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Pavel [41]
3 years ago
10

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:
melamori03 [73]3 years ago
6 0
Use slope formula to find slope <span><span><span>−9−5</span><span>−/12−8</span></span>=<span><span>−14/</span><span>−20</span></span>=<span>7/10</span></span> 7/10 = m = slope  
y= mx + b
y= (7/10)x + b
(8,5)
(5) = (7/10)(8) + b
b = -3/5

So.... <span>y=(<span>7/10)</span>x −<span>35

</span></span> multiply by 10

<span>10y=7x−6</span><span> --> -7x + 10y = -6
 --> 7x - 10y = 6
 So, if 7x - 10y = 3 a good model?</span>
miskamm [114]3 years ago
6 0

Answer:

If the model is good, then both points will check in the equation. Substituting 8 for x and 5 for y 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 y-intercept should be –3/5. Hope this helps!!! :)

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The daily mean temperature in a particular place is 83. how many cooling-degree days were accumulated?
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Degree days are the difference between the daily temperature mean, (high temperature plus low temperature divided by two) and 65°F. 65°F because we assume that at this temperature we do not need cooling or heating. 
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4 0
3 years ago
Need help with this one.
ella [17]

Answer:

The answer should be 55

Step-by-step explanation:

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Which second-degree polynomial function has a leading coefficient of –1 and root 5 with multiplicity 2?
PIT_PIT [208]
A second degree polynomial function has the general form:

                  \displaystyle{f(x)=ax^2+bx+c, where a\neq0.

The leading coefficient is a, so we have a=-1.

5 is a double root means that :

i) f(5)=0,
ii) the discriminant D is 0, where D=b^2-4ac.

Substituting x=5, we have

                                        f(5)=a(5)^2+b(5)+c,

and since f(5)=0, and a is -1 we have:

                                          0=-25+5b+c
thus c=25-5b.


By ii) \displaystyle{b^2-4ac=0.

Substituting a with -1 and c with 25-5b we have:
                                     
          \displaystyle{b^2-4ac=0
          \displaystyle{b^2-4(-1)(25-5b)=0
          \displaystyle{b^2+4(25-5b)=0
          \displaystyle{b^2-20b+100=0
          \displaystyle{(b-10)^2=0
          \displaystyle{b=10 


Finally we find c: c=25-5b=25-50=-25

Thus the function is        \displaystyle{f(x)=-x^2+10x-25


Remark: It is also possible to solve the problem by considering the form

f(x)=-1(x-5)^2 directly.

In general, if a quadratic function has leading coefficient a, and has a root r of multiplicity 2, then its form is f(x)=a(x-r)^2
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2 years ago
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nalin [4]

Answer:

c. 45d+.25m

Step-by-step explanation:

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