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castortr0y [4]
3 years ago
10

Which of the following equations is written in vertex form?

Mathematics
1 answer:
ivanzaharov [21]3 years ago
3 0
The third one is correct.
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Geometry 100 pts plz help!!! screenshot attatched
nirvana33 [79]

Answer:

1. A

2. D

3. C

4. E

5. B

Step-by-step explanation:

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Please help me with the attachment
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Answer:

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Stephan added 26 new contacts to his phone list. He had to delete 16 old contacts before he could add any more. Stefan now has a
bezimeni [28]

Answer: 26-16= c

c=10

Step-by-step explanation:

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One of the roots of the quadratic equation dx^2+cx+p=0 is twice the other, find the relationship between d, c and p
scZoUnD [109]

Answer:

c^2 = 9dp

Step-by-step explanation:

Given

dx^2 + cx + p = 0

Let the roots be \alpha and \beta

So:

\alpha = 2\beta

Required

Determine the relationship between d, c and p

dx^2 + cx + p = 0

Divide through by d

\frac{dx^2}{d} + \frac{cx}{d} + \frac{p}{d} = 0

x^2 + \frac{c}{d}x + \frac{p}{d} = 0

A quadratic equation has the form:

x^2 - (\alpha + \beta)x + \alpha \beta = 0

So:

x^2 - (2\beta+ \beta)x + \beta*\beta = 0

x^2 - (3\beta)x + \beta^2 = 0

So, we have:

\frac{c}{d} = -3\beta -- (1)

and

\frac{p}{d} = \beta^2 -- (2)

Make \beta the subject in (1)

\frac{c}{d} = -3\beta

\beta = -\frac{c}{3d}

Substitute \beta = -\frac{c}{3d} in (2)

\frac{p}{d} = (-\frac{c}{3d})^2

\frac{p}{d} = \frac{c^2}{9d^2}

Multiply both sides by d

d * \frac{p}{d} = \frac{c^2}{9d^2}*d

p = \frac{c^2}{9d}

Cross Multiply

9dp = c^2

or

c^2 = 9dp

Hence, the relationship between d, c and p is: c^2 = 9dp

8 0
3 years ago
A random sample of 10 parking meters in a beach community showed the following incomes for a day. Assume the incomes are normall
Vlad1618 [11]

Answer: (2.54,6.86)

Step-by-step explanation:

Given : A random sample of 10 parking meters in a beach community showed the following incomes for a day.

We assume the incomes are normally distributed.

Mean income : \mu=\dfrac{\sum^{10}_{i=1}x_i}{n}=\dfrac{47}{10}=4.7

Standard deviation : \sigma=\sqrt{\dfrac{\sum^{10}_{i=1}{(x_i-\mu)^2}}{n}}

=\sqrt{\dfrac{(1.1)^2+(0.2)^2+(1.9)^2+(1.6)^2+(2.1)^2+(0.5)^2+(2.05)^2+(0.45)^2+(3.3)^2+(1.7)^2}{10}}

=\dfrac{30.265}{10}=3.0265

The confidence interval for the population mean (for sample size <30) is given by :-

\mu\ \pm t_{n-1, \alpha/2}\times\dfrac{\sigma}{\sqrt{n}}

Given significance level : \alpha=1-0.95=0.05

Critical value : t_{n-1,\alpha/2}=t_{9,0.025}=2.262

We assume that the population is normally distributed.

Now, the 95% confidence interval for the true mean will be :-

4.7\ \pm\ 2.262\times\dfrac{3.0265}{\sqrt{10}} \\\\\approx4.7\pm2.16=(4.7-2.16\ ,\ 4.7+2.16)=(2.54,\ 6.86)

Hence, 95% confidence interval for the true mean= (2.54,6.86)

7 0
3 years ago
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