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nordsb [41]
3 years ago
11

The graph of a quadratic function contains (4,-64) and it's vertex is at the origin.write an equation for this function.

Mathematics
1 answer:
lesya692 [45]3 years ago
5 0

Answer:

Step-by-step explanation:

Not sure what form you need this in, but it really doesn't matter, as you'll see in the final equation.  I used the vertex form and solved for a:

y=a(x-h)^2+k

We are given the vertex (h, k) as the origin (0, 0), and we have a point that the graph goes through as (4, -64).  That's our x and y.  Plugging in what we have:

-64=a(4-0)^2+0 gives us

-64 = 16a and

a = -4.  That means that the quadratic equation is

y=-4x^2  which is both vertex form and standard form here, no difference.

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Katyanochek1 [597]

Write tan in terms of sin and cos.

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\displaystyle \lim_{x\to0}\frac{\sin(x)}x = 1

Rewrite and expand the given limand as the product

\displaystyle \lim_{t\to0}\frac{\sin(6t)}{\sin(2t)\cos(6t)} = \lim_{t\to0} \frac{\sin(6t)}{6t} \times \frac{2t}{\sin(2t)} \times \frac{6t}{2t\cos(6t)} \\\\ = \left(\lim_{t\to0} \frac{\sin(6t)}{6t}\right) \times \left(\lim_{t\to0}\frac{2t}{\sin(2t)}\right) \times \left(\lim_{t\to0}\frac{3}{\cos(6t)}\right)

Then using the known limit above, it follows that

\displaystyle \left(\lim_{t\to0} \frac{\sin(6t)}{6t}\right) \times \left(\lim_{t\to0}\frac{2t}{\sin(2t)}\right) \times \left(\lim_{t\to0}\frac{3}{\cos(6t)}\right) = 1 \times 1 \times \frac3{\cos(0)} = \boxed{3}

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Answer:

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Step-by-step explanation:

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2 years ago
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Answer:

Step-by-step explanation:

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2 years ago
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laila [671]

Answer:

The value of angle is 56.9°

Step-by-step explanation:

First, you have to find the length of hypotenuse using Pythagoras' Theorem, c² = a² + b² :

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let \: a = 8.25 \\ let \: b = 6.45

{c}^{2}  =  {8.25}^{2}  +  {6.45}^{2}

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4 0
2 years ago
A nationwide study of American homeowners revealed that 65% have one or more lawn mowers. A lawnequipment manufacturer, located
GenaCL600 [577]

Answer:

z=\frac{0.684 -0.65}{\sqrt{\frac{0.65(1-0.65)}{497}}}=1.589  

p_v =P(z>1.589)=0.056  

If we compare the p value obtained with the significance level assumed \alpha=0.05 we see that p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion of homes in Omaha with one or more lawn mowers is not ignificantly higher than 0.65

Step-by-step explanation:

Data given and notation

n=497 represent the random sample taken

X=340 represent the homes in Omaha with one or more lawn mowers

\hat p=\frac{340}{497}=0.684 estimated proportion of homes in Omaha with one or more lawn mowers

p_o=0.65 is the value that we want to test

\alpha represent the significance level

z would represent the statistic (variable of interest)

p_v represent the p value (variable of interest)  

Concepts and formulas to use  

We need to conduct a hypothesis in order to test the true proportion of homes in Omaha with one or more lawn mowers is higher than 0.65.:  

Null hypothesis:p\leq 0.65  

Alternative hypothesis:p > 0.65  

When we conduct a proportion test we need to use the z statistic, and the is given by:  

z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}} (1)  

The One-Sample Proportion Test is used to assess whether a population proportion \hat p is significantly different from a hypothesized value p_o.

Calculate the statistic  

Since we have all the info requires we can replace in formula (1) like this:  

z=\frac{0.684 -0.65}{\sqrt{\frac{0.65(1-0.65)}{497}}}=1.589  

Statistical decision  

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.  

The next step would be calculate the p value for this test.  

Since is a right tailed test the p value would be:  

p_v =P(z>1.589)=0.056  

If we compare the p value obtained with the significance level assumed \alpha=0.05 we see that p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion of homes in Omaha with one or more lawn mowers is not ignificantly higher than 0.65

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3 years ago
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