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When the sum of two numbers is 22, and their difference is 12, the lesser of the two numbers is 5.

An equation is a formula in mathematics that expresses the equality of two expressions by connecting them with the equals sign =. In its most basic form, an equation is a mathematical statement that shows that two mathematical expressions are equal. 3x + 5 = 14, for example, is an equation in which 3x + 5 and 14 are two expressions separated by a 'equal' sign. A mathematical equation that depicts the relationship between two expressions on opposite sides of the sign. It mostly consists of one variable and one equal to symbol. 2x - 4 = 2 is an example.

Here,

Let numbers be a and b.

a+b=22

a=22-b

b-a=12

b-(22-b)=12

2b-22=12

2b=34

b=17

a=5

The lesser of the two numbers is 5 when the sum of two numbers is 22, and their difference is 12.

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Answer:

B. Seraphina is not correct. The sum of the squares of the two legs of the triangle is not equal to the square of the hypotenuse.

Step-by-step explanation:

If KLM is a right triangle, then 12^2+16^2=19^2,

144+256 = 361

400 = 361 false

so, KLM is not a right triangle as the sum of the squares of the two legs are not equal to the square of the hypotenuse.

The sum of the squares of the two legs of the triangle is not equal to the square of the hypotenuse. Seraphina is incorrect.

If ABC is a triangle with AC as the hypotenuse and angle B with 90 degrees then we have:

|AC|^2 = |AB|^2 + |BC|^2    

where |AB| = length of line segment AB. (AB and BC are the rest of the two sides of that triangle ABC, AC being the hypotenuse).

Seraphina says that ΔKLM is a right triangle.

Let us check whether triangle ΔKLM is a right triangle or not.

By the Pythagoras theorem, we have

19² = 16² + 12²

361 = 256 + 144

361 ≠ 400

Thus, the triangle is not a right triangle.

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Answer: correct me if I’m wrong

Step-by-step explanation:

The range of the function is all real numbers.

In mathematics, a function is an expression, rule, or law that specifies a relationship between two variables (the independent variable and the dependent variable). Functions are common in mathematics and are required for the formulation of physical relationships in the sciences.

We know,

The x-values (Input values) on the graph define the domain of the function.

and, The y-values (output values) on the graph determine the range of a function.

From the graph attached,

Here the x value start from +1 to ∞.

Thus, the range of the function is all real numbers.

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Answer:

c.the mean of each data set

Answer:

A

Step-by-step explanation:

Answer: x>-1

Step-by-step explanation:

Isolate the variable by dividing each side by factors that don't contain the variable.

The property applied is the associative property

There are several algebraic properties in mathematics; each of which have their own applications

Some of these properties are:

The mathematical expression is given as:

\(13 \cdot (20 \cdot 5)\)

And it is calculated as follows:

\(13 \cdot (20 \cdot 5) = 13 \cdot 100\)

\(13 \cdot (20 \cdot 5) = 13 00\)

The property applied above is the associative property

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The distance from the north most ship to the observation is 2.9 miles and the distance from the south most ship to the observation is 9.9 miles.

The bearing is defined as the angle measured clockwise from the north direction. A ship is anchored off a long straight shoreline that runs north and south.

From two observation points miles apart on shore, the bearings of the ship are 52 degrees and 134 degrees respectively. To find the distance from the ship to each of the observation points, we can use trigonometry.

Let's call the distance from the north observation point to the ship x, and the distance from the south observation point to the ship y. The bearings from the north and south observation points can be drawn as follows:

[asy]

unitsize(1cm);

pair O = (0,0), N = (-2,0), S = (2,0), A = (-2,1), B = (2,-1);

draw(O--N--A,Arrow);

draw(O--S--B,Arrow);

\(label("$52^\circ$", N + 0.3*dir(52), NE);\)

\(label("$134^\circ$", S + 0.3*dir(134), SE);\)

draw(O--A--B--cycle,dashed);

\(label("$x$", (N+A)/2, W);\)

\(label("$y$", (S+B)/2, E);\)

[/asy]

We can use the tangent function to find x and y, since we have the angle and opposite side.

From the north observation point, we have:

\($$\tan(52^\circ) = \frac{x}{2}$$$$x = 2\tan(52^\circ) \approx 2.95$$\)

From the south observation point, we have:

\($$\tan(134^\circ) = \frac{y}{2}$$$$y = 2\tan(134^\circ) \approx 9.88$$\)

Therefore, the distance from the north most ship to the observation is 2.9 miles and the distance from the south most ship to the observation is 9.9 miles.

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Answer:

2

Step-by-step explanation:

The common difference is the difference between two consecutive terms in an arithmetic progression

Rounding the answer to two decimal places, the relative frequency of students with red hair among the sample population of students with gray eyes is approximately 0.63. Option D

To find the relative frequency of students with red hair among the sample population of students with gray eyes, we need to divide the number of students with gray eyes and red hair by the total number of students with gray eyes.

From the given table, we can see that there are 22 students with gray eyes and red hair.

The total number of students with gray eyes is the marginal total for gray eyes, which is 35.

To find the relative frequency, we divide the number of students with gray eyes and red hair by the total number of students with gray eyes:

Relative frequency = Number of students with gray eyes and red hair / Total number of students with gray eyes

Relative frequency = 22 / 35

Simplifying the fraction, we have:

Relative frequency = 0.6286

Rounding the answer to two decimal places, the relative frequency of students with red hair among the sample population of students with gray eyes is approximately 0.63.

Therefore, the correct answer is option OD) 0.63.

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In mathematics, a sequence is an ordered list of elements, often denoted by brackets or parentheses.

The term "a sequence of length n" refers to a sequence that contains a specific number of elements, where the number of elements is denoted by "n."

When the domain of a function is the set of natural numbers (N), it is called an infinite sequence. An infinite sequence can be thought of as an ordered list that continues indefinitely.

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\(6i~~ + ~~2\sqrt{3}j\implies < \stackrel{a}{6}~~,~~\stackrel{b}{2\sqrt{3}} > \\\\[-0.35em] ~\dotfill\\\\ \stackrel{magnitude}{\sqrt{a^2+b^2}}\implies \sqrt{6^2+(2\sqrt{3})^2}\implies \sqrt{36+(2^2\cdot 3)}\implies \sqrt{36+12}\implies \sqrt{48} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{direction}{tan^{-1}\left( \cfrac{b}{a}\right)}\implies tan^{-1}\left( \cfrac{2\sqrt{3}}{6} \right)\implies tan^{-1}\left( \cfrac{\sqrt{3}}{3} \right) \\\\\\ tan^{-1}\left( \cfrac{1}{\sqrt{3}} \right)\implies 30^o\)

a) The cumulative distribution function F of X is: F(x) = P(X ≤ x) = 1

b) The mean E(X) is: E(X) = 0.5 hours.

c) The variance Var(X) is: the variance of X is approximately 0.0833 hours².

To find the cumulative distribution function (CDF) of X, we need to consider two cases: when the boy comes back inside before 45 minutes and when the boy is brought inside by his mother after 45 minutes.

(a) Cumulative Distribution Function (CDF) F of X:

For 0 ≤ x < 0.75 (when the boy comes back inside before 45 minutes):

Since the boy comes back inside uniformly distributed on the interval [0, 1], the CDF is simply the probability that the boy comes back inside before time x.

The probability of the boy coming back inside before x is equal to x, as it is a uniform distribution.

F(x) = P(X ≤ x) = x

For 0.75 ≤ x ≤ 1 (when the mother brings the boy inside after 45 minutes):

If the boy is not back inside at 45 minutes (0.75 hours), his mother brings him inside.

Therefore, the probability of the boy coming back inside before x in this case is 1.

F(x) = P(X ≤ x) = 1

(b) Mean (Expected Value) E(X):

The expected value of X, denoted by E(X), is the average time the boy spends outside before coming back inside.

For a uniform distribution on the interval [0, 1], the expected value is the midpoint of the interval.

Therefore, E(X) = (1 + 0) / 2 = 0.5 hours.

(c) Variance Var(X):

The variance of X, denoted by Var(X), measures the spread or variability of the random variable X around its mean.

For a uniform distribution on the interval [0, 1], the variance is given by:

Var(X) = (1² - 0²) / 12 = 1 / 12 ≈ 0.0833 hours²

So, the variance of X is approximately 0.0833 hours².

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To determine whether the Law of Sines or the Law of Cosines is needed to solve the triangle, we can compare the given information with the formulas for each law.

The Law of Sines states:

a / sin(A) = b / sin(B) = c / sin(C)

The Law of Cosines states:

c^2 = a^2 + b^2 - 2ab * cos(C)

In this case, we are given the lengths of all three sides of the triangle (a = 6.0, b = 7.7, c = 13.6). Therefore, we have enough information to use the Law of Cosines to solve the triangle.

Using the Law of Cosines, we can find the measures of the angles:

c^2 = a^2 + b^2 - 2ab * cos(C)

(13.6)^2 = (6.0)^2 + (7.7)^2 - 2 * 6.0 * 7.7 * cos(C)

184.96 = 36 + 59.29 - 92.4 * cos(C)

184.96 = 95.29 - 92.4 * cos(C)

92.67 = -92.4 * cos(C)

cos(C) ≈ -1

Since the cosine of an angle cannot be greater than 1 or less than -1, it is not possible for the given triangle to have an angle with a cosine of -1. Therefore, the triangle is not solvable with the given side lengths.

In this case, the Law of Cosines is needed, but the triangle cannot be solved with the given information.

The value of x using the theorem of intersecting secants is 10

From the question, we have the following parameters that can be used in our computation:

intersecting secants

Using the intersecting secants equation, we have

8 * (8 + x) = 6 * (6 + 18)

Evaluate the like terms

So, we have

8 * (8 + x) = 6 * 24

Divide both sides by 8

8 + x = 6 * 3

So, we have

8 + x = 18

Subtract 8 from both sides

x = 10

Hence, the value of x is 10

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Answer:

the answer to pt 1 is D. x=1.5 x=0 or x=6

and the answer to part 2 is C. x=1±I√29/5

Step-by-step explanation:

1. The solution of this equation is x = 0 or x = 6 which is option D.

2. The solution of this equation is \(x = \dfrac{1 \pm i\sqrt{29}}{5}\) which is option C.

Use the concept of quadratic equation defined as:

The definition of a quadratic as a second-degree polynomial equation demands that at least one squared term must be included. It also goes by the name quadratic equations.

The quadratic equation has the following generic form:

ax² + bx + c = 0

1. The given quadratic equation is,

3x = 0.5x²

Rearranging it,

0.5x² - 3x = 0

Common out x,

x(0.5x - 3) = o

Then we have,

x = 0 or,  

0.5x - 3 = 0

Add 3 on both sides,

0.5x  = 3

Divide both sides by 0.5,

x = 3/0.5

x = 6

Hence the solution of this quadratic equation is,

x = 0 and x = 6 which is option D.

2. The given quadratic equation is:

0 = 5x² - 2x + 6

It can be written as,

5x² - 2x + 6 = 0

Since we know the quadrature formula for ax² + bx + c = 0 is,

\(x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

Here we have,

a = 5

b = -2

c = 6

Substitute values into the formula,

\(x = \dfrac{2 \pm \sqrt{4 - 120}}{10}\\\\x = \dfrac{1 \pm i\sqrt{29}}{5} \text{ }\text{ \text{ }} \text{\text{ }\ since } \sqrt{-1} = i\)

Hence, option C is correct.

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The complete question is:

Solve each of the quadratic equations.

PT. 1

3x = 0.5x²

A. x = -6 or x = 0

B. x = -4 or x = 3

C. x = -2 or x = 1.5

D. x = 0 or x = 6

PT.2

0 = 5x² - 2x + 6

A.  \(x = \dfrac{1 \pm 3i}{5}\)

B. \(x = \dfrac{1 \pm i\sqrt{11}}{5}\)  

C. \(x = \dfrac{1 \pm i\sqrt{29}}{5}\)

The value of x is 4.

To determine the value of x, we can use the concept of similarity between shapes.

Similar shapes have corresponding sides that are proportional to each other.

Given the dimensions of the first shape as 5, x, and 5, and the dimensions of the second shape as 30, 24, and 30, we can set up the following proportion:

5/x = 30/24

To solve for x, we can cross-multiply:

30 · x = 5 · 24

30x = 120

Dividing both sides of the equation by 30:

x = 120 / 30

x = 4

Therefore, the value of x is 4.

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Answer:

x ≈ 8098.08

Step-by-step explanation:

using the rule of logarithms

\(log_{b}\) x = n ⇒ x = \(b^{n}\)

note that ln represents \(log_{e}\)

given

9 = ln(x + 5) or

ln(x + 5) = 9 , or \(log_{e}\) (x + 5) = 9 , then

x + 5 = \(e^{9}\) ( subtract 5 from both sides )

x = \(e^{9}\) - 5 ≈ 8103.08 - 5 = 8098.08 ( to the nearest hundredth )

Answer: its -28

Step-by-step explanation:

i dont know i used a app

Therefore the percentage increase is 72

By using the formula for the distance we will see that the correct option is B:

√52

We will use the general formula to find the distance between two points (a, b) and (c, d)

it is:

distance = √( (a - c)^2 + (b - d)^2)

Here we want to find the distance between the points (-2,-3) and (4,1), if we use the above formula we will get:

distance = √( (-2 - 4)^2 + (-3 - 1)^2) = √( 36 + 16)

distance = √52

The correct option is B.

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A quadrilateral is a polygon with four sides and four vertices. It is a closed two-dimensional shape that can be classified into different types based on the lengths of its sides and the measures of its angles. Some common types of quadrilaterals include squares, rectangles, etc.

The statement is false, and a counterexample can be constructed. Consider a quadrilateral with opposite angles of 120° and 60°, such as an isosceles trapezoid with legs of length 1 and bases of length 2. The sum of the opposite angles is 180°, so they are supplementary. However, it can be shown that no circle can be circumscribed about this quadrilateral. Therefore, the correct answer is D: False, because the opposite angles form a linear pair.

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Answer:

The height of the flagpole is about 25.6 feet.

Step-by-step explanation:

The figure is omitted--please sketch it.

Put your calculator in degree mode.

Let h be the height of the flagpole.

\( \tan(52) = \frac{h}{20} \)

\(h = 20 \tan(52) = 25.6\)

The rate (in km/h) at which the distance from the plane to the radar station is increasing a minute later is 0 km/h (rounded to the nearest whole number).

To solve this problem, we can use the concepts of trigonometry and related rates.

Let's denote the distance from the plane to the radar station as D(t), where t represents time. We want to find the rate at which D is changing with respect to time (dD/dt) one minute later.

Given:

The plane is flying with a constant speed of 360 km/h.

The plane passes over the radar station at an altitude of 1 km.

The plane is climbing at an angle of 30°.

We can visualize the situation as a right triangle, with the ground radar station at one vertex, the plane at another vertex, and the distance between them (D) as the hypotenuse. The altitude of the plane forms a vertical side, and the horizontal distance between the plane and the radar station forms the other side.

We can use the trigonometric relationship of sine to relate the altitude, angle, and hypotenuse:

sin(30°) = 1/D.

To find dD/dt, we can differentiate both sides of this equation with respect to time:

cos(30°) * d(30°)/dt = -1/D^2 * dD/dt.

Since the plane is flying with a constant speed, the rate of change of the angle (d(30°)/dt) is zero. Thus, the equation simplifies to:

cos(30°) * 0 = -1/D^2 * dD/dt.

We can substitute the known values:

cos(30°) = √3/2,

D = 1 km.

Therefore, we have:

√3/2 * 0 = -1/(1^2) * dD/dt.

Simplifying further:

0 = -1 * dD/dt.

This implies that the rate at which the distance from the plane to the radar station is changing is zero. In other words, the distance remains constant.

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Answer:

2 dollars

Step-by-step explanation:

24-10=14

7 pencils cost 14 dollars

1 pencil costs 2 dollars

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Sequence= 4, 16

Difference=12

16+12=28

Answer is...

33+33=28

By completing the square, the vertex form of quadratic equation is (x - 1)² - 3.

In this question we find a quadratic equation in standard form, which must be modified into vertex form by completing the square, which consists in modifying part of the equation into a perfect square trinomial. The vertex form of the quadratic equation is:

y = C · (x - h)² + k

First, write the polynomial in standard form:

x² - 2 · x - 2

Second, use algebra properties to expand the expression:

(x² - 2 · x - 2) + 0

(x² - 2 · x - 2) + 3 - 3

(x² - 2 · x + 1) - 3

Third, use the definition of perfect square trinomial:

(x - 1)² - 3

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