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3 years ago

Pls help. i will give brainliest to the most helpful, detailed answer

Mathematics

2 answers:

Rasek [7]3 years ago

5 0

Number 3 using table is 0.59, 3.41
Graphed on calculator and changed the table set to increments of .1 all the way to .000001 to estimate where the zeros where.

Number 1,2,4,5 are displayed in attachment.

Nikolay [14]3 years ago

5 0

Answer:

the answers are

Step-by-step explanation:

1)a.-3,-8

2)c.-6, -5/2

3)A. 0.59, 3.41

4)D. 5,1

5)D. 3.74 seconds

quadratic equation quiz check unit 6 lesson 5

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Two friends share 1/3 of a pitcher of

Oxana [17]

Answer:

⅙ pitcher

Step-by-step explanation:

½ of ⅓ pitcher = ½ × ⅓ pitcher = ⅙ pitcher

7 0

3 years ago

Verify that y1(t) = 1 and y2(t) = t ^1/2 are solutions of the differential equation:

Papessa [141]

Answer: it is verified that:

* y1 and y2 are solutions to the differential equation,

* c1 + c2t^(1/2) is not a solution.

Step-by-step explanation:

Given the differential equation

yy'' + (y')² = 0

To verify that y1 solutions to the DE, differentiate y1 twice and substitute the values of y1'' for y'', y1' for y', and y1 for y into the DE. If it is equal to 0, then it is a solution. Do this for y2 as well.

Now,

y1 = 1

y1' = 0

y'' = 0

So,

y1y1'' + (y1')² = (1)(0) + (0)² = 0

Hence, y1 is a solution.

y2 = t^(1/2)

y2' = (1/2)t^(-1/2)

y2'' = (-1/4)t^(-3/2)

So,

y2y2'' + (y2')² = t^(1/2)×(-1/4)t^(-3/2) + [(1/2)t^(-1/2)]² = (-1/4)t^(-1) + (1/4)t^(-1) = 0

Hence, y2 is a solution.

Now, for some nonzero constants, c1 and c2, suppose c1 + c2t^(1/2) is a solution, then y = c1 + c2t^(1/2) satisfies the differential equation.

Let us differentiate this twice, and verify if it satisfies the differential equation.

y = c1 + c2t^(1/2)

y' = (1/2)c2t^(-1/2)

y'' = (-1/4)c2t(-3/2)

yy'' + (y')² = [c1 + c2t^(1/2)][(-1/4)c2t(-3/2)] + [(1/2)c2t^(-1/2)]²

= (-1/4)c1c2t(-3/2) + (-1/4)(c2)²t(-3/2) + (1/4)(c2)²t^(-1)

= (-1/4)c1c2t(-3/2)

≠ 0

This clearly doesn't satisfy the differential equation, hence, it is not a solution.

6 0

3 years ago

Help me please! i have to redo a quiz, but i need answers, no links

Galina-37 [17]

Answer:

Step-by-step explanation:

This is because it got 2 times bigger

7 0

3 years ago

100 POINTS !! WILL MARK BRANILEST !! PLS HURRY

oee [108]

Answer:

log4x+log4(x-6)=2

place under single log using multiplication rule

log4(x(x-6))=2

convert to exponential form:(base(4) raised to log of number(2)=number(x(x-6)

4^2=x(x-6)

16=x^2-6x

x^2-6x-16=0

(x-8)(x+2)=0

x=8

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Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

What is the formula for a confidence interval?

ikadub [295]

Answer:

a) The formula is given by mean $\pm$ the margin of error. Where the margin of error is the product between the critical value from the normal standard distribution at the confidence level selected and the standard deviation for the sample mean.

b) $\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

If the distribution for X is normal or if the sample size is large enough we know that the distribution for the sample mean $\bar X$ is given by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

Part a

The formula is given by mean $\pm$ the margin of error. Where the margin of error is the product between the critical value from the normal standard distribution at the confidence level selected and the standard deviation for the sample mean.

Part b

The confidence interval for the mean is given by the following formula:

$\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$

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4 years ago