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Angelina_Jolie [31]

3 years ago

13

Complete the point-slope equation of the line through (3,6)(3,6)left parenthesis, 3, comma, 6, right parenthesis and (5,-8)(5,−8

)left parenthesis, 5, comma, minus, 8, right parenthesis.

1 answer:

ArbitrLikvidat [17]3 years ago

5 0

Answer:

The required equation of line in point-slope form is:

$y-6 = -7(x-3)$

Step-by-step explanation:

Given points are:

(x1,y1) = (3,6)

(x2,y2) = (5,-8)

The point-slope form of an equation is given by:

$y-y_1 = m(x-x_1)$

Here m is slope of the line and (x1,y1) is a point on line

The slope is calculated using the formula:

$m = \frac{y_2-y_1}{x_2-x_1}$

Putting the values we get

$m = \frac{-8-6}{5-3}\\m=\frac{-14}{2}\\m= -7$

Putting in the equation

$y-y_1 = -7(x-x_1)$

Now we have to put a point in the equation, putting (3,6) in the equation

$y-6 = -7(x-3)$

Hence,

The required equation of line in point-slope form is:

$y-6 = -7(x-3)$

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The water works commission needs to know the mean household usage of water by the residents of a small town in gallons per day.

valina [46]

Answer:

A sample of 179 is needed.

Step-by-step explanation:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1 - 0.85}{2} = 0.075$

Now, we have to find z in the Ztable as such z has a pvalue of $1 - \alpha$ .

That is z with a pvalue of $1 - 0.075 = 0.925$ , so Z = 1.44.

Now, find the margin of error M as such

$M = z\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

A previous study found that for an average family the variance is 1.69 gallon?

This means that $\sigma = \sqrt{1.69} = 1.3$

If they are using a 85% level of confidence, how large of a sample is required to estimate the mean usage of water?

A sample of n is needed, and n is found for M = 0.14. So

$M = z\frac{\sigma}{\sqrt{n}}$

$0.14 = 1.44\frac{1.3}{\sqrt{n}}$

$0.14\sqrt{n} = 1.44*1.3$

$\sqrt{n} = \frac{1.44*1.3}{0.14}$

$(\sqrt{n})^2 = (\frac{1.44*1.3}{0.14})^2$

$n = 178.8$

Rounding up

A sample of 179 is needed.

7 0

3 years ago

Write explicit formula for a1=3, r=-2; then generate 1st 5 terms

Crazy boy [7]

Answer:

an = 3(-2)^(n-1)
3, -6, 12, -24, 48

Step-by-step explanation:

These variable names, a1, r, are commonly used in relationship to geometric sequences. We assume you want the terms of a geometric sequence with these characteristics.

a1 is the first term. r is the ratio between terms, so is the factor to find the next term from the previous one.

a1 = 3 (given)

a2 = a1×r = 3×(-2) = -6

a3 = a2×r = (-6)(-2) = 12

a4 = a3×r = (12)(-2) = -24

a5 = a4×r = (-24)(-2) = 48

The first 5 terms are 3, -6, 12, -24, 48.

__

The explicit formula for the terms of a geometric sequence is ...

an = a1×r^(n -1)

Using the given values of a1 and r, the explicit formula for this sequence is ...

an = 3(-2)^(n -1)

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3 years ago

The numerator and the denominator of a certain fraction are consecutive odd numbers. If nine is subtracted frm the numerator, th

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The denominator of a fraction is 1 more than 3 times the numerator. If the denominator is doubled and the numerator is increased by 2, the value of the resulting fraction is 1/4. Find the original fraction.

4 0

4 years ago

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GEOMETRY are these congruent or not?

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3 years ago

Measure the lengths of the sides of ∆ABC in GeoGebra, and compute the sine and the cosine of ∠A and ∠B. Verify your calculations

marusya05 [52]

Answer:

$Sin \angle A =0.80$

$Cos \angle A=0.60$

$Sin \angle B =0.60$

$Cos \angle B=0.80$

Step-by-step explanation:

Given

I will answer this question using the attached triangle

Solving (a): Sine and Cosine A

In trigonometry:

$Sin \theta =\frac{Opposite}{Hypotenuse}$ and

$Cos \theta =\frac{Adjacent}{Hypotenuse}$

So:

$Sin \angle A =\frac{BC}{BA}$

Substitute values for BC and BA

$Sin \angle A =\frac{8cm}{10cm}$

$Sin \angle A =\frac{8}{10}$

$Sin \angle A =0.80$

$Cos \angle A=\frac{AC}{BA}$

Substitute values for AC and BA

$Cos \angle A=\frac{6cm}{10cm}$

$Cos \angle A=\frac{6}{10}$

$Cos \angle A=0.60$

Solving (b): Sine and Cosine B

In trigonometry:

$Sin \theta =\frac{Opposite}{Hypotenuse}$ and

$Cos \theta =\frac{Adjacent}{Hypotenuse}$

So:

$Sin \angle B =\frac{AC}{BA}$

Substitute values for AC and BA

$Sin \angle B =\frac{6cm}{10cm}$

$Sin \angle B =\frac{6}{10}$

$Sin \angle B =0.60$

$Cos \angle B=\frac{BC}{BA}$

Substitute values for BC and BA

$Cos \angle B=\frac{8cm}{10cm}$

$Cos \angle B=\frac{8}{10}$

$Cos \angle B=0.80$

Using a calculator:

$A = 53^{\circ}$

So:

$Sin(53^{\circ}) =0.7986$

$Sin(53^{\circ}) =0.80$ -- approximated

$Cos(53^{\circ}) = 0.6018$

$Cos(53^{\circ}) = 0.60$ -- approximated

$B = 37^{\circ}$

So:

$Sin(37^{\circ}) = 0.6018$

$Sin(37^{\circ}) = 0.60$ --- approximated

$Cos(37^{\circ}) = 0.7986$

$Cos(37^{\circ}) = 0.80$ --- approximated

8 0

3 years ago

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