An object with reflectional symmetry can be created by reflecting it about an axis called the _____.
A. Point of symmetry
B. Line of symmetry
C. Symmetrical half
D. Equator of symmetry

Answer:

(5,-2).

Step-by-step explanation:

First, let's find the original center of the circle, we have

[tex]x^2 - 4x + y^2 + 2y = 4[/tex]

we are going to complete square adding and subtracting 4 for the x terms and 1 for the y terms

[tex]x^2 - 4x+4-4 + y^2 + 2y+1-1 = 4[/tex]

[tex](x-2)^2 - 4 + (y+1)^2 - 1 = 4[/tex]

[tex](x-2)^2+ (y+1)^2 - 5 = 4[/tex]

[tex](x-2)^2+ (y+1)^2 = 4+5[/tex]

[tex](x-2)^2+ (y+1)^2 = 9.[/tex]

The canonical formula of a circumference is [tex](x-h)^2+(y-k)^2=r^2[/tex]

Then, we have a circle with [tex]r^2 =9[/tex] and center (h,k)=(2,-1).

Now, if we translate the circle 3 units to right and 1 unit down, then all the points in the circle will be translated including the center. Especifically, the x values will be added 3 units and the y-vaues will be subtracted 1 unit, then the new center will be

(2+3,-1-1) = (5,-2).

Let assume that [tex]\sqrt7[/tex] is a rational number. Therefore it can be expressed as a fraction [tex]\dfrac{a}{b}[/tex] where[tex]a,b\in\mathbb{Z}[/tex] and [tex]\text{gcd}(a,b)=1[/tex].

[tex]\sqrt7=\dfrac{a}{b}\\\\7=\dfrac{a^2}{b^2}\\\\a^2=7b^2[/tex]

This means that [tex]a^2[/tex] is divisible by 7, and therefore also [tex]a[/tex] is divisible by 7.

So, [tex]a=7k[/tex] where [tex]k\in\mathbb{Z}[/tex]

[tex](7k)^2=7b^2\\\\49k^2=7b^2\\\\7k^2=b^2[/tex]

Analogically to [tex]a^2=7b^2[/tex] ------- [tex]b^2[/tex] is divisible by 7 and therefore so is [tex]b[/tex].

But if both numbers [tex]a[/tex] and [tex]b[/tex] are divisible by 7, then [tex]\text{gcd}(a,b)=7[/tex] which contradicts our earlier assumption that [tex]\text{gcd}(a,b)=1[/tex].

Therefore [tex]\sqrt7[/tex] is an irrational number.

Final answer:

To prove that √7 is irrational, we assume the opposite and show that it leads to a contradiction. We start by assuming that √7 is rational and can be expressed as a fraction. By squaring both sides of the equation and simplifying, we get an equation that leads to p and q being divisible by 7, which contradicts our initial assumption. Therefore, √7 must be irrational.

Explanation:

To prove that √7 is irrational using the method of contradiction, we assume the opposite, which is that √7 is rational. This means that it can be expressed as a fraction p/q, where p and q are integers with no common factors other than 1. We can then square both sides of the equation (√7)² = (p/q)² and simplify to get the equation 7 = p²/q². Rearranging, we have p²= 7q².

From this equation, we can deduce that p² is divisible by 7, which means p must also be divisible by 7. Let's represent p as 7r, where r is another integer. Substituting back into the equation, we get (7r)² = 7q², which simplifies to 49r² = 7q². Dividing both sides by 7, we have 7r² = q². This implies that q² is also divisible by 7, so q must also be divisible by 7.

However, if both p and q are divisible by 7, this contradicts our initial assumption that p/q has no common factors other than 1. Therefore, our assumption that √7 is rational must be false, and hence √7 is irrational.

Answer:

5

Step-by-step explanation:

Answer:

11 1/3

Step-by-step explanation:

Answer:

C.Apex(105)

Step-by-step explanation:

The transformation that could NOT have been applied to the original figure to create an image with the same area is:

C. Dilation by a scale factor of 0.5.

To determine which transformation could NOT have been applied to the original figure to create the image with the same area, let's analyze each option:

A. Rotation of 90 degrees: When a figure is rotated by 90 degrees, the area remains the same.

This transformation is possible.

B. Reflection across a horizontal line: A reflection across a horizontal line (a horizontal flip) preserves the area of the figure.

This transformation is possible.

C. Dilation by a scale factor of 0.5: Dilation involves scaling a figure by a certain factor.

If a figure is dilated by a scale factor of 0.5, both its length and width are reduced to half, which means the area is reduced to one-fourth of the original area.

This transformation is not possible because it changes the area.

D. Translation down 4 and then right 1: A translation moves a figure without changing its size or shape. A translation down 4 and then right 1 would not change the area of the figure, as it simply shifts the figure's position.

This transformation is possible.

So, the transformation that could NOT have been applied to the original figure to create an image with the same area is:

C. Dilation by a scale factor of 0.5

Dilation by a scale factor of 0.5 would change the area, making it incompatible with the condition that the areas of the original figure and its image are the same.

For similar question on transformation.

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Using translation concepts, it is found that the equation of the resulting graph is given by:

[tex]y = \sqrt[3]{x + 4} - 5[/tex]

A translation is represented by a change in the function graph, according to operations such as multiplication or sum/subtraction in it's definition.

In this problem, the changes are given as follows:

Hence the equation of the resulting graph is given by:

[tex]y = \sqrt[3]{x + 4} - 5[/tex]

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The equation of the graph \\sqrt[3]{x} shifted down by 5 units and to the left by 4 units is \\sqrt[3]{x + 4} - 5.

The transformation of the graph of y = \\sqrt[3]{x} down by 5 units and to the left by 4 units is represented by the function y = \\sqrt[3]{x + 4} - 5. When a graph is shifted to the left by 'a' units, we replace 'x' in the equation with (x + a). Similarly, when a graph is shifted down by 'b' units, we subtract 'b' from the function, resulting in f(x) - b. Therefore, incorporating both transformations into the original function, we get y = \\sqrt[3]{x + 4} - 5.

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Answer:

- 1/2

Step-by-step explanation:

I just did it

Answer: Option 'B' is correct.

Step-by-step explanation:

Since we have given that

[tex]5x^2+14x-3[/tex]

As we know "Split the middle term":

[tex]5x^2+15x-x-3\\\\=5x(x+3)-1(x+3)\\\\=(5x-1)(x+3)[/tex]

Since it is quadratic equation so, it has 2 roots.

So, the roots are (x+3) and (5x-1).

Hence, Option 'B' is correct.

Answer:

1) Option B-  $728

2) Kara's annual premium is $62.5.

Step-by-step explanation:

1) Given : Daniel, a 37-year-old male, bought a $160,000, 10-year life insurance policy.

To find : What is Daniel's annual premium?

Solution :

In the table given,

Annual premium is per $1000 of coverage.

Daniel's life insurance policy buy is $160,000  

Annual premium per $1000 coverage Daniel buy at $160

37 year old male 10 year cost is $4.55

(marked in the attached table below)

Daniel's annual premium is [tex]4.55\times 160[/tex]

[tex]=\frac{455}{100}\times 160[/tex]

[tex]=728[/tex]

Daniel's annual premium is $728

Therefore, Option B is correct.

2) Given : Kara, a 25-year-old female, bought a $50,000, 10-year life insurance policy through her employer. Kara is paid semimonthly.

To find : How much is deducted from each of her paychecks for life insurance?

Solution :

In the table given,

Annual premium is per $1000 of coverage.

Kara's life insurance policy buy is $50,000  

Annual premium per $1000 coverage Kara buy at $50

25 year old female 10 year cost is $2.50

(marked in the attached table below)

Kara is paid semimonthly

So, cost is[tex]\frac{2.50}{2}=1.25[/tex]

Kara's annual premium is [tex]1.25\times 50[/tex]

[tex]=\frac{1.25}{100}\times 50[/tex]

[tex]=62.5[/tex]

Kara's annual premium is $62.5.

to determine if it is even replace x with -x and see if the answer is identical.

 In this case, this function is even

Answer:

The function [tex]f(x)=x^6-x^4[/tex] is:

Even

Step-by-step explanation:

A function f(x) is even if:

f(-x)= f(x)

A function f(x) is odd if:

f(-x)= -f(x)

Here, we are given a function f(x) as:

[tex]f(x)=x^6-x^4[/tex]

[tex]f(-x)=(-x)^6-(-x)^4\\\\ =x^6-x^4\\\\=f(x)[/tex]

f(-x)=f(x)

Hence, the function [tex]f(x)=x^6-x^4[/tex] is:

Even

Answer:

$66.97 is his first payment went toward reduction.

Step-by-step explanation:

Given : Jesse took out a 30-year loan for $85,000 at 7.2% interest, compounded monthly. If his monthly payment on the loan is $576.97.

To find : How much of his first payment went toward note reduction(reducing principal)?  

Solution :

First we find the interest of 1 month on a loan of $85,000 at 7.2% interest.

[tex]I=85000\times \frac{7.2}{12\times 100}[/tex]

[tex]I=85000\times 0.006[/tex]

[tex]I=510[/tex]

Interest of 1 month is $510.

Monthly payment = $576.97

Now, first payment or reducing principal is given by

F= monthly payment - interest of 1 month

F=$576.97- $510

F=$66.97

Therefore, $66.97 is his first payment went toward reduction.

The apothem (a), radius (r), and perimeter (p) of a regular figure are defined as follows: the apothem is the perpendicular distance from the center of the figure to one of its sides, the radius is the distance from the center of the figure to any point on its circumference, and the perimeter is the total length of all its sides.

The apothem (a) of a regular figure is the perpendicular distance from the center of the figure to one of its sides.

The radius (r) of a regular figure is the distance from the center of the figure to any point on its circumference.

The perimeter (p) of a regular figure is the total length of all its sides.

Point P is on the unit circle U. The values of trigonometric functions at t are [tex]sin\theta=\dfrac{8}{17},\;cos\theta=-\dfrac{15}{17}[/tex][tex],tan\theta=-\dfrac{8}{15},\;cot \theta=-\dfrac{15}{8},\;cosec\theta=\dfrac{17}{8},\;sec\theta=-\dfrac{17}{15}[/tex].

Given:

From the given figure, the coordinate of point P is [tex](-\dfrac{15}{17}, \dfrac{8}{17})[/tex].

The point P is present on a unit circle U. So, in general, the coordinates of a point on the circle will be [tex](x,y)\equiv (cos\theta, sin\theta)[/tex].

By comparing the given coordinate with the general expression, the value sine and cosine function will be,

[tex]sin\theta=\dfrac{8}{17}\\cos\theta=-\dfrac{15}{17}[/tex]

Now, the other trigonometric functions will be,

[tex]tan\theta=\dfrac{sin\theta}{cos\theta}\\tan\theta=-\dfrac{8}{15}\\cot \theta=-\dfrac{15}{8}\\cosec\theta=1/sin\theta=\dfrac{17}{8}\\sec\theta=1/cos\theta=-\dfrac{17}{15}[/tex]

Therefore, the values of trigonometric functions at t are [tex]sin\theta=\dfrac{8}{17},\;cos\theta=-\dfrac{15}{17}[/tex][tex],tan\theta=-\dfrac{8}{15},\;cot \theta=-\dfrac{15}{8},\;cosec\theta=\dfrac{17}{8},\;sec\theta=-\dfrac{17}{15}[/tex].

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