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

The coterminal angle to -ㅠ/2 in the range between 0 and 2ㅠ radians is 3ㅠ/2.

For any angle A, all the coterminal angles are of the form:

B = A + n*(2ㅠ).

Where n is an integer different than zero.

In this case, the angle is A = -ㅠ/2

So the coterminal angles are of the form:

B = -ㅠ/2 + n*(2ㅠ)

We want it to be between 0 and 2ㅠ, so we can define n = 1, then we get:

B = -ㅠ/2 + (2ㅠ) = -ㅠ/2 + ((4/2)ㅠ) = 3ㅠ/2

Then we can conclude that the correct option is b.

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The statement that is false is 15% of her students are in biology, the correct option is A.

Suppose two dimensions are there, viz X and Y. Some values of X are there as X_1, X_2, ... , X_n and some values of Y are there as Y_1, Y_2, ... , Y_n

List them in title of the rows and left to the columns. There will be n \times k table of values will be formed(excluding titles and totals), such that:

Value(ith row, jth column) = Frequency for intersection of  X_i  and Y_j  (assuming X values are going in rows, and Y values are listed in columns).

Then totals for rows, columns, and whole table are written on bottom and right margin of the final table.

For n = 2, and k = 2, the table would look like:

[tex]\begin{array}{cccc}&Y_1&Y_2&\rm Total\\X_1&n(X_1 \cap Y_1)&n(X_1\cap Y_2)&n(X_1)\\X_2&n(X_2 \cap Y_1)&n(X_2 \cap Y_2)&n(X_2)\\\rm Total & n(Y_1) & n(Y_2) & S \end{array}[/tex]

where S denotes total of totals, also called total frequency.

n is showing the frequency of the bracketed quantity, and intersection sign in between is showing occurrence of both the categories together.

We are given that;

The relative frequency table

Now,

This is because the relative frequency table shows that 0.15 of her students are freshmen in biology and 0.2 of her students are sophomores in biology. Therefore, the total percentage of her students in biology is 0.15 + 0.2 = 0.35, which is 35%, not 15%.

The other statements are true based on the relative frequency table.

Therefore, by the two-way table the answer will be 15% of her students are in biology.

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mean is average, add all numbers and divide by quantity of numbers.

Median is the middle value, list the numbers from smallest to largest and find the middle number.

mode is the number that appears the most.

1. Mean = 6

Median = 5

Mode = 6

2. Mean = 86.1

Median = 86.5

Mode = 85

Answer:

h(x) = 3x(x - 2) - 4

Step-by-step explanation:

A quadratic function is a polynomial having a degree of 2.

==============================================================

On expanding h(x) :

⇒ h(x) = 3x (x - 2) - 4

⇒ h(x) = 3x² - 6x - 4

=============================================================

Hence, h(x) is a quadratic function.

Answer:

h(x) = 3x(x - 2) - 4

Step by-step explanation

A quadratic function has a degree of 2.

Firstly, we expand h(x):

h(x) = 3x (x - 2) - 4

OR h(x) = 3x squared - 6x - 4
Therefore, it is h(x)

The expression that is equivalent to (-11/15)⁵ is (-11)⁵/(15)⁵

Given the expression as shown below

(-11/15)⁵

This expression means that we are to raise the value -11 and 14 raised to the power of 5 individually as shown:

(-11/15)⁵ = (-11)⁵/(15)⁵

Hence the expression that is equivalent to (-11/15)⁵ is (-11)⁵/(15)⁵

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Answer:

Never

Step-by-step explanation:

If p is true and ~ q is false, then p -> ~ q is ____ false

This translates to:

true -> true

So it will never be false.

Answer:

its the Pentagon, the other shaps have 2 lines if symmetry

The error in the distance if the uppermost portion of a 3. 0m long stadia rod is inclined 14 cm toward the observer and the rod intercept is 1. 75m on a horizontal sight is 8.167cm.

The tangent or tanθ in a right angle triangle is the ratio of its perpendicular to its base. it is given as,

Tan(θ) = Perpendicular/Base

where,

θ is the angle,

Perpendicular is the side of the triangle opposite to the angle θ,

The base is the adjacent smaller side of the angle θ.

Given the uppermost portion of a that is 3.0m long stadia rod is inclined 14 cm toward the observer, therefore, the angle of inclination, θ can be written as

tan(θ)=0.14/3

θ = 2.672°

Also, the rod intercept is 1.75m, therefore, t the error in distance in the distance is,

tanθ=x/1.75

x = 1.75 tan(θ)

x = 1.75 × tan(2.672°)

x = 0.08167 m = 8.167 cm

Hence, the error in the distance if the uppermost portion of a 3. 0m long stadia rod is inclined 14 cm toward the observer and the rod intercept is 1. 75m on a horizontal sight is 8.167cm.

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[tex]\text{Apply Pythagorean theorem,}\\\\~~~~~~~~\text{Hypotenuse}^2=\text{Base}^2 + \text{Perpendicular}^2\\\\\implies 9^2 = \text{Base}^2 + \left(2\sqrt{14} \right)^2\\\\\implies 81 = \text{Base}^2 + 56\\\\\implies \text{Base}^2 = 81-56\\\\\implies \text{Base}^2 = 25\\\\\implies \text{Base} = \sqrt{25}\\\\\implies \text{Base} = 5\\\\\text{The length of the third side is 5 units.}[/tex]

Answer:

(4, 2)

Step-by-step explanation:

the midpoint of points (x1, y1) and (x2, y2) is ([tex]\frac{x_{1} + x_{2}}{2}[/tex],[tex]\frac{y_{1} + y_{2}}{2}[/tex])

plug in points (-2, -4) and (10, 8) to get ((10+(-2))/2), (8+(-4))/2) = (4, 2)

The distance travelled from the starting point is 29 km , the bearing is 344 degree

The missing diagram is attached with the answer.

It is the horizontal angle between the object and another object.

It is given that the car starts from point travels for 30km on a bearing of 145, then 10 km on a bearing of 250

(a) the distance travelled from the starting point

It can be understood from the diagram that

To determine the distance , b is to determined4

By applying Cosine rule

b² = a² +c²-2ac cos[tex]\rm \theta[/tex]

b² = 10²  +30²  - 2* 30*10 cos75

b² = 100+900 -600* 0.2588

b² = 100-155.2914

b² = 844.7

b = 29 km

Therefore the distance travelled from the starting point is 29 km.

(b) the bearing of the starting point from the car.

To determine the bearing ,

angle C has to be determined

By sine rule

sin C / c = sin B / b

sin C = sin 75 * 30 / 29

sin C = 0.9970

C = sin⁻¹ 0.9970

C = 85.59 degree

angle C = α +70 degree

α  = 15.59 degree

Bearing = 70 + 20 + 9 0 + 90+74 = 244 degree

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The measure of angle that is coterminal with 425° is 425° + (1440n)°

Coterminal angles are angles in multiples of the angles in standard positions examples(360°).

It is mostly gotten by adding  360° or multiples of 360 to the angle.

One of the unique behavior of coterminal angles is that their sine, cosine and tangent are equal.

Analysis:

From the options, the only one with an addition of 360 or multiples of 360 is 425 + 1440n

If we put n = 1, the coterminal angle gotten is 1865.

if we find the sine of 425°, we get 0.9063 which is same as the sine of 1865°.

In conclusion, the coterminal angle to 425° is 425° + (1440n)°.

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Answer:

B. 425° – (840n)°, for any integer n

[tex]\begin{array}{|c|ll} \cline{1-1} \textit{a\% of b}\\ \cline{1-1} \\ \left( \cfrac{a}{100} \right)\cdot b \\\\ \cline{1-1} \end{array}~\hspace{5em}\stackrel{\textit{20\% of 20000}}{\left( \cfrac{20}{100} \right)20000}\implies 4000\qquad \impliedby downpayment[/tex]

When the sum of two angles measures up to 180° then these angles are known as supplementary angles of each other. The supplementary angles are 153° and 27°. The Complementary angles are 34° and 56°.

When the sum of two angles measures up to 180° then these angles are known as supplementary angles of each other.

for example, ∠x + ∠y = 180°, therefore, the ∠x and ∠y are the supplementary angles of each other.

When the sum of two angles measures up to 90° then these angles are known as complementary angles of each other.

for example, ∠x + ∠y = 90°, therefore, the ∠x and ∠y are the complementary angles of each other.

The supplementary angles are 153° and 27°.

The Complementary angles are 34° and 56°.

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By the remainder theorem of polynomial division, the complete equation is f(-3) = 11

The equation is given as:

f(x)/x + 3

Set the divisor to 0

x + 3 = 0

Solve for x

x = -3

Given that the quotient has a remainder of 11.

It means that:

f(-3) = 11

Hence, the complete equation is f(-3) = 11

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Answer:

B

Step-by-step explanation:

(f + g)(x)

= f(x) + g(x)

= 2x + 1 + x² - 7 ← collect like terms

= x² + 2x - 6 ← in standard form

The value of (f + g)(x) is x² + 2x - 6.

Arithmetic operations are a set of four basic operations to be performed to add, subtract, multiply or divide two or more quantities. They include the study of numbers including order of operations which are useful in all the other parts of mathematics such as algebra, data handling, and geometry.

Given are two functions, f(x) = 2x+1 and g(x) = x²-7, we are asked to find

(f + g)(x),

So we know,

(f + g)(x) = f(x) + g(x)

Therefore,

(f + g)(x)

= f(x) + g(x)

= 2x + 1 + x² - 7

= x² + 2x - 6

Hence the value of (f + g)(x) is x² + 2x - 6.

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Answer:

x = 3

Step-by-step explanation:

2x + 12 = 18

2x = 18-12

2x = 6

x = 3

give brainliest please!

hope htis helps :)

Answer:6

Step-by-step explanation:

0x+12+x=18

0x+x=18-12

x=6

Answer:

Yes.

Step-by-step explanation:

When [tex]x=15[/tex],

[tex]x+9=15+9=24 > 21[/tex].

Answer:

Yes this is true

Step-by-step explanation:

24 is greater than 21; which means this statement is always true.

The answers of the given geometrical shapes are:

1. 226.29 cubic yard

2. 46.1 cubic inches

3. 2413.71 cubic inches

4. 234.67 cubic millimeter

5. 340.48 cubic yards

6. 113.14 cubic feet

7. 528 cubic centimeter

8. 905.14 cubic centimeter

9. 56.57 cubic feet

Mensuration is the branch of geometry that deals with measurement of volumes and areas of geometrical shapes.

Analysis:

1. volume of cylinder =π [tex]r^{2}[/tex]h. where r = 6/2 = 3 yard, h = 8 yard

                                  =  [tex]\frac{22}{7}[/tex] x [tex]3^{2}[/tex] x 8 = 226.29 cubic yard

2. volume of cone = [tex]\frac{1}{3}[/tex] x π  [tex]r^{2}[/tex]h. where r = 2inches, h = 11

                              =  [tex]\frac{1}{3}[/tex] x [tex]2^{2}[/tex] x   [tex]\frac{22}{7}[/tex]  x 11 = 46.1 cubic inches

3. volume of cylinder = π [tex]r^{2}[/tex]h. where r = 8, h = 12

                                   =   [tex]\frac{22}{7}[/tex] x [tex]8^{2}[/tex] x 12 = 2413.71 cubic inches

4. volume of cone =  [tex]\frac{1}{3}[/tex] x π  [tex]r^{2}[/tex]h. where r = 4, h = 14

                               =  [tex]\frac{1}{3}[/tex]  x  [tex]\frac{22}{7}[/tex] x [tex]4^{2}[/tex] x 14 = 234.67 cubic millimeter

5. volume of cone =  [tex]\frac{1}{3}[/tex] x π  [tex]r^{2}[/tex]h. where r = 5, h = 13

                              =  [tex]\frac{1}{3}[/tex] x  [tex]\frac{22}{7}[/tex] x [tex]5^{2}[/tex] x 13 = 340.48 cubic yards

6. volume of cylinder = π [tex]r^{2}[/tex]h. where r = 3, h = 4

                                   = [tex]\frac{22}{7}[/tex] x [tex]3^{2}[/tex] x 4 = 113.14 cubic feet

7. volume of cone = [tex]\frac{1}{3}[/tex] x π  [tex]r^{2}[/tex]h.  r = 6, h = 14

                               =  [tex]\frac{1}{3}[/tex] x  [tex]\frac{22}{7}[/tex] x [tex]6^{2}[/tex] x 14 = 528 cubic centimeter

8. volume of cylinder = π [tex]r^{2}[/tex]h. where r = 6, h = 8

                                   =   [tex]\frac{22}{7}[/tex] x [tex]6^{2}[/tex] x 8 = 905.14 cubic centimeter

9. volume of cylinder =  π [tex]r^{2}[/tex]h. where r = 1.5, h = 4

                                    =   [tex]\frac{22}{7}[/tex]  x [tex]1.5^{2}[/tex] x 4 = 56.57 cubic feet

In conclusion, the volume of cylinders and cones for the given dimensions are:

1. 226.29 cubic yard

2. 46.1 cubic inches

3. 2413.71 cubic inches

4. 234.67 cubic millimeter

5. 340.48 cubic yards

6. 113.14 cubic feet

7. 528 cubic centimeter

8. 905.14 cubic centimeter

9. 56.57 cubic feet

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The number of potential customers that the company has reached is 10,000  e--mail addresses.

A word problem is a real-life problem and it can be approached with the use of a mathematical model, interpretation into variables, and arithmetic operations usage.

Given that:

The number of potential customers that the company has reached is:

= 2500  3500 + 4000 e--mail addresses.

= 10,000  e--mail addresses.

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The correct solution to the limits of x in the tiles can be seen below.

The limits of x approaching a given number of a quadratic equation can be determined by knowing the value of x at that given number and substituting the value of x into the quadratic equation.

From the given diagram, we have:

1.

[tex]\mathbf{ \lim_{x \to 9^+} (\dfrac{|x-9|}{-x^2-34+387}) }[/tex]

So, x - 9 is positive when x → 9⁺. Therefore, |x -9) = x - 9

[tex]\mathbf{ \lim_{x \to 9^+} (\dfrac{x-9}{-x^2-34+387}) }[/tex]

Simplifying the quadratic equation, we have:

[tex]\mathbf{ \lim_{x \to 9^+} (-\dfrac{1}{x+43}) }[/tex]

Replacing the value of x = 9

[tex]\mathbf{ = (-\dfrac{1}{9+43}) }[/tex]

[tex]\mathbf{ = -\dfrac{1}{52} }[/tex]

2.

[tex]\mathbf{ \lim_{x \to 8^-} (\dfrac{8-x}{|-x^2-63x+568|}) }[/tex]

Thus |-x²-63x+568| = -x²-63x+568

[tex]\mathbf{ \lim_{x \to 8^-} (\dfrac{1}{x+71}) }[/tex]

[tex]\mathbf{=\dfrac{1}{8+71} }[/tex]

[tex]\mathbf{=\dfrac{1}{79} }[/tex]

3.

[tex]\mathbf{ \lim_{x \to 7^+} (\dfrac{|-x^2-17x+168| }{x-7}) }[/tex]

[tex]\mathbf{ \lim_{x \to 7^+} (\dfrac{-x^2-17x+168 }{x-7}) }[/tex]

[tex]\mathbf{ \lim_{x \to 7^+} (-x-24)}[/tex]

[tex]\mathbf{ \lim_{x \to 7^+} (-7-24)}[/tex]

= -31

4.

[tex]\mathbf{ \lim_{x \to 6^-} (\dfrac{|x-6| }{-x^2-86x+552}) }[/tex]

[tex]\mathbf{ \lim_{x \to 6^-} (\dfrac{-x+6 }{-x^2-86x+552}) }[/tex]

[tex]\mathbf{ \lim_{x \to 6^-} (\dfrac{1}{x+92}) }[/tex]

[tex]\mathbf{ \lim_{x \to 6^-} (\dfrac{1}{6+92}) }[/tex]

[tex]\mathbf{ =\dfrac{1}{98}}[/tex]

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Answer:

The Ganges

Step-by-step explanation:

Answer:

Ganga

Step-by-step explanation:

Answer:

Negative; it describes a violent reaction based on rage.

Step-by-step explanation:

Answer:

False

Step-by-step explanation:

Protons have positive charge.

Answer:

Protons have a positive charge.

Step-by-step explanation:

The domain of the function when Toni participates in a 10-mile charity walk will be [0, 9, 18, 27, 36, 45, 54, 63, 72, 81, 90]

It should be noted that a function simply means a relation that's used to express a variable with another variable.

In this case, it should be noted that a domain simply means the set of possible values of the independent variable. The function here is 9x. The domain is all x-values or inputs of a function. Therefore, the domain is depicted above.

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[tex]\dfrac{7x^2+14}{(x^2+3)^2}=\dfrac{7x^2+21-7}{(x^2+3)^2}=\dfrac{7x^2+21}{(x^2+3)^2}-\dfrac{7}{(x^2+3)^2}=\dfrac{7(x^2+3)}{(x^2+3)^2}-\dfrac{7}{(x^2+3)^2}=\\=\dfrac{7}{x^2+3}-\dfrac{7}{(x^2+3)^2}[/tex]

Answer:

18

Step-by-step explanation:

Mode =  modal class (Recurring)
The number/value that appears the most which in this case is 18 because it can be seen twice.