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The solution to the equation r(1 - 2cosФ) = 1 is given as x² + y² - 4x√(x² + y²)  + 4x² = 1

An equation is an expression that shows the relationship between two or more variables and numbers.

In polar form:

r = √(x² + y²) and cosФ = x / √(x² + y²)

Hence:

r(1 - 2cosФ) = 1

√(x² + y²) [1 - 2(x / √(x² + y²))] = 1

√(x² + y²) - 2x = 1

Take square of both sides:

x² + y² - 4x√(x² + y²)  + 4x² = 1

The solution to the equation r(1 - 2cosФ) = 1 is given as x² + y² - 4x√(x² + y²)  + 4x² = 1

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Answer:

option B is correct

To write any equations in the form of matrix

first make a matrix of all coefficients of that equations.. here coefficients of first equation is 4 and - 2 and that of second is 0 ( as there is no value of X) and 3

then make second matrix of X and y

last of answer of the equations, here (-7 ) and 5

The asymptotes of the graph of [tex]f(x) = \frac{5}{x - 1}[/tex] are as follows:

In this problem, the function is:

[tex]f(x) = \frac{5}{x - 1}[/tex]

For the vertical asymptote, we have that:

[tex]x - 1 = 0 \rightarrow x = 1[/tex].

For the horizontal asymptote, we have that:

[tex]y = \lim_{x \rightarrow \infty} f(x) = \frac{5}{\infty - 1} = 0[/tex].

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Answer:

3x - 7

Step-by-step explanation:

This questions asks about the subtraction of functions. To find the answer, we can subtract the same way we would subtract polynomials.

Subtracting Expressions

First, set up the subtraction problem by taking the expression for f(x) and subtracting the expression equal to g(x) from it.

This looks like:

Next, distribute the negative to the second binomial

Then, combine like terms

Since we are not given a x-value, this is the final answer.

Solving for an X-Value

If you are given an x-value, the best way to solve the equation is to first solve each function and then subtract.

For example, using the same rules for each function let's solve (f-g)(2).

First, solve for f(2) and g(2)

Then, subtract the 2 values

Answer:

A = 6

B = 1

C = 14

Step-by-step explanation:

A = 25 - 19 = 6

B = 6 - 5 = 1

C = 19 - 5 = 14

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Answer:

Define different tipes of discontuinity of a function .also,write the condition for increasing, decreasing and concave of function

Answer:

what is the control ofprice floir or price celling then scplain one advantage an one dissvantage

Answer: 0.625

Step-by-step explanation:

Answer:

x = [tex]\frac{7}{5}[/tex]

Step-by-step explanation:

To solve for x, we need to isolate it [get it alone on one side]

We should start by simplifying (combining the like-terms of our equation):

5 + 5 = 5x + 5 - 2

10 = 5x + 3

- 3         - 3        (subtract 3 from both sides to isolate x)

7 = 5x

÷5  ÷5   (divide both sides by 5 to find "1"x)

[tex]\frac{7}{5}=x[/tex]  

So, x = [tex]\frac{7}{5}[/tex]

(x = 7/5)

hope this helps!!

5 + 5 = 5x + 5 - 2

Answer :

[tex]x = \frac{7}{5} [/tex]

Step-by-step explanation:

5 + 5 = 5x + 5 - 2

10 = 5x + 3

Take 3 to the other side

10 - 3 = 5x

7 = 5x

[tex] \frac{7}{5} = x[/tex]

[tex]x = \frac{7}{5} [/tex]

The sides length of the cubical box is 3.8 cm if the volume of a cubical box is 54.872 cm³ option second is correct.

It is defined as three-dimensional geometry that has six square faces and eight vertices.

We have a volume of a cubical box is 54.872 cm³

V = 54.872 cm³

As we know the volume of the cube:

V = side³

54.872 = side³

Taking cube root on both sides:

side = 3.8 cm

Thus, the sides length of the cubical box is 3.8 cm if the volume of a cubical box is 54.872 cm³ option second is correct.

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The system of the linear inequality that fulfill the condition of the graph will be y ≥ 2x – 1 and y ≤ 1/3 x + 1. Then the correct option is A.

The complete question is attached below.

Inequality is defined as an equation that does not contain an equal sign. Inequality is a term that describes a statement's relative size and can be used to compare these two claims.

The system of the linear inequality that fulfill the condition of the graph will be

y ≥ 2x – 1

y ≤ 1/3 x + 1

Then the correct option is A.

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Answer: 13/69

Step-by-step explanation:

13 students are female and got a B out of a total of 69 students, so the probability is 13/69

Answer:

C.6

Step-by-step explanation:

Answer:

[tex]x=\dfrac{-3\pm3\sqrt{5}}{2}[/tex]

Step-by-step explanation:

Given equation:

[tex]x^2+3x-9=0[/tex]

Completing the square

Move the constant to the right side by adding 9 to both sides:

[tex]\implies x^2+3x=9[/tex]

Add the square of half the coefficient of x to both sides:

[tex]\implies x^2+3x+\left(\dfrac{3}{2}\right)^2=9+\left(\dfrac{3}{2}\right)^2[/tex]

[tex]\implies x^2+3x+\dfrac{9}{4}=\dfrac{45}{4}[/tex]

Factor the trinomial on the left side:

[tex]\implies \left(x+\dfrac{3}{2}\right)^2=\dfrac{45}{4}[/tex]

Square root both sides:

[tex]\implies x+\dfrac{3}{2}=\pm\sqrt{\dfrac{45}{4}}[/tex]

[tex]\implies x+\dfrac{3}{2}=\pm\dfrac{\sqrt{45}}{\sqrt{4}}}[/tex]

[tex]\implies x+\dfrac{3}{2}=\pm\dfrac{\sqrt{9 \cdot 5}}{2}[/tex]

[tex]\implies x+\dfrac{3}{2}=\pm\dfrac{\sqrt{9} \sqrt{5}}{2}[/tex]

[tex]\implies x+\dfrac{3}{2}=\pm\dfrac{3\sqrt{5}}{2}[/tex]

Subtract 3/2 from both sides:

[tex]\implies x=\pm\dfrac{3\sqrt{5}}{2}-\dfrac{3}{2}[/tex]

[tex]\implies x=\dfrac{-3\pm3\sqrt{5}}{2}[/tex]

The length of C from A is 3.5 miles if AB=3.5 miles and BC is 6 miles .

Given The length of triangle ABC AB=3.5 miles , BC=6miles and AC=d

We have to find the value of d

First of all we have to draw a perpendicular from point A on side BC .

It forms two triangles ABE and AEC if E is the point where perpendicular stands on BC.

In triangle BEA

AE=[tex]\sqrt{3.5^{2} -3^{2} }[/tex]

=[tex]\sqrt{12.25-9}[/tex]

=[tex]\sqrt{3.25}[/tex]

=1.8

Now in triangle AEC

AC=[tex]\sqrt{1.8^{2} +3^{2} }[/tex]

=[tex]\sqrt{3.24+9}[/tex]

=[tex]\sqrt{12.24}[/tex]

=[tex]\sqrt{12.24}[/tex]

=3.49 MILES

By rounding off to the nearest tenth we will get 3.4 miles

Hence the value of d is 3.4 miles.

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Answer:its 7.7

Step-by-step explanation: i did it on edge

Answer:

No

Step-by-step explanation

The square root of 10 is irrational because it isn't a fraction

The most appropriate Venn diagram is known to be option b in the image shown.

A Common multiples are known to be the multiples that are said to be common to two or that have more numbers in them.

The Least Common Multiple ( LCM ) is known to be a term that connote  the Lowest Common Multiple ( LCM ) and also that has the Least Common Divisor ( LCD).

From the image shown the lowest multiples are 5², 2⁶, 3 while the greatest common factor is 2⁶.

Option B is correct because the greatest common factor which is  2⁶ can only be found in the middle and this was gotten when you add:

2³ . 2³

Collect like terms

2 ³⁺³

2⁶

Therefore, the greatest common factor which is  2⁶

To get the  Least Common Multiple, one need to add all the factors in the Venn diagram which are therefore:

5², 2³ . 2³, 3

=5², 2 ³⁺³, 3

=5², 2⁶, 3

Therefore, The most appropriate Venn diagram is known to be option b in the image shown.

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Given that f(x) = x/(x - 3) and g(x) = 1/x and the application of function operators, f ° g (x) = 1/(1 - 3 · x) and the domain of the resulting function is any real number except x = 1/3.

Let be f and g functions. Composition is a binary function operation where the variable of the former function (f) is substituted by the latter function (g). If we know that f(x) = x/(x - 3) and g(x) = 1/x, then the composed function is:

[tex]f\,\circ\,g \,(x) = \frac{\frac{1}{x} }{\frac{1}{x}-3}[/tex]

[tex]f\,\circ\,g\,(x) = \frac{\frac{1}{x} }{\frac{1-3\cdot x}{x} }[/tex]

[tex]f\,\circ\,g\,(x) = \frac{1}{1-3\cdot x}[/tex]

The domain of the function is the set of x-values such that f ° g (x) exists. In the case of rational functions of the form p(x)/q(x), the domain is the set of x-values such that q(x) ≠ 0. Thus, the domain of f ° g (x) is [tex]\mathbb{R} - \{\frac{1}{3} \}[/tex].

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The analytical approach is preferable because the primary issue with Euclidean geometry is that it is not enough to support all of the theorems that he claims to establish.

It is defined as the branch of mathematics that is concerned with the size, shape, and orientation of two-dimensional figures.

As we have given about the Euclidean and analytical geometry.

Without using any numerical measurements, we have investigated the connections between points, lines, and planes in Euclidean geometry.

In analytical geometry, we've looked at how the locations of points in a Cartesian coordinate system related to algebra and geometry.

The analytical approach is preferable because the primary issue with Euclidean geometry is that it is not enough to support all of the theorems that he claims to establish.

If the Euclidean geometry can be more advantageous in particular circumstances

Example: If using terrain or building chart),

Thus, the analytical approach is preferable because the primary issue with Euclidean geometry is that it is not enough to support all of the theorems that he claims to establish.

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Answer:

See below ~

Step-by-step explanation:

1) a) 50 km / 2 h

⇒ 25 km/h

b) 540 kg / 3 days

⇒ 180 kg/day

c) 125 m / 5 min.

⇒ 25 m/min.

A unit rate means a rate for one of something. We write this as a ratio with a denominator of one. For example, if you ran 70 yards in 10 seconds, you ran on average 7 yards in 1 second. Both of the ratios, 70 yards in 10 seconds and 7 yards in 1 second, are rates, but the 7 yards in 1 second is a unit rate…

a) 50km/2h

=> 25 kilometers per hour

b) 540kg/3d

=> 180 kilogram per day

c) 125m/5min

=> 25 meter per minute

The transformation of a function may involve any change. When the graph of f(x) is transformed into 5 units to the right and one unit up, then the function g(x) is obtained.

The transformation of a function may involve any change.

Usually, these can be shifted horizontally (by transforming inputs) or vertically (by transforming output), stretched (multiplying outputs or inputs) etc.

If the original function is y = f(x), assuming the horizontal axis is the input axis and the vertical is for outputs, then:

Horizontal shift (also called phase shift):

Vertical shift

Stretching:

Given the function f(x)=x², which is transformed to g(x)=(x-5)²+1, therefore, the graph of both the functions are given below.

When the graph of f(x) is transformed into 5 units to the right and one unit up, then the function g(x) is obtained.

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Answer:

(3, 2)

Step-by-step explanation:

To be considered part of a solution set for the inequality provided, a given point must lie within the shaded region that is graphed. Since we want to find the point that is not included in the solution set for the inequality, we need to find the point that does not lie in the shaded region.

Let's test each point to see which doesn't lie in the region. Refer to the image below if you're having trouble graphing the points:

Hope this helps!

Answer:

[tex]360xy^2|xz^3|[/tex]

Step-by-step explanation:

Given expression:

[tex]12x \sqrt{900x^2y^4z^6}[/tex]

[tex]\textsf{Apply radical rule} \quad \sqrt{ab}=\sqrt{a}\sqrt{b}:[/tex]

[tex]\implies 12x \sqrt{900}\sqrt{x^2}\sqrt{y^4}\sqrt{z^6}[/tex]

Replace 900 with 30² :

[tex]\implies 12x \sqrt{30^2}\sqrt{x^2}\sqrt{y^4}\sqrt{z^6}[/tex]

[tex]\textsf{Apply radical rule} \quad \sqrt{a^2}=a, \quad a \geq 0:[/tex]

[tex]\implies 12x \cdot 30|x|\sqrt{y^4}\sqrt{z^6}[/tex]

[tex]\implies 360x|x|\sqrt{y^4}\sqrt{z^6}[/tex]

(We need to use the absolute value of √x² since the x term was originally to the power of 2, which means the value of x² is always positive since the exponent is even).

[tex]\textsf{Apply exponent rule} \quad \sqrt{a^m}=a^{\frac{m}{2}}:[/tex]

[tex]\implies 360x|x|\cdot y^{\frac{4}{2}}\cdot z^{\frac{6}{2}}[/tex]

Simplify:

[tex]\implies 360xy^2|xz^3|[/tex]

(We need to use the absolute value of z³ since the z term was original to the power of 6, which means the value of z⁶ is always positive since the exponent is even).

P(x) goes to infinity as x goes to negative infinity, and P(x) goes to negative infinity as x goes to infinity , Option B is the correct answer.

A polynomial is an expression which consists of variables , Coefficient , indeterminate and mathematical operations.

The polynomial given has  lead term -x⁷

So when x --> ∞  , -x⁷---> - ∞

and when x ---> - ∞  , -x⁷ ---->  ∞

Similarly for the function P(x)

So when x --> ∞  , P(x)---> - ∞

and when x ---> - ∞  ,  P(x) ---->  ∞

P(x) goes to infinity as x goes to negative infinity, and P(x) goes to negative infinity as x goes to infinity

Therefore Option B is the correct answer.

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11/36 would be the probability of getting 4 , when two dice are thrown simultaneously.

As there are 6 faces of die,
So Probability of Not getting four in one dice = [tex]5/6[/tex]

As both events are mutually exclusive, the probability of not getting four in two dice = [tex](5/6) * (5/6) = 25/36[/tex]

So probability of getting at least one four is = [tex]1-(25/36) = 11/36[/tex]

So probability = [tex]11/36[/tex]

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Answer:

b

Step-by-step explanation:

Answer: 80%

Step-by-step explanation:

Since only 2 out of 10 women dropped out, 10 - 2 = 8

One woman = 10%, 8 x 10% = 80%

The percentage of the women who moved to the next stage is 80%.

Percentage can be described as a fraction out of an amount that is usually expressed as a number out of hundred.

Women who moved to the next stage: 10 - 2 = 8

(8/10) x 100 = 80%

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[tex]\huge\text{Hey there!}[/tex]

[tex]\mathsf{(7\times 10^5) + (4 \times 10^2)}\\\mathsf{= 7\times 10^5 + 4 \times 10^2}[/tex]

[tex]\large\textsf{Let's solve for the LEFT SIDE first}[/tex]

[tex]\mathsf{7\times 10^5}\\\\\mathsf{10^5}\\\mathsf{= 10\times10\times10\times10\times10}\\\mathsf{= 100 \times 100\times 10}\\\mathsf{= 10,000 \times 10}\\\mathsf{= 100,000}\\\\\mathsf{7\times100,000}\\\mathsf{= 700,000}[/tex]

[tex]\large\textsf{NEW EQUATION: }\bold{700,000} + {4}\times 10^2[/tex]

[tex]\large\textsf{Now, let's solve for the RIGHT SIDE last}[/tex]

[tex]\mathsf{4\times 10^2}\\\\\mathsf{10^2}\\\mathsf{= 10\times10}\\\mathsf{= 100}\\\\\mathsf{4\times100}\\\mathsf{= 400}[/tex]

[tex]\large\text{NEW EQUATION: 700,000 + \bf 400}[/tex]

[tex]\large\text{700,000 + 400}\\\\\large\text{SIMPLIFY IT!}\\\\\large\text{= 700,400}[/tex]

[tex]\huge\text{Therefore, the answer should be: \boxed{\mathsf{700,400}}}\huge\checkmark[/tex]

[tex]\huge\text{Good luck on your assignment \& enjoy your day!}[/tex]

~[tex]\frak{Amphitrite1040:)}[/tex]