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mote1985 [20]
3 years ago
8

Devon wants to write an equation for a line that passes through 2 of the data points he has collected. The points are (8, 5) and

(–12, –9). He writes the equation 7x – 10y = 3. Is this a good model? Explain your reasoning.
Mathematics
2 answers:
iren2701 [21]3 years ago
7 0
<span> If the model is good, then both points will check in the equation. Substituting 8 for </span>x <span>and 5 for </span>y<span> in the equation results in 56 – 50 = 3, which is not true. Therefore, the model is not good. Using (–12, –9) as a check results in –84 + 90 = 3. The constant value in the equation should be 6, not 3. In slope-intercept form, the </span>y<span>-intercept should be –3/5.</span>
Nuetrik [128]3 years ago
6 0

Answer:

The points do not make the equation true.

The correct equation in standard form is

7x – 10y = 6.

The y-intercept should be –3/5.

Step-by-step explanation:

sample response

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<img src="https://tex.z-dn.net/?f=cos%20%5B%20arctan%28%5Cfrac%7B12%7D%7B5%7D%29%20-%20arcsin%20%28%5Cfrac%7B-3%7D%7B5%7D%29%5D"
qwelly [4]

Answer: -16/65

Step-by-step explanation:

Drawing the right triangle (as attached) gives us that \arctan \left(\frac{12}{5} \right)=\arcsin \left(\frac{12}{13} \right)

Also, -\arcsin \left(-\frac{3}{5} \right)=\arcsin \left(\frac{3}{5} \right)

This means our original expression is equal to:

\cos \left[\arcsin \left(\frac{12}{13} \right)+\arcsin \left(\frac{3}{5} \right) \right]

Using the cosine addition formula, which states \cos(a+b)=\cos a \cos b-\sin a \sin b, we get this itself is equal to:

\cos \left(\arcsin \left(\frac{12}{13} \right) \right)\cos \left(\arcsin \left(\frac{3}{5} \right)\right)-\sin \left(\arcsin \left(\frac{12}{13} \right) \right)\sin \left(\arcsin \left(\frac{3}{5} \right)\right)

Since \sin^{2} \theta+\cos^{2} \theta=1, we know that:

\sin^{2} \left(\arcsin \left(\frac{12}{13} \right)\right)+\cos^{2} \left(\arcsin \left(\frac{12}{13} \right)\right)=1\\\\\frac{144}{169} +\cos^{2} \left(\arcsin \left(\frac{12}{13} \right)\right)=1\\\\cos^{2} \left(\arcsin \left(\frac{12}{13} \right)\right)=\frac{25}{169}\\\\cos \left(\arcsin \left(\frac{12}{13} \right)\right)=\frac{5}{13}

Similarly, cos(arcsin(3/5))=4/5.

This means the given expression is equal to:

\left(\frac{5}{13} \right) \left(\frac{4}{5} \right)-\left(\frac{12}{13} \right) \left(\frac{3}{5} \right)\\\\\frac{20}{65}-\frac{36}{65}=\boxed{-\frac{16}{65}}

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2 years ago
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Elenna [48]

Answer:

False.

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4 0
3 years ago
A chef is making half of the frosting recipe. How much cream of tartar is needed? How much granulated sugar is needed?
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Stan is purchasing sub-flooring for a kitchen he is remodeling. The area of the floor is 180 ft' and
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Answer:

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Step-by-step explanation:

The area of a rectangle is:

 

Area = Length x Width

 

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Area = Length x Width

180 = Length x 12

 

Divide both sides of the equation by 12:

 

180/12 = Length

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