Solve the inequality. 8x-5>_27. A.x>_4. B.x>_11/4. C.x<_4. D.x<_11/4

Answer:

16x^4 - 96x^3y + 216x^2y^2 - 216xy^3 + 81y^4

Step-by-step explanation:

(2x - 3y)^4

Fifth line on a Pascal Triangle

1, 4, 6 4, 1

(1) 2x^4

2^4 = 16

2x^4 = 16x^4

16x^4

(4) 2x^3 (-3y)^1

2^3 = 8

-3^1 = -3

8 times -3 times 4 = -96

-96x^3y

(6) 2x^2 (-3y)^2

2^2 = 4

-3^2 = 9

4 times 9 times 6 = 216

216x^2y^2

(4) 2x^1 (-3y)^3

2^1 = 2

-3^3 = -27

2 times - 27 times 4 = -216

-216xy^3

(1) (-3y)^4

-3^4 = 81

81y^4

16x^4 - 96x^3y + 216x^2y^2 - 216xy^3 + 81y^4

Answer:

81^1/12

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The expression ''4 square root of 81³'' can be rewritten as,

⇒ [tex]81^{\frac{3}{4} }[/tex]

Mathematical expression is defined as the collection of the numbers variables and functions by using operations like addition, subtraction, multiplication, and division.

Given that;

The expression to write is,

⇒ 4 square root of 81³

Now, It can be written as;

⇒ 4 square root of 81³

⇒ [tex]\sqrt[4]{81^{3} }[/tex]

By rule of exponent we get;

⇒  [tex]81^{\frac{3}{4} }[/tex]

Thus, The expression ''4 square root of 81³'' can be rewritten as,

⇒ [tex]81^{\frac{3}{4} }[/tex]

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To find the area of the region completely bounded by the curve y=-x^2+x+6, you can integrate the equation with respect to x and evaluate it between the x-values where the curve intersects the x-axis. By solving the quadratic equation -x^2+x+6=0, you can determine the x-values. Then, evaluate the definite integral between these x-values to find the area.

The area of the region completely bounded by the curve y=-x^2+x+6 can be found by integrating the equation with respect to x and evaluating it between the appropriate bounds. The integral of the given equation is ∫(-x^2+x+6) dx. To find the area, we need to find the definite integral between the x-values where the curve intersects the x-axis. First, set the equation equal to zero and solve for x:

-x^2+x+6=0

This quadratic equation can be factored as: (x-2)(x+3). Therefore, the curve intersects the x-axis at x=2 and x=-3.

By evaluating the definite integral between x=-3 and x=2, we can find the area of the region:

Area = ∫-32 (-x^2+x+6) dx

Integrating this equation will give you the area of the region bounded by the curve y=-x^2+x+6.

Answer:

b and c

Step-by-step explanation:

2020 edge assignment

Answer:

Step-by-step explanation:

option b

Given a polynomial with rational coefficients, if √3 is a root, its irrational conjugate -√3 must also be a root due to the Irrational Conjugates Theorem, ensuring that the polynomial maintains rational coefficients.

The subject of this question is a polynomial function with rational coefficients, which is encountered in mathematics, particularly in algebra. When a polynomial has rational coefficients and one of its roots is an irrational number, such as the square root of 3 (√3), the Irrational Conjugates Theorem states that its conjugate, in this case, the negative square root of 3 (-√3), must also be a root of the polynomial. Therefore, the polynomial function in question must include both √3 and -√3 as roots to have rational coefficients.

Given that the function already has -2 as a root, we know that (x + 2) is a factor of the polynomial. Additionally, because √3 is a root, the factors (x - √3) and (x + √3) will also be part of the polynomial. To find the roots of a quadratic equation or higher-order polynomials, one can set the function equal to zero and solve for x, either by factoring, applying the quadratic formula, or other algebraic methods.

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The equation of the line parallel to y=-2/3x+4 and passing through (-2, -2) is y = -2/3x - 10/3.

To find the equation of a line parallel to y = -2/3x + 4 and passing through the point (-2, -2), we need to use the fact that parallel lines have the same slope. The given equation has a slope of -2/3, so the parallel line will also have a slope of -2/3. Using the point-slope form of a line, we can write the equation as:

y - y1 = m(x - x1)

where (x1, y1) is the given point and m is the slope. Plugging in the values, we have:

y - (-2) = -2/3(x - (-2))

Simplifying the equation, we get:

y - (-2) = -2/3(x + 2)

y + 2 = -2/3x - 4/3

y = -2/3x - 4/3 - 2

y = -2/3x - 4/3 - 6/3

y = -2/3x - 10/3

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Final answer:

Carol practices for 8.5 hours per week, so in a 5-week period, she would practice for a total of 42.5 hours.

Explanation:

The student asked how many hours Carol practices her culinary skills in a 5-week period, if she practices for 17 hours in a 2-week period. To find the answer, we calculate how many hours Carol practices per week by dividing the total hours she practices in two weeks by two. Then we multiply the weekly hours by the number of weeks in question, which is five.

The calculation is as follows: Carol practices for 17 hours / 2 weeks = 8.5 hours per week. Then, 8.5 hours/week x 5 weeks = 42.5 hours in total for a 5-week period.

Answer:  Second option is correct.

Step-by-step explanation:

Since we have given that

ΔABC has measures a=2, b=2, m∠A=30⁰

As we know the "Law of sines " i.e.

[tex]\frac{a}{\sin A}=\frac{b}{\sin B}\\[/tex]

so, we put the given values in above formula:

[tex]\frac{2}{\sin 30\textdegree}=\frac{2}{\sin B}\\\\\implies \sin 30\textdegree=\sin B\\\\\implies B=30\textdegreee[/tex]

Hence, Second option is correct.

The velocity of the plane is 200 mph due north and the velocity of the wind is 23 mph due west. The velocity vector of the plane is 200 mph due north minus 23 mph due west. The ground speed of the plane can be found using the Pythagorean theorem.

Part I: The velocity of the plane can be represented as 200 mph due north, and the velocity of the wind can be represented as 23 mph due west.

Part II: To find the velocity vector of the plane, we subtract the velocity of the wind from the velocity of the plane. The resultant velocity vector of the plane is 200 mph due north minus 23 mph due west.

Part III: The ground speed of the plane is the magnitude of the resultant velocity vector of the plane. We can calculate it using the Pythagorean theorem: ground speed = square root of (200^2 + 23^2).

Final answer:

To prove that the diagonals of a square are congruent and perpendicular, label the vertices of a square on a coordinate grid and calculate the slopes and lengths using the slope formula and distance formula respectively. The diagonals have slopes of +1 and -1, proving they are perpendicular, and they have equal lengths, proving they are congruent.

Explanation:

To prove that the diagonals of a square are congruent and perpendicular, we place a square with its vertices on a coordinate grid and label each vertex with variables.

Let's consider a unit square where c = 1 for simplicity, which means the length of each side is 1 unit. Place the square so that one vertex is at the origin (0,0), and label the vertices A(0,0), B(1,0), C(1,1), and D(0,1).

The diagonal AC will have endpoints at A(0,0) and C(1,1), and diagonal BD will have endpoints at B(1,0) and D(0,1). The slope of diagonal AC is (1 - 0)/(1 - 0) = 1, and the slope of diagonal BD is (1 - 0)/(0 - 1) = -1. Since the product of their slopes is -1 (1 * -1 = -1), this proves that they are perpendicular to each other.

To show they are congruent, we calculate their lengths using the distance formula: the distance between two points (x1,y1) and (x2,y2) is √[(x2 - x1)² + (y2 - y1)²]. Applying this to AC and BD reveals both lengths to be √[(1-0)² + (1-0)²] = √[1 + 1] = √2, proving the diagonals are congruent.

Yes, it is possible to have more than one bisector in a line segment.

Bisector is a line that divides a line or an angle in to two equivalent parts. There are two types of Bisectors based on what geometrical shape it bisects.

 In general 'to bisect' something means to cut it into two equal parts. The bisector  is the one that doing the cutting process.

With a line bisector, we cut a line segment into two equal parts with another line - the bisector. Just imagine the line PQ is being cut into two equal lengths (PF and FQ) by the bisector line AB.

Whenever AB intersects at a right angle, it is called the "perpendicular bisector" of PQ. If it crosses at any other angle it is simply called a bisector. Drag the points A or B and see both types.

For obvious reasons, the point F is called the midpoint of the line PQ,

To solve this problem, we have to manually solve for the value of x for each choices or equations. The correct equation will give a value of -1 since the linear equations intersects at point (-1, -4).

1st: 7x + 3 = x + 3

7x – x = 3 – 3

6x = 0

x = 0                (FALSE)

2nd: 7x − 3 = x – 3

7x – x = 3 – 3

6x = 0

x = 0                (FALSE)

3rd: 7x + 3 = x − 3

7x – x = - 3 – 3

6x = -6

x = -1               (TRUE)

4th: 7x − 3 = x + 3

7x – x = 3 + 3

6x = 6

x = 1                (FALSE)

Therefore the answer is:

7x + 3 = x − 3

In this exercise, we are going to solve using our knowledge of systems and in this way we will find that the equation that satisfies the points.

As we know that the equation that will satisfy will have to have the values ​​of X=-1, we will solve each one of the alternatives as:

[tex]7x + 3 = x + 3\\6x = 0\\x = 0[/tex]

We realize that the value of x is not what we want so it doesn't satisfy us.

[tex]7x - 3 = x - 3\\7x- x = 3- 3\\6x = 0\\x = 0[/tex]

We realize that the value of x is not what we want so it doesn't satisfy us.

[tex]7x + 3 = x − 3\\6x = -6\\x = -1[/tex]

[tex]7x − 3 = x + 3\\7x – x = 3 + 3\\6x = 6\\x = 1[/tex]

We realize that the value of x is not what we want so it doesn't satisfy us.

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The area of the composite figure is 6π + 16 cm²

Composite figures are composed of different dimensional figures. The area of a composite figure is the sum of the whole 2 dimensional figures that forms the composite figure.

Therefore, the figure above has 3 semi circle and 1 square.

Therefore, the area can be calculated as follows;

area = 1 / 2 πr² + 1 / 2 πr² + 1 / 2 πr² + L²

area = 3 / 2 (πr²) + L²

where

r = 2 cm

L = 4 cm

Therefore,

area of the composite figure = 3 / 2(π × 4) + 4²

area of the composite figure = 3 / 2(4π) + 16

area of the composite figure =  6π + 16 cm²

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