

39+4

Hope this helped!

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oksian1 [2.3K]

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As you can notice, they all follow the same steps: dividing by the coefficient of the variable both sides, so you can the variable alone on the first side of the equation, and its value on the second side.

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3 years ago

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aksik [14]

Answer:

Step-by-step explanation:

The most reasonable one is B because its 7 for side B and the rest for B are either bigger or equal to the given number

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2 years ago

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A cooler has a temperature of 32 degrees Fahrenheit. A bottled drink is placed in the cooler with an initial temperature of

FinnZ [79.3K]

Answer:

$k \approx 0.44$

Step-by-step explanation:

Given function:

$f(t) = (ce)^{-kt}+32$

As per question statement:

Initial temperature of bottle is 70 $^\circ F$ .

i.e. when time = 0 minutes, f(t) = 70 $^\circ F$

$70 = ce^{-k\times 0}+32\\\Rightarrow 38 = ce^{0}\\\Rightarrow c = 38$

After t = 3, f(t) = 42 $^\circ F$

$42 = 38 \times e^{-k\times 3}+32\\\Rightarrow 42-32 = 38 \times e^{-3k} \\\Rightarrow 10 = 38 \times e^{-3k} \\\Rightarrow e^{3k} = \dfrac{38}{10}\\\Rightarrow e^{3k} = 3.8\\\\\text{Taking } log_e \text{both the sides:}\\\\\Rightarrow log_e {e^3k} = log_e {3.8}\\\Rightarrow 3k \times log_ee=log_e {3.8} (\because log_pq^r=r \times log_pq)\\\Rightarrow 3k \times 1=log_e {3.8}\\\Rightarrow 3k = 1.34\\\Rightarrow k = \dfrac{1.34}{3}\\\Rightarrow k \approx 0.44$

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3 years ago

10. Write an inequality relating the measure of angle 1 to the measure of angle 2

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2 years ago

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Answer:

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3 years ago

The question is below:​

The question is below: