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Anarel [89]
3 years ago
14

What is the answer to number 8.

Mathematics
1 answer:
Viktor [21]3 years ago
4 0

Its 20. |-2|+4-(-2)×(7). This simplifies to 2+4+(2×7)

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A volcano erupted
IgorC [24]
Percentage decrease = d
h1 = 2950
h2 = 2550

d/100 = (h2-h1)/h1

d/100 = (2950-2550)/2950
d=13.56
3 0
3 years ago
If <img src="https://tex.z-dn.net/?f=tan%20%28x%29%20%3D%20%5Cfrac%7B5%7D%7B12%7D" id="TexFormula1" title="tan (x) = \frac{5}{12
Alekssandra [29.7K]

Explanation:

First, we need to find the values of the sine and cosine of x knowing the value of tan x and x being in the 3rd quadrant. Since tan x = 5/12, using Pythagorean theorem, we know that

\sin x = -\frac{5}{13}\;\;\text{and}\;\;\cos x = -\frac{12}{13}

Note that both sine and cosine are negative because x is in the 3rd quadrant.

Recall the addition identities listed below:

\sin(\alpha + \beta) = \sin\alpha\sin\beta + \cos\alpha\cos\beta

\Rightarrow \sin(180+x) = \sin180\sin x + \cos180\cos x

\;\;\;\;\;\;= -\sin x = \dfrac{5}{13}

\cos(\alpha - \beta) = \cos\alpha \cos\beta + \sin\alpha \sin\beta

\Rightarrow \cos(180 - x) = \cos180\cos x + \sin180\sin x

\;\;\;\;\;\;=-\cos x = \dfrac{12}{13}

\tan(\alpha - \beta) = \dfrac{\tan\alpha - \tan\beta}{1 + \tan\alpha\tan\beta}

\Rightarrow \tan(360 - x) = \dfrac{\tan 360 - \tan x}{1 + \tan 360 \tan x}

\;\;\;\;\;\;= -\tan x = -\dfrac{5}{12}

Therefore, the expression reduces to

\sin(180+x) + \tan(360-x) + \frac{1}{\cos(180-x)}

\;\;\;\;\;= \left(\dfrac{5}{13}\right) + \left(\dfrac{5}{12}\right) + \dfrac{1}{\left(\frac{12}{13}\right)}

\;\;\;\;\;= \dfrac{49}{26}

5 0
3 years ago
Would I need Circumference or Area to calculate the distance around a circular track?
Hatshy [7]

Answer:

The circumference because it's not about the interior space of the circular track but the track itself.

Hope this helped!

4 0
3 years ago
Read 2 more answers
Hey can you please help me posted picture of question
GREYUIT [131]
To solve this problem you must apply the proccedure shown below:

 1. You have the following expression:

 (3+3i)-(13+15i)

 2. If you want to substract both terms, you  need to substract the real numbers and the complex numbers. Then, you obtain:

 3+3i-13-15i
 (3-13)+(3i-15i)

 3. Then, you obtain the following result:

 -10-12i

 4. Therefore, as you can see, the correct answer is the last option (option D), which is:
 D. -10-12i
4 0
4 years ago
Calculate the total heat required to completely turn 20g of ice at -20°c to steam​
PtichkaEL [24]

Answer:

The total heat required to completely turn 20 g of ice at -20°C to steam, is 65880 J

Step-by-step explanation:

The given parameters are;

The mass of the ice = 20 g

The initial temperature of the ice = -20°C

The specific heat capacity of the ice, ΔH(fusion) = 2.03 J/(g·°C)

The latent heat of fusion of ice = 334 J/g

The specific heat capacity of the water = 4.184 J/(g·°C)

The latent heat of evaporation of water = 2501 J/g

The heat required to raise the temperature of the ice to from -20°C to 0°C, Q₁, is given as follows;

Q₁ = 20 × 2.03 × (0 - (-20)) = 812

Q₁ = 812 J

The latent heat required to melt the 20 g of ice at 0°C is given as follows;

Q₂ = ΔH(fusion) × mass, m = 334 × 20 = 6680

Q₂ = 6680 J

The heat, Q₃, required to raise the temperature of the ice to 100°C (boiling point temperature) is given as follows;

Q₃ = 20 × 4.184 × (100 - 0) = 8368

Q₃ = 8368 J

The heat, Q₄, required to evaporate the 20 g of water at 100°C is given as follows

Q₄ = 2501 × 20 = 50020

Q₄ =  50020 J

The total heat, Q_{total} required to completely turn 20 g of ice at -20°C to steam, is given as follows;

Q_{total} = Q₁ + Q₂ + Q₃ + Q₄ =  812 + 6680 + 8368 + 50020 = 65880

Q_{total} = 65880 J.

6 0
3 years ago
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