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3 years ago

What does it mean when we say that a formula models real-world phenomena?

1 answer:

7 0

A formula models real-world phenomena when it describes the relationship between the variables of a real life situation. We use formulas in our every day life, but maybe are not aware of it. Some examples of using math formulas in the real world are: - the most obvious example is that we use formulas in the grocery store (multiplication, estimation, percentages,...

- we use formulas while baking (measuring ingredients, understanding ratios and proportions,<span> converting metrics,...)</span>

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There was a country concert held at the park. For every 3 men there were 5 women that went to the concert. If 30 more women atte

8090 [49]

Answer:

3m == 5w, that means for such unit, there is a 2 person difference

30 person difference = 30/2 = 15units

and again one unit means 3m and 5w

so 15*3men, and 15*5women

and you can check the difference is 30

8 0

3 years ago

Which of the following are solutions to the equation below? Check all that apply. 15x2 - 44x + 32 = 0

krek1111 [17]

If you want to find the solutions for this you have to factor it. Since it's a second degree polynomial, you'll have 2 solutions. Factoring this using the quadratic formula, you'll get factors of (5x-8)(3x-4). Solving these for x you get x = 8/5 and x = 4/3.

6 0

3 years ago

Please help me for the love of God if i fail I have to repeat the class

Elena-2011 [213]

$\theta$ is in quadrant I, so $\cos\theta>0$ .

$x$ is in quadrant II, so $\sin x>0$ .

Recall that for any angle $\alpha$ ,

$\sin^2\alpha+\cos^2\alpha=1$

Then with the conditions determined above, we get

$\cos\theta=\sqrt{1-\left(\dfrac45\right)^2}=\dfrac35$

and

$\sin x=\sqrt{1-\left(-\dfrac5{13}\right)^2}=\dfrac{12}{13}$

Now recall the compound angle formulas:

$\sin(\alpha\pm\beta)=\sin\alpha\cos\beta\pm\cos\alpha\sin\beta$

$\cos(\alpha\pm\beta)=\cos\alpha\cos\beta\mp\sin\alpha\sin\beta$

$\sin2\alpha=2\sin\alpha\cos\alpha$

$\cos2\alpha=\cos^2\alpha-\sin^2\alpha$

as well as the definition of tangent:

$\tan\alpha=\dfrac{\sin\alpha}{\cos\alpha}$

Then

1. $\sin(\theta+x)=\sin\theta\cos x+\cos\theta\sin x=\dfrac{16}{65}$

2. $\cos(\theta-x)=\cos\theta\cos x+\sin\theta\sin x=\dfrac{33}{65}$

3. $\tan(\theta+x)=\dfrac{\sin(\theta+x)}{\cos(\theta+x)}=-\dfrac{16}{63}$

4. $\sin2\theta=2\sin\theta\cos\theta=\dfrac{24}{25}$

5. $\cos2x=\cos^2x-\sin^2x=-\dfrac{119}{169}$

6. $\tan2\theta=\dfrac{\sin2\theta}{\cos2\theta}=-\dfrac{24}7$

7. A bit more work required here. Recall the half-angle identities:

$\cos^2\dfrac\alpha2=\dfrac{1+\cos\alpha}2$

$\sin^2\dfrac\alpha2=\dfrac{1-\cos\alpha}2$

$\implies\tan^2\dfrac\alpha2=\dfrac{1-\cos\alpha}{1+\cos\alpha}$

Because $x$ is in quadrant II, we know that $\dfrac x2$ is in quadrant I. Specifically, we know $\dfrac\pi2$ , so $\dfrac\pi4$ . In this quadrant, we have $\tan\dfrac x2>0$ , so