(Geometry course question) Explain why it is not possible to construct an equilateral triangle that has
three vertices with integer coordinates. Show your work by providing an example and
including a graph. Consider an equilateral triangle whose base lies along the x-axis.

Answer:

Create a table.

Plug values into the equation to complete the table.

Use the table to plot points on the graph.

Draw lines through the points on your graph.

The graph should look linear because there is an x that doesn't have a power greater than or less than 1.

Answer: 2000 meters

Step-by-step explanation: A kilometer means 1,000 meters a kilo means 1000. So, 2 x1000 = 2000 meters.

1. The value of the linear correlation coefficient r is given as 0.8159. The correct option is a. There is insufficient evidence to support the claim of a nonlinear correlation between the two variables

2. The scatterplot reveals a distinct pattern that is not a straight-line pattern.

This is the data that we get when we try to get the existing relationship that is in existence between the variables that are used in a regression equation.

The value of r was calculated using a statistical tool online. The graph showed us it was not a straight line. The value of the r = 0.8159

The results of the test is in the attachment

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Answer:

Step-by-step explanation:

[tex]\dfrac{ 1 }{ 2 } \times 12\times 15\times \sin ( 110 )[/tex]

==> 84.57

Answer:

-1 and 1

Step-by-step explanation:

Reflecting D over the y-axis maps it onto (1,2).

This is supposed to map onto F(0,3), so this is a shift 1 unit left and 1 unit up.

So, the blanks are -1 and 1, respectively.

Answer: (x + [-1], y + [1])

Step-by-step explanation:

      See attached. We can draw, or picture it in our heads, what the reflection would look like. Then we can pick one (or multiple to test) points and see the translation.

      We can also test with a set of points. B', (2, 4) becomes G in the transformation. G is at (1, 5)

      (1 - 2, 5 - 4) -> (-1, 1)

Answer: (-5, -5)

Step-by-step explanation:

(-4, 0)

Here, -4 is our x value and 0 is our y value

The x value moves horizontal (left and right) while the y value moves vertical (up and down)

If we move 1 unit left, we change x by -1 so, -4 -1 = -5

If we move 5 units down, we change y by -5 so, 0 -5 = -5

Therefore,

our new endpoint is: (-5, -5)

Answer:

C. slope

Greetings !

The symbol for slope is "m". Slope can be calculated by taking any two locations along a line and calculating the ratio of the RISE to the RUN (the ratio of the difference between vertical positions of the locations to the difference between horizontal positions of the locations).

Answer: slope.

Step-by-step explanation:

The number of people who must be included in the survey, that is, the sample size should be 340.

We assume the sample size to be n.

The confidence interval required is 99%.

The Z-score corresponding to the 99% confidence interval is (Z) 2.58.

The true proportion (p) of the sample is not given, so we assume it to be 50% or 0.5.

The margin of error for the sample (E) is given to be 7% or 0.07.

By the formula of margin of error, we know that:

E = Z√[{p(1 - p)}/n].

Putting in the values, we have, we get:

0.07 = 2.58√[{0.5*0.5}/n],

or, (0.07/2.58)² = 0.25/n,

or, n = 0.25/{(0.07/2.58)²} = 339.6122449 ≈ 340.

Thus, the number of people who must be included in the survey, that is, the sample size should be 340.

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The vertical asymptote of f(x) is (A) x = 0, –9.

To find the vertical asymptote of f(x):

The vertical asymptotes of a function are the zeroes of the denominator of a rational function

The function is given as: [tex]f(x) = \frac{(x-9)}{(x^{3} -81x)}[/tex]

Set the denominator to 0:

Factor out x:

Express 81 as 9^2:

Express the difference between the two squares:

Split, [tex]x=0[/tex] or [tex]x=-9[/tex] or [tex]x+9=0[/tex].

Solve for x:

Therefore, the vertical asymptote of f(x) is (A) x = 0, –9.

(See attachment for the graph of f(x))

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The complete question is given below:

Consider the function f(x)=(x-9)/(x^3-81x) . find the vertical asymptote(s) of f(x).

A) x = 0, –9

B) x = –9

C) x = 0, 9

D) x = 9

According to the developed scale, a radiant day is 16 times brighter than a dim day.

We assume the brightness of a dim day to be x.

According to the developed scale, the brightness of an illuminated day will be 4 times that of a dim day.

Thus, the brightness of an illuminated day = 4*the brightness of a dim day = 4x.

According to the developed scale, the brightness of a radiant day will be 4 times that of an illuminated day.

Thus, the brightness of a radiant day = 4*the brightness of an illuminated day = 4*4x = 16x.

Now, the ratio of the brightness of a radiant day to the brightness of a dim day = 16x:x = 16x/x = 16:1.

Thus, according to the developed scale, a radiant day is 16 times brighter than a dim day.

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The question provided is incomplete. The complete question is:

"Suppose you have developed a scale that indicates the brightness of sunlight. Each category in the table is 4 times brighter than the next lower category. For example, a day that is dazzling is 4 times brighter than a day that is radiant. How many times brighter is a radiant day than a dim day?

Dim=2

Illuminated=3

Radiant=4

Dazzling=5"

Answer: x= 3.4

Step-by-step explanation:

Solve for x by simplifying both sides of the equation, then isolating the variable.

Answer:

x = 3.4

Step-by-step explanation:

[tex]10x - 4.5 + 3x = 12x -1.1 \\ 13x - 4.5 = 12x - 1.1 \\ 13x - 12x - 4.5 = 12x - 12x - 1.1 \\ x - 4.5 = - 1.1 \\ x - 4.5 + 4.5 = - 1.1 + 4.5 \\ x = 3.4[/tex]

Step-by-step explanation:

log 6 = log (2×3) = log 2 + log 3 = 0.3010+0.4771

=0.7781

Answer:

[tex]\sf \log_{10}6=0.7781[/tex]

Step-by-step explanation:

Given:

[tex]\sf \log_{10} 2 = 0.3010[/tex]

[tex]\sf \log_{10} 3 = 0.4771[/tex]

To find log₁₀ 6, first rewrite 6 as 3 · 2:

[tex]\sf \implies \log_{10}6=\log_{10}(3 \cdot 2)[/tex]

[tex]\textsf{Apply the log product law}: \quad \log_axy=\log_ax + \log_ay[/tex]

[tex]\implies \sf \log_{10}(3 \cdot 2)=\log_{10}3+\log_{10}2[/tex]

Substituting the given values for log₁₀ 3 and log₁₀ 2:

[tex]\begin{aligned} \sf \implies \log_{10}3+\log_{10}2 & = \sf 0.4771+0.3010\\ & = \sf 0.7781 \end{aligned}[/tex]

Therefore:

[tex]\sf \log_{10}6=0.7781[/tex]

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The desired measures for the data-set is given by:

The five number summary is composed by the measures explained below, except the IQR.

In this problem, we have that:

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The measure of the angle rst is [tex]105^{o}[/tex].

According to the theorem, a quadrilateral can be inscribed by a circle if and only if its opposing angles are supplementary. A quadrilateral that may be encircled by a circle is known as a cyclic quadrilateral.

The sum opposite angles of a quadrilateral is 180 degrees.

We can see that the angles rst and angle tqr are opposite angles of quadrilateral qrst, hence supplementary angles.

Therefore,

<RST + <TQR = 180

5x+15 + 4x+ 3 = 180

9x + 18 = 180

9x = 180 - 18

9x = 162

x = 162/9

x = 18

Since <RST = 5x+15

<RST = 5(18) + 15

<RST = 90 + 15

<RST = 105degrees

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The value of the Population Variance is 100.

According to the statement

we have given that the A population of N = 10 scores has μ = 50 and σ = 5. And we have to find the population variance.

So, For this purpose,

we know that the population variance can be defined as the average of the distances from each data point in a particular population to the mean squared, and it indicates how data points are spread out in the population.

Population variance = mean /(standard deviation/number of samples)

Population variance = μ / (σ /N )

Substitute the values in it and then

Population variance = 50 / (5 /10 )

Population variance = 50 / (0.5 )

Population variance = 100.

So, The value of the Population Variance is 100.

Disclaimer: This question was incomplete. Please find the full content below.

Question:

A population of N = 10 scores has μ = 50 and σ = 5. What is the population variance?

a. 10

b. the square root of 5

c. the square root of 50

d. 25

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Answer:

11/20

Step-by-step explanation:

You could turn them all into equivalent fractions, but the common denominator would be 1,763,580.  Yuck.

Look at the relationship for the numerators to the denominators.  The only numerator that is more than half of its denominator is 11/20.  Half of 20 would be 10 and 11 has a larger value than 10, so as a decimal, it would be more than .5.  All of the other numerators are less than half, so that they have to less than .5.

If you changed all of the numbers to decimals, you would see that 11/20 has the largest value.

Answer:

(x+5, y-3)

Step-by-step explanation:

I would say this because when a plane is translated up it would be positive and when it's translated to the left it would be negative,

Answer:

Read explanation

Step-by-step explanation:

I answered another one of your questions just like this in detail so I'll keep this one shorter.

Domain: (-∞, 0]

Range: (-3,∞)

x-intercept: Approximately -3.75

y=intercept: -3

The capital borrowed, or the principal when the interest was Q468.75 is Q7500.

The principal P, amount borrowed, at a certain rate of interest R%, for  a time of T years, giving an interest of I, can be calculated using the formula:

P = (I * 100)/(R * T).

In the question, we are asked to find the capital borrowed, that is, the principal, when the user pays Q468.75 after 15 months at 5% of income.

Thus, Interest (I) = Q468.75, rate of interest (R) = 5%, and time (T) = 15 months = 15/12 years = 1.25 years.

Thus, the principal P, can be calculated by substituting the values in the formula: P = (I * 100)/(R * T).

P = (468.75*100)/(5*1.25),

or, P = 46875/6.25,

or, P = 7500.

Thus, the capital borrowed, or the principal when the interest was Q468.75 is Q7500.

The provided question is in Spanish. The question in English is:

"After 15 months of having borrowed capital at 5% of income, I have to pay Q468.75 in interest. What capital has been borrowed?"

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we conclude that the distance to the other side is 467 feet.

We know that the width of the river is 602ft. This means that the distance from shore to shore is 602 feet.

If the boy is at a distance of 135 from the shore, then the distance to the other side is given by the difference:

d = 602ft - 135ft = 467ft

Then, we conclude that the distance to the other side is 467 feet.

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Answer:

someone please help me check my page

someone please help me check my pageStep-by-step explanation:

Let hats be x

Let shirts be y

We can set up a system of equations to figure out the cost of one hat.

5x+3y = 176 and 3x+5y = 208

Since we are finding the cost of the hats, we will need to isolate x from the equation. We can first get rid of y. One way we can do that is by multiplying the first equation with 5 and the second one with -3, effectively making y add up to 0 when we solve for x.

5(5x+3y = 176) --> 25x+15y = 880

-3(3x+5y = 208) --> -9x-15y = -624

We can now add the 2 equations together and solve for x.
25x+15y = 880

-9x-15y = -624

+>>>>>>>>>>>>>>>

16x = 256

x = 16

So the the cost of one hat is 16 dollars.

Answer:

$16

Step-by-step explanation:

Define the variables:

Create two equations using the defined variables and the given information.

Equation 1

Five hats and three shirts cost $176.

⇒ 5x + 3y = 176

Equation 2

Three hats and five shirts cost $208.

⇒ 3x + 5y = 208

Rewrite Equation 2 to make y the subject:

[tex]\implies \sf 3x+5y-3x=208-3x[/tex]

[tex]\implies \sf 5y=208-3x[/tex]

[tex]\implies \sf \dfrac{5y}{5}=\dfrac{208-3x}{5}[/tex]

[tex]\implies \sf y=\dfrac{208-3x}{5}[/tex]

Substitute the found expression for y into Equation 1 and solve for x:

[tex]\implies \sf 5x+3\left(\dfrac{208-3x}{5} \right)=176[/tex]

[tex]\implies \sf 5x+\dfrac{624-9x}{5}=176[/tex]

[tex]\implies \sf 5x+124.8-1.8x=176[/tex]

[tex]\implies \sf 3.2x+124.8=176[/tex]

[tex]\implies \sf 3.2x+124.8-124.8=176-124.8[/tex]

[tex]\implies \sf 3.2x=51.2[/tex]

[tex]\implies \sf \dfrac{3.2x}{3.2}=\dfrac{51.2}{3.2}[/tex]

[tex]\implies \sf x=16[/tex]

Therefore, the cost of one hat is 16 dollars.

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Rotating 90 degrees counterclockwise: [tex](x,y) \longrightarrow (-y,x)[/tex]

[tex]A(-4,4) \longrightarrow A'(-4, -4)\\\\B(-3, 4) \longrightarrow B'(-4, -3)\\\\C(-3, 1) \longrightarrow C'(-1, -3)\\\\D(-4, 1) \longrightarrow D'(-1, -4)[/tex]

Translate along [tex]\langle 3, 4 \rangle[/tex]: [tex](x,y) \longrightarrow (x+3, y+4)[/tex]

[tex]A'(-4, -4) \longrightarrow A''(-1, 0)\\\\B'(-4, -3) \longrightarrow B''(-1, 1)\\\\C'(-1, -3) \longrightarrow C''(2, 1)\\\\D'(-1, -4) \longrightarrow D''(2, 0)[/tex]

The total surface area of the figure is 92 m².

The surface area of a figure is the sum of the whole sides of the figure.

Therefore,

surface area = 4 × 4 + 4 × 4 + 6 × 4 + 2 × 4 + 1 / 2 (2 + 6)3.5 +  1 / 2 (2 + 6)3.5

Hence,

surface area = 16 + 16 + 24 + 8 + 14 + 14

Therefore,

surface area = 32 + 24 + 8 + 28

surface area = 56 + 8 + 28

surface area = 92 m²

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Answer:

176 square feet

Step-by-step explanation:

add the areas of the rectangle and triangle

rectangle area is

16 times 8 which is

128

rectangle area is

1/2 times 12 times 8 which is

48

then add them

128 plus 48 is

176 square feet

The length of the function y = 3x over the given interval [0, 2] is 3.2 units

For given question,

We have been given a function  y = 3x

We need to find the length of the function on the interval x = 0 to x = 2.

Let f(x) = 3x where f(x) = y

We have  f'(x) = 3,  so  [f'(x)]² = 9.  

Then the arc length is given by,

[tex]\int\limits^a_b {\sqrt{1+[f'(x)]^2} } \, dx\\\\= \int\limits^2_0 {\sqrt{1+9} }\, dx\\\\=\sqrt{10}\\\\ =3.2[/tex]

This means, the arc length is 3.2 units.

Therefore, the length of the function y = 3x over the given interval [0, 2] is 3.2 units

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Based on the given data; the table represent an inverse variation and the equation that model the data in the table is y = 8/x

y = k × x

4 = k × 2

4 = 2k

k = 4/2

k = 2

when y = 2 and x = 4

y = k × x

= 2 × 4

y = 8

y = k/x

4 = k / 2

8 = k

when y = 2 and x = 4

y = k/x

2 = k/4

2 × 4 = k

k = 8

So,

y = k/x

y = 8/x

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Answer:

[tex]e^{4x}=e^4+4e^4(x-1)+8e^4(x-1)^2+...[/tex]

[tex]\displaystyle e^{4x}=\sum^{\infty}_{n=0} \dfrac{4^ne^4}{n!}(x-1)^n[/tex]

Step-by-step explanation:

Taylor series expansions of f(x) at the point x = a

[tex]\text{f}(x)=\text{f}(a)+\text{f}\:'(a)(x-a)+\dfrac{\text{f}\:''(a)}{2!}(x-a)^2+\dfrac{\text{f}\:'''(a)}{3!}(x-a)^3+...+\dfrac{\text{f}\:^{(r)}(a)}{r!}(x-a)^r+...[/tex]

This expansion is valid only if [tex]\text{f}\:^{(n)}(a)[/tex] exists and is finite for all [tex]n \in \mathbb{N}[/tex], and for values of x for which the infinite series converges.

[tex]\textsf{Let }\text{f}(x)=e^{4x} \textsf{ and }a=1[/tex]

[tex]\text{f}(x)=\text{f}(1)+\text{f}\:'(1)(x-1)+\dfrac{\text{f}\:''(1)}{2!}(x-1)^2+...[/tex]

[tex]\boxed{\begin{minipage}{5.5 cm}\underline{Differentiating $e^{f(x)}$}\\\\If $y=e^{f(x)}$, then $\dfrac{\text{d}y}{\text{d}x}=f\:'(x)e^{f(x)}$\\\end{minipage}}[/tex]

[tex]\text{f}(x)=e^{4x} \implies \text{f}(1)=e^4[/tex]

[tex]\text{f}\:'(x)=4e^{4x} \implies \text{f}\:'(1)=4e^4[/tex]

[tex]\text{f}\:''(x)=16e^{4x} \implies \text{f}\:''(1)=16e^4[/tex]

Substituting the values in the series expansion gives:

[tex]e^{4x}=e^4+4e^4(x-1)+\dfrac{16e^4}{2}(x-1)^2+...[/tex]

Factoring out e⁴:

[tex]e^{4x}=e^4\left[1+4(x-1)+8}(x-1)^2+...\right][/tex]

Taylor Series summation notation:

[tex]\displaystyle \text{f}(x)=\sum^{\infty}_{n=0} \dfrac{\text{f}\:^{(n)}(a)}{n!}(x-a)^n[/tex]

Therefore:

[tex]\displaystyle e^{4x}=\sum^{\infty}_{n=0} \dfrac{4^ne^4}{n!}(x-1)^n[/tex]

The general term for the arithmetic sequence; -13, -11, -9, -7 is Tn = -15 - 2n.

-13, -11, -9, -7

Common difference, d = Second term - first term

= -11 - (-13)

= -11 + 13

= 2

Therefore,

Tn = a + (n - 1)d

Where,

Tn = a + (n - 1)d

= -13 + (n - 1)2

= -13 - 2n - 2

Tn = -15 - 2n

So therefore, the general term for the arithmetic sequence; -13, -11, -9, -7 is Tn = -15 - 2n.

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