Which expression is equivalent to √2+4?
OA. (2+4)²
OB. (2+4)²
OC. 6
D. 62

The simplest form of the expression given in the task content is; 5.5x +7.42.

The expression given in the task content is;

5.5(x+6 1/2)-(x+9 1/3) - (19-x)

Hence, the expression can be written in its simplest form as follows;

5.5x + 35.75 -x -9.33 -19 +x

= 5.5x +7.42.

Remarks: The complete task requires that the expression be written in its simplest form.

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Applying the angles of intersecting secants theorem, the measure of arc XY is: 32°.

According to the angles of intersecting secants theorem, m∠XZY = 1/2(arc VW - arc XY).

Plug in the values:

39 = 1/2(110 - arc XY)

2(39) = 110 - arc XY

78 = 110 - arc XY

78 - 110 = - arc XY

-32 = - arc XY

Arc XY = 32°

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Using the binomial distribution, the probabilities are given as follows:

a) 0.37 = 37%.

b) 0.5065 = 50.65%.

c) 0.3260 = 32.60%.

The formula is:

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]

The parameters are:

For this problem, the fixed parameter is:

p = 0.37.

Item a:

The probability is P(X = 1) when n = 1, hence:

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]P(X = 1) = C_{1,1}.(0.37)^{1}.(0.63)^{0} = 0.37[/tex]

Item b:

The probability is P(X = 3) when n = 3, hence:

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]P(X = 3) = C_{3,3}.(0.37)^{3}.(0.63)^{0} = 0.5065[/tex]

Item c:

The probability is P(X = 2) when n = 4, hence:

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]P(X = 2) = C_{4,2}.(0.37)^{2}.(0.63)^{2} = 0.3260[/tex]

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The dimensions of the length and width are 25 meters and 12. 5 meters respectively.

Area of a rectangle

area of rectangle = l × w

where

l = length

w = width

Note that,

Perimeter = 2w+ l

50 = 2w + l

l = 50  - 2w

Hence,

area = w(50 - 2w)

288 = 50w - 2w²

-2w² + 50w - 288 = 0

-w² + 25w - 144 = 0

The dimension w can be found as follows;

w = - b / 2a

where

a = -1

b = 25

w = - 25 / 2 ×  -1

w = 12.5 meters

Then, we have that

l = 50  - 2w

Substitute the value of w

l = 50 - 2(12.5)

l = 50 - 25

I = 25 meters

Thus, the dimensions of the length and width are 25 meters and 12. 5 meters respectively.

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In order to have a greater fertilizer coverage, you need to increase flow rate and: A. increase speed to approximately 7.1 mph so that you cover the field more quickly.

A flow rate is an average rate and it can be defined as the number of flow units that are created per unit time such as fertilizer coverage on a farm.

In this context, we can infer and logically deduce that you have to increase flow rate and your increase speed to approximately 7.1 mph, so that you can cover the field more quickly and have a greater fertilizer coverage.

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Complete Question:

The soil samples for the next field indicate that fertilizer coverage needs to be greater. To achieve this, you need to increase flow rate. How would you achieve this?

A. Increase speed to approximately 7.1 mph so that you cover the field more quickly

B. Increase the engine speed to approximately 2,000 rpm

C. Decrease speed to approximately 6.0 mph so that you cover the field more slowly

D. Shift to second gear so that the engine speed slows

Answer:

D

Step-by-step explanation:

It pass the vertical line test.

Hope this helps <3

As per the given graph, the graph fails the vertical line test. The correct option is B.

A graph is a visual representation of data that uses lines, bars, or other symbols to show how different values are related to each other.

Graphs are often used to display information in a way that is easy to understand and interpret.

The vertical line test is a way to determine whether a given graph represents a function.

According to this test, if any vertical line intersects the graph at more than one point, then the graph does not represent a function.

In the given graph, we can see that there are some vertical lines that intersect the graph at more than one point, such as the vertical line passing through x = 2.

Therefore, the graph does not pass the vertical line test and does not represent a function.

Thus, the correct answer is B.

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Answer:

12 years old

Step-by-step explanation:

Set up the following equation:

x + 2 = 2(x - 5) (distribute the 2)

x + 2 = 2x - 10 (subtract 2 from both sides)

x = 2x - 12 (subtract 2x from both sides)

-x = -12 (multiply both sides by negative 1)

x = 12

Brainliest, please :)

Answer: 12

Step-by-step explanation:

Let's algebraically represent this situation. Say his current age is x. We can rephrase the sentence to say that in two years, his age is the same as twice his age 5 years ago.

The phrase "in two years" means when his age is 2 more than his current age, or x + 2.

The phrase "same as" represents equality. The expressions on both sides have the same value.

The phrase "twice his age 5 years ago" is his age minus 5, multiplied by 2. We can write this as 2*(x-5).

Now that we have "translated" the equation from words to numbers, let's put it all in one equation and solve:

[tex]x+2=2(x-5)[/tex]

[tex]x+2=2x-10[/tex] [Distributive Property]

[tex]x+2-2x-2=2x-10-2x-2[/tex] [Adding -2x-2 to both sides]

[tex]-x=-12[/tex] [Combining like terms and simplifying]

[tex]x=12[/tex] [Multiplying both sides by -1 to remove the negative]

Hence, Billy's age is 12.

The equation of a hyperbola is  [tex]x^{2}[/tex]/9 - [tex]y^{2}[/tex]/4 = 1

What are the steps to Hyperbola equation ?

To find the equation of hyperbola, the following steps must be taken. You need to identify;

The general equation of a hyperbola can be expressed as

[tex]x^{2}[/tex]/[tex]a^{2}[/tex] - [tex]y^{2}[/tex]/[tex]b^{2}[/tex] = 1

From the graph, we have the following parameters

a = 3

[tex]a^{2}[/tex] = [tex]3^{2}[/tex] = 9

b = 2

[tex]b^{2}[/tex] = [tex]2^{2}[/tex] = 4

The equation of a hyperbola can be expressed as

[tex]x^{2}[/tex]/9 - [tex]y^{2}[/tex]/4 = 1

The vertices = V(+/-a,0) = (+/-3,0)

The center = C(0,0)

The focus = F(+/-C, 0)

Where [tex]C^{2}[/tex] = [tex]a^{2}[/tex]  + [tex]b^{2}[/tex]

C = [tex]\sqrt{9 + 4}[/tex]

C = [tex]\sqrt{13}[/tex]

Focus = F(+/- [tex]\sqrt{13}[/tex], 0)

The general equation for asymptote = +/- b/a X

= +/-2/3X

Therefore, the equation of a hyperbola can be expressed as

[tex]x^{2}[/tex]/9 - [tex]y^{2}[/tex]/4 = 1

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Answer:

3

Step-by-step explanation:

Numbers of beads used in making triangles:

3, 6, 10, 15, 21, 28, 36, ...

We need to pick a few numbers from those above that add to 41.

3 + 10 + 28 = 41

Answer: 3 triangles

Answer:

244 ft

Step-by-step explanation:

Scale: 1/2 in = 16 ft

Scale length = 7 5/8 inches

Real length = ?

We can use a proportion.

1/2 in : 16 ft = 7 5/8 in. : x

(1/2)/16 = (7 5/8)/x

(1/2)x = 16 × 7 5/8

(1/2)x = 122

x = 244

Answer: C. 244 ft  

The volume of the glasses is as follows;

The volume of cylinder glass 1 = πr²h

The volume of cylinder glass 1 = 3.14 × 3² × 6

The volume of cylinder glass 1 = 3.14 × 9 × 6

The volume of cylinder glass 1 = 169.56 cm³ = 170 cm³

The volume of the glass 2 = 2 / 3 π r³ + πr²h

The volume of the glass 2 = 2 / 3 × 3.14 × 3³ + 3.14 × 3² × 3

The volume of the glass 2 = 56.52 + 84.78

The volume of the glass 2 = 141.3 cm³ = 141 cm³

The volume of conical glass 3 = 1 / 3 πr²h

The volume of conical glass 3 = 1 / 3 × 3.14 × 3² × (√7² - 3² = √40)

The volume of conical glass 3 = 9.42 × √40

The volume of conical glass 3 = 59.5773111176

The volume of conical glass 3 = 60 cm³

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The value of x is -23 and the values of AB, BC, and AC are -33, 12 and -21

The equations are given as

AB = 13 + 2x

BC = 12

AC = x + 2

Point B is between A and C on Line AC.

So, we have

AC = AB + BC

This gives

x + 2 = 13 + 2x + 12

Evaluate the like terms

x - 2x = 13 + 12 - 2

This gives

-x = 23

Divide by -1

x =-23

Substitute x =-23 in AB = 13 + 2x and AC = x + 2

AB = 13 - 2 * 23 = -33

AC = -23 + 2 = -21

Hence, the value of x is -23 and the values of AB, BC, and AC are -33, 12 and -21

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Answer:

0

Step-by-step explanation:

sin²t + cos²t = 1, so since cos²t = 1, sin²t = 0, and thus sint = y = 0.

Answer:

A = 55

Step-by-step explanation:

take the two identical triangles and join them so that you get a rectangle wih sides 5 and 11. now multiply them togeter to get the are of the polynom

The area of triangle is 55 square centimeters.

Area is the amount of space occupied by a two-dimensional figure.

The formula for the area of triangle is

[tex]Area = \frac{1}{2} \times base\times height[/tex]

According to the given question.

We have a figure of a quadrilateral which is made up of four right angled tringles.

Now,

The area of a triangle with base 5cm and height 7 cm

= [tex]\frac{1}{2} \times 5\times 7[/tex]

[tex]=\frac{35}{2}[/tex]

[tex]= 17.5[/tex] square centimeters

And, the area of triangle with base 4cm and height 5cm

= [tex]\frac{1}{2} \times 4\times 5[/tex]

[tex]= 10[/tex] square centimeters

Since, the quadrilateral is made up of 2 right angled triangle with base 5cm and height 7cm and 2 right angled triangle with base 4cm and height 5cm.

Therefore, the area of quadrilateral or the given fiure

= 2( area of triangle with height 7cm and base 5cm + area of triangle with base 4cm and height 5cm )

= 2(17.5 + 10) square centimeter

= 2 × 27.5

= 55 square centimeters

Hence, the area of triangle is 55 square centimeters.

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The geometric center of the graph under the function is 5/3

The equation of the function is given as:

f(x) = 3 - |x - 2|

The interval is given as:

x ∈ [0, 3]

The geometric center (gc) of the graph under the function is calculated using

gc = ∫x f(x) dx/∫f(x) dx

Substitute the known values in the above equation

gc = ∫x * (3 - |x - 2|) dx/∫3 - |x - 2| dx

Integrate the numerator and the denominator of the above equation

gc = [-1/6(x - 2)((2|x -2| - 9)x + 2|x - 2| - 18)]/[3x - 1/2[|x - 2|(x - 2)]]

Recall that the interval is given as x ∈ [0, 3]

Substitute the interval values in the above equation.

The equation is then simplified using a graphing calculator.

So, we have

gc = (65/6)/(13/2)

Express the quotient expression as a product

gc = (65/6) * (2/13)

Divide 65 by 13

gc = 5/6 * 2

Divide 2 by 6

gc = 5/3

Hence, the geometric center of the graph under the function is 5/3

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The function that will have a horizontal asymptotes are function B, C and D

A function is said to have a horizontal asymptotes of the degree of the numerator is less or equal to that of the denominator.

From the rational expression given, the function A and E wil have no horizontal asymptotes since the power of the numerator is greater abd equal to the denominator,

The function that will have a horizontal asymptotes are function B, C and D

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Firstly changing mixed fraction.

[tex] \frac{3}{2} - \frac{5}{4} + \frac{23}{4} [/tex]

By taking the LCM

[tex] \frac{6 - 5 + 23}{4} [/tex]

-5 + 23 (-) (+) = (-)

[tex] \frac{6 + 18}{4} [/tex]

[tex] \frac{24}{4} [/tex]

[tex]6[/tex]

Hope it helps you, any confusions you may ask!

Answer:

6 (work below)

Step-by-step explanation:

1 1/2 = 3/2

1 1/4 = 5/4

5 3/4 = 23/4

The least common multiple of the denominators is 4, so 3/2 will become 6/4.

6/4 - 5/4 = 1/4 + 23/4 = 24/4 = 6

Brainliest, please :)

Using a system of equations, it is found that her normal week is of €285.

A system of equations is when two or more variables are related, and equations are built to find the values of each variable.

For this problem, the variables are given as follows:

The first equation, considering the first sentence, is given by:

x + 8y = 375.

The second equation, considering the second sentence, is given by:

x + 4y = 330.

Then the system is:

Multiplying the second equation by -2, the system is:

Adding the equations, we have that her normal week is given by:

x = 660 - 375 = €285.

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Answer: [tex]x=\frac{\log (3/5)}{\log 5}[/tex]

Step-by-step explanation:

[tex]3=5^{x+1}\\\\\log_{5} 3=x+1\\\\x=\log_{5} 3-1\\\\x=\log_{5} 3-\log_{5} 5\\\\x=\log_{5} \left(\frac{3}{5} \right)\\\\x=\frac{\log (3/5)}{\log 5}[/tex]

The sum of the exponential expression is e^8x + e^2x

Exponential functions are inverse of logarithmic function

Given the exponential function below;

(e^4x)² + e^2x

Expand

e^8x + e^2x

Hence the sum of the exponential expression is e^8x + e^2x

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Answer:

all real values of x where x < -1

The correct statement is all real values of x where x < -1.

Option A is the correct answer.

We have,

To determine the values for which the graph of the function

f(x) = (x - 3)(x + 1) is positive and decreasing, we need to analyze the behavior of the function.

Let's consider the factors individually:

(x - 3): This factor is positive for x > 3 and negative for x < 3.

(x + 1): This factor is positive for x > -1 and negative for x < -1.

To determine the overall sign of the function, we need to consider the signs of both factors together.

When both factors are positive or both factors are negative, the function is positive.

When the factors have opposite signs, the function is negative.

From the above analysis, we can conclude the following:

When x > 3, both factors are positive, so the function is positive.

When -1 < x < 3, the factor (x - 3) is negative, while the factor (x + 1) is positive. Therefore, the function is negative.

When x < -1, both factors are negative, so the function is positive again.

Since we are looking for values where the function is positive and decreasing, we can eliminate the options that include positive and increasing regions (such as x > 3).

Thus,

The correct statement is all real values of x where x < -1.

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1. Consult the linked question. [tex]n(A\cap B\cap W) = \boxed{3}[/tex].

2. We have

[tex]n(A\cap B) = n(A\cap B \cap W) + n(A\cap B \cap W') \\\\ \implies 11 = 3 + n(A\cap B \cap W') \\\\ \implies n(A\cap B\cap W') = 8[/tex]

Similarly we can find

[tex]n(A\cap B' \cap W) = 3[/tex]

[tex]n(A'\cap B\cap W) = 6[/tex]

[tex]n(A\cap B'\cap W') = 16[/tex]

[tex]n(A'\cap B\cap W') = 8[/tex]

[tex]n(A'\cap B'\cap W) = 12[/tex]

Then the total number of students that applied to at least two of the universities is

[tex]\underbrace{n(A\cap B\cap W') + n(A\cap B'\cap W) + n(A'\cap B\cap W)}_{\text{only 2}} + \underbrace{n(A\cap B\cap W)}_{\text{all 3}} = \boxed{20}[/tex]

3. There's a small ambiguity here. Are we interested in students that applied to zero universities? If so, there are

[tex]\underbrace{n(U\setminus(A\cup B\cup W))}_{\text{none}} + \underbrace{n(A\cap B'\cap W') + n(A'\cap B\cap W') + n(A'\cap B'\cap W)}_{\text{only 1}} = \boxed{40}[/tex]

If we want students that applied to at least 1 school, we omit the first term and get a total of 36.

4. Split this set of students into those that also applied to Wachemo and those that did not.

[tex]n(A \cap B') = n(A\cap B' \cap W) + n(A\cap B'\cap W') = \boxed{19}[/tex]

Answer:

[tex]\bf m = - 1[/tex]

Step-by-step explanation:

[tex]\bf m= −(4+m)+2[/tex]

First of all, let's Rearrange the given terms :-

[tex]\bf m = - ( m +4) + 2[/tex]

Now, let's distribute:-

[tex]\bf m = - m - 4 + 2[/tex]

Add numbers -4 + 2 = -2

[tex]\bf m = - m - 2[/tex]

Add m to both sides, and combine like terms :-

[tex]\bf 2m = - 2[/tex]

Divide both sides by 2 :-

[tex]\bf m = - 1[/tex]

Answer:

Step-by-step explanation:

hello : here is a solution

The area of the interior above the polar axis is -0.858 square units

The area bounded by a polar curve between θ = θ₁ and θ = θ₂ is given by

[tex]A = \int\limits^{\theta_{2} }_{\theta_{1} } {\frac{1}{2}r^{2} } \, d\theta[/tex]

Now, since we have the curve r = 1 - sinθ and we want to find the area of the interior above the polar axis, we integrate from θ = 0 to θ = π, since this is the region above the polar axis.

So, [tex]A = \int\limits^{\theta_{2} }_{\theta_{1} } {\frac{1}{2}r^{2} } \, d\theta\\= \int\limits^{\pi}_{0} {\frac{1}{2}(1 - sin\theta)^{2} } \, d\theta\\= \int\limits^{\pi}_{0} {\frac{1}{2}[1 - 2sin\theta + (sin\theta)^{2}] } \, d\theta\\= \int\limits^{\pi}_{0} {\frac{1}{2}\, d\theta - \int\limits^{\pi}_{0} 2sin\theta \, d\theta+ \int\limits^{\pi}_{0} (sin\theta)^{2} } \, d\theta\\[/tex]

[tex]A = \int\limits^{\pi}_{0} {\frac{1}{2}\, d\theta - 2\int\limits^{\pi}_{0} sin\theta \, d\theta+ \int\limits^{\pi}_{0} \frac{(1 - cos2\theta)}{2} } \, d\theta\\= \int\limits^{\pi}_{0} {\frac{1}{2}\, d\theta - 2\int\limits^{\pi}_{0} sin\theta \, d\theta+ \int\limits^{\pi}_{0} \frac{1}{2} } \, d\theta -\int\limits^{\pi}_{0} \frac{cos2\theta}{2} } \, d\theta\\[/tex]

[tex]A = \int\limits^{\pi}_{0} \, d\theta - 2\int\limits^{\pi}_{0} sin\theta \, d\theta -\int\limits^{\pi}_{0} \frac{cos2\theta}{2} } \, d\theta\\= [\theta]_{0}^{\pi} - 2[-cos\theta]^{\pi} _{0} - [\frac{sin2\theta}{4}]_{0}^{\pi} \\= [\theta]_{0}^{\pi} + 2[cos\theta]^{\pi} _{0} - [\frac{sin2\theta}{4}]_{0}^{\pi}\\= [\pi - 0] + 2[cos\pi - cos0] - \frac{ [sin2\pi - sin0]}{4}\\= \pi + 2[-1 - 1] - \frac{ [0 - 0]}{4}\\= \pi + 2[-2] - \frac{ [0]}{4}\\= \pi - 4 - 0\\= \pi - 4\\= 3.142 - 4\\= -0.858[/tex]

So, the area of the interior above the polar axis is -0.858 square units

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Answer:

8.1

Step-by-step explanation:

[tex]\sqrt{(-1-0)^2 + (5-(-3))^2}=\sqrt{65} \approx 8.1[/tex]

Answer:

A = -5

B = 1

Step-by-step explanation:

Hello!

You simply plug in the given x-value into the equation to find the missing y-value.

Plug in -1 for x into y = 2x - 3.

Plug in 2 for x into y = 2x - 3.

He pays $0.24 less when using his loyalty card.

We know that, without the card, he would pay $1.679 dollars per liter of gasoline.

With his card, he would pay $1.439

To find the relation between the two prices, we just need to take the difference between the two values (a difference is a subtraction).

d = $1.679 - $1.439 = $0.24

He pays $0.24 less when using his loyalty card.

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