A system of linear equations is shown below, where A and B are real numbers.

3x + 4y = A

Bx – 6y = 15

What values could A and B be for this system to have no solutions?

Answer:[tex]\Large\boxed{First~choice.~y\leq \frac{1}{2}x+2 }[/tex]

Step-by-step explanation:

Given function form

Point-Slope form : y = mx + b

Given the information from the graph

Since the graph crosses (0, 2), b = 2

Find the slope

Choose any two points from the graph

(x₁, y₁) = (0, 2)

(x₂, y₂) = (2, 3)

Slope = (y₂ - y₁) / (x₂ - x₁)

Slope = (3 - 2) / (2 - 0)

Slope = 1 / 2

m = 1 / 2

Therefore, the equation form of the inequality is y = (1/2)x + 2

Since it is a solid line, the values on the line are also included. Thus, it must be either greater than or equal to, or less than or equal to.

The shaded region is below the solid line, therefore, it is ≤

Therefore, the inequality is [tex]\Large\boxed{y\leq \frac{1}{2}x+2 }[/tex]

To double-check the sign, we can substitute (0, 0) into the inequality to see whether it is true.

y ≤ (1/2) x + 2

0 ≤ (1/2) 0 + 2

0 ≤ 2              TRUE

Hope this helps!! :)

Please let me know if you have any questions

Answer:

Step-by-step explanation:

The domain is an x thing. We are being asked to determine what x values to function takes on. We see that the blue dot on the left is at the x-value of 2. It doesn't go any farther to the left of 2. Since the dot is closed, the inequality symbol will have the "equal to" line underneath.

The function then continues along the x axis and stops at 5, closed hole. This means that the function does not go past the x value of 5 but it is included in our domain.

2 ≤ x ≤ 5. That is choice B.

Step-by-step explanation:

The turn of the 20th century = 1900

So it is 1900 + 107 years ago, hence, 2007

The solution to the system of equations is (-2, 3)

The order of the stages of recession will be 4, 2, 3, 1, 3.

What happens in a recession?

Who benefits in a recession?

Rental agents, landlords, and property management companies can thrive during a recession when renting is likely to become a more appealing option, if not the only one available.

So, these stages are occur -

4. peak production

2. falling production

3. falling demand (occurs twice)

1. job loss

3. falling demand (occurs twice)

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The complete question is-

There are five stages in a recession.

1. job loss

2. falling production

3. falling demand (occurs twice)

4. peak production

In which order do these stages occur?

4, 3, 2, 1, 3

3, 2, 1, 3, 4

4, 3, 2, 3, 1

4, 2, 3, 1, 3

4, 2, 3, 1, 3.Answer:4, 2, 3, 1, 3.

4, 2, 3, 1, 3.Step-by-step explanation:4, 2, 3, 1, 3.

The nurse should dispose of the mask. This exists because the safety mechanisms of the mask may be compromised and therefore pose a risk to the patient that exists undergoing surgery.

A surgical mask exists primarily utilized to protect patients and healthcare employees from people who may contain a respiratory infection or to protect sterilized or disinfected medical devices and supplies. The nurse should dispose of the mask. This exists because the safety mechanisms of the mask may be compromised and therefore pose a risk to the patient that exists undergoing surgery. It exists still suggested that medical personnel err on the side of warning when it arrives to the managing of equipment, whether it be a significant piece of equipment or even a small one like a surgical mask.

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Applying properties of exponents, the quotient is given as follows:

[tex]2\sqrt[3]{3} - \sqrt[3]{18}[/tex]

The expression is given by:

[tex]\frac{6 - 3\sqrt[3]{6}}{\sqrt[3]{9}}[/tex]

The 3 can be inserted into the cube root, as follows:

[tex]\frac{6 - 3\sqrt[3]{6}}{\sqrt[3]{9}} = \frac{6 - \sqrt[3]{6 \times 3³}}{\sqrt[3]{9}}[/tex]

Applying the subtraction, we have that the expression is:

[tex]\frac{6}{\sqrt[3]{9}} - \frac{\sqrt[3]{162}}{\sqrt[3]{9}} = \frac{6}{\sqrt[3]{9}} - \sqrt[3]{\frac{162}{9}} = \frac{6}{\sqrt[3]{9}} - \sqrt[3]{18}[/tex]

The denominator can be simplified as follows:

[tex]\sqrt[3]{9} = \sqrt[3]{3^2} = 3^{\frac{2}{3}}[/tex]

Then:

[tex]\frac{6}{\sqrt[3]{9}} = \frac{2 \times 3}{3^{\frac{2}{3}}} = 2 \times 3^{1 - \frac{2}{3}} = 2 \times 3^{\frac{1}{3}} = 2\sqrt[3]{3}[/tex]

Hence the quotient is given by:

[tex]2\sqrt[3]{3} - \sqrt[3]{18}[/tex]

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Answer:

Option C: x = 6

Step-by-step explanation:

Hi There!

7 + 5 (x - 3) = 22

=> 7 + 5x - 15 = 22

=> 5x = 22 - 7 + 15

=> 5x = 30

=> x = 30/5

=> x = 6

x = 6 makes the statement true.

Thank you,

Eddie

The system of equations to find the value of x, the first number, and y, the second number are;

y = -x + 84

y = x² + 6

The system of equation are equations that contains more than two equation and unknown.

Let the two unknown numbers x and y

If the sum of two numbers is 84, hence;

x +y = 84

y = -x + 84

If the square of the first number is 6 more than the second number, hence;

x² = y + 6

Substitute equation 2 into 2 to have:

x² = y + 6

x² = (-x+84) + 6

x² = -x+84 + 6

x² +x+84 + 6  = 0

x² +x + 90 = 0

Hence the system of equations to find the value of x, the first number, and y, the second number are;

y = -x + 84

y = x² + 6

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the sum between negative 4 and 13, this can be tought as:

The expression is just the sum:

-4 + 13

So we have the sum between negative 4 and 13, this can be tought as:

Now, none of the written options (a, b, and c) depict any of this, so I assume that the correct option is the one that you did not write, which is option D.

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The formula for moose population is P = 4760 + (number of years x 200).

The moose population in 2003 would be 6760.

When a population increases linearly, it means that it increases by the same amount each year.  

Rate of linear increase: (population in 1999 - population in 1993) / difference in years

(5960 - 4760) / (1999 - 1993)

1200 / 6 = 200

Linear function : initial population + (rate of increase x number of years)

Population in 2003: 4760 + (200 x 10)

4760 + 2000 = 6760

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Missing data:

Viewer's Age Group  Excellent Good Average Poor Marginal Total

16–25                          52                 42       12       7           113

26–35                          33                50         5       9          97

36–45                          58                 12       28      34         132

46–55                          25                17       22       12            76

56 +                               12                 5          3         8             28

Marginal Total             180               126      70      70           446

A rating of good or excellent indicates the audience liked the movie, while a rating of poor indicates the audience disliked the movie.

A movie producer accomplished a survey behind a preview screening of her latest movie to estimate how the film would be accepted by viewers from various age groups. The table displays the numbers of viewers in various age groups who ranked the film excellent, good, average, and poor.

25/446 = 0.05605

0.05605 [tex]*[/tex] 100% = 5.605%

Out of the entire respondents, the percentage of respondents from the 46–55 age group who ranked the film excellent exists at 5.605%.

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Answer:

[tex]x=\frac{P^2t}{(m-t^3)}[/tex]

Step-by-step explanation:

[tex]P^2=\frac{mx}{t} -t^2x\\\\P^2+t^2x=\frac{mx}{t}\\\\t(P^2+t^2x)=mx\\\\P^2t+t^3x=mx\\\\P^2t=mx -t^3x\\\\P^2t=x(m-t^3)\\\\\frac{P^2t}{(m-t^3)} =\frac{x(m-t^3)}{(m-t^3)} \\\\\frac{P^2t}{(m-t^3)}=x[/tex]

Answer: (5/2, 3)

Step-by-step explanation:

Substituting in x=5/2,

[tex]2(5/2)+5y=20\\\\5+5y=20\\\\5y=15\\\\y=3[/tex]

So, the point is (5/2, 3)

The probability that it also rained that day is to be considered as the 0.30 and the same is to be considered.

The extent to which an event is likely to occur, measured by the ratio of the favorable cases to the whole number of cases possible.

The probability that the temperature is lower than 80°F and it rained can be measured by determining the number at the intersection of a temperature that less than 80°F and rain.

So, This number is 0.30.

Hence, we can say that it was less than 80°F on a given day, the probability that it also rained that day is 0.30.

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The above question is incomplete.

The conditional relative frequency table was generated using data that compared the outside temperature each day to whether it rained that day. A 4-column table with 3 rows titled weather. The first column has no label with entries 80 degrees F, less than 80 degrees F, total. The second column is labeled rain with entries 0.35, 0.3, nearly equal to 0.33. The third column is labeled no rain with entries 0.65, 0.7, nearly equal to 0.67. The fourth column is labeled total with entries 1.0, 1.0, 1.0. Given that it was less than 80 degrees F on a given day, what is the probability that it also rained that day?

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Answer: Its 0.3 so the first one

Step-by-step explanation:

Im 101% sure

You can use the well-known binomial series,

[tex]\displaystyle (1+x)^\alpha = \sum_{k=0}^\infty \binom \alpha k x^k[/tex]

where

[tex]\dbinom \alpha k = \dfrac{\alpha(\alpha-1)(\alpha-2)\cdots(\alpha-(k-1))}{k!} \text{ and } \dbinom \alpha0 = 1[/tex]

Let [tex]\alpha=\frac12[/tex] and replace [tex]x[/tex] with [tex]3x[/tex]; then the series expansion is

[tex]\displaystyle (1+3x)^{1/2} = \sum_{k=0}^\infty \binom{\frac12}k (3x)^k[/tex]

and the first 4 terms in the expansion are

[tex]\sqrt{1+3x} \approx 1 + \dfrac{\frac12}{1!}(3x) + \dfrac{\frac12\cdot\left(-\frac12\right)}{2!}(3x)^2 + \dfrac{\frac12\cdot\left(-\frac12\right)\cdot\left(-\frac32\right)}{3!}(3x)^3[/tex]

which simplify to

[tex]\sqrt{1+3x} \approx \boxed{1 + \dfrac32 x - \dfrac98 x^2 + \dfrac{27}{16} x^3}[/tex]

You can also use the standard Maclaurin coefficient derivation by differentiating [tex]f[/tex] a few times.

[tex]f(x) = (1+3x)^{1/2} \implies f(0) = 1[/tex]

[tex]f'(x) = \dfrac32 (1+3x)^{-1/2} \implies f'(0) = \dfrac32[/tex]

[tex]f''(x) = -\dfrac94 (1+3x)^{-3/2} \implies f''(0) = -\dfrac94[/tex]

[tex]f'''(x) = \dfrac{81}8 (1+3x)^{-5/2} \implies f'''(0) = \dfrac{81}8[/tex]

Then the 3rd order Maclaurin polynomial is the same as before,

[tex]\sqrt{1+3x} \approx f(0) + \dfrac{f'(0)}{1!} x + \dfrac{f''(0)}{2!} x^2 + \dfrac{f'''(0)}{3!} x^3 = 1 + \dfrac32 x - \dfrac98 x^2 + \dfrac{27}{16} x^3[/tex]

Now,

[tex]\sqrt{1.3} = \sqrt{1+3x} \bigg|_{x=\frac1{10}} \\\\ ~~~~~~~~ \approx 1 + \dfrac32 \left(\dfrac1{10}\right) - \dfrac98 \left(\dfrac1{10}\right)^2 + \dfrac{27}{16} \left(\dfrac1{10}\right)^3 \\\\ ~~~~~~~~ = \dfrac{18,247}{16,000} \approx \boxed{1.14044}[/tex]

Compare to the actual value which is closer to 1.14018.

[tex]\sqrt{1+3x}=1+\frac{3}{2} x-\frac{9}{8} x^{2} + \frac{81}{8}x^{3}[/tex] is the maclaurin polynomial and estimate value of [tex]\sqrt{1.3}[/tex] is 1.14. This can be obtained by using the formula to find the maclaurin polynomial.

Given the polynomial,

[tex]f(x)=\sqrt{1+3x}=(1+3x)^{\frac{1}{2} }[/tex]

The formula to find the maclaurin polynomial,

[tex]f(0)+\frac{f'(0)}{1!}x+\frac{f''(0)}{2!}x^{2} + \frac{f'''(0)}{3!}x^{3}[/tex]

Next we have to find f'(x), f''(x) and f'''(x),

By putting x = 0 , we get,

Therefore the maclaurin polynomial by using the formula will be,

[tex]\sqrt{1+3x}=f(0)+\frac{f'(0)}{1!}x+\frac{f''(0)}{2!}x^{2} + \frac{f'''(0)}{3!}x^{3}[/tex]

[tex]\sqrt{1+3x}=1+\frac{3}{2} x-\frac{9}{8} x^{2} + \frac{81}{8}x^{3}[/tex]

To find the value of [tex]\sqrt{1.3}[/tex]  we can use the maclaurin polynomial,

[tex]\sqrt{1.3}[/tex] is  [tex]\sqrt{1+3x}[/tex] with x = 1/10,

[tex]\sqrt{1+3(1/10)}=1+\frac{3}{2} (1/10)-\frac{9}{8} (1/10)^{2} + \frac{81}{8}(1/10)^{3}[/tex]

[tex]\sqrt{1+3(1/10)}=\frac{18247}{16000} = 1.14[/tex]

Hence [tex]\sqrt{1+3x}=1+\frac{3}{2} x-\frac{9}{8} x^{2} + \frac{81}{8}x^{3}[/tex] is the maclaurin polynomial and estimate value of [tex]\sqrt{1.3}[/tex] is 1.14.

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To find the area, we take into account that your data is:

To calculate the area of the circle, we use the following formula:

Area: π * r²

We substitute the data in the formula, and we begin to solve.

taking the square root

Multiply the two numbers

Therefore, the area of the circle is 113.04 cm². The correct alternative is option "A".

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In the question, it is given that a circle has a radius of 6 cm , we have to find the Area.

To Find the area of the circle , we must know this formula:

[tex] \quad\quad\quad \longrightarrow\underline{\boxed{\color{violet}{A=\pi*r^2}}}[/tex]

Now, we will substitute the values in the formula:

[tex]\rightarrow\sf{A=\pi*r^2} \: \: \: \: \: \: \: ( \red{\pi = 3.14})[/tex]

[tex]\rightarrow\sf{A = 3.14 * 6^2}[/tex]

[tex]\rightarrow\sf{A = 3.14 * 36}[/tex]

[tex]\rightarrow\sf{A = 113.0}[/tex]

The area of circle is 113.0 approx .

[tex] \color{blueviolet}{ \mathfrak{llNotYourll}}[/tex]

Answer:

1/15 of the dishes on the menu are vegetarian pasta.

Step-by-step explanation:

1/10 of 2/3 of all dishes are vegetarian pasta.

1/10 of 2/3 of all = 1/10 * 2/3 * 1 = 2/30 = 1/15

Answer:296 m

I’m assuming “breadth” is width and garderean is just the park.

(25+25+12+12) times 4.

A is never defined

we can see that the center is (-3, 3) and the radius is 9 units.

The general circle equation, for a circle with a center (a, b) and radius R is given by:

(x - a)^2 + (y - b)^2 = R^2

Here we have the equation:

x^2 + y^2 + 6x = 6y + 63

Let's complete squares:

x^2 + y^2 + 6x - 6y =  63

(x^2 + 6x) + (y^2 - 6y) = 63

(x^2 + 2*3x) + (y^2 - 2*3y) = 63

Now we can add and subtract 9, (two times) so we get:

(x^2 + 2*3x + 9) - 9 + (x^2 - 2*3x + 9) - 9 = 63

(x + 3)^2 + (y - 3)^2 = 63 + 9 + 9 = 81 = 9^2

(x + 3)^2 + (y - 3)^2 = 9^2

Comparing with the general circle equation, we can see that the center is (-3, 3) and the radius is 9 units.

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Answer:

n = 2

Step-by-step explanation:

cross multiply

-5 times -4 = 20

10n = 20

n = 2

[tex]\bf{Move \ the \ negative \ symbol \ to \ the \ left. }[/tex]

[tex]\boldsymbol{\sf{-\dfrac{10n}{5}=-4 \ \ \longmapsto \ \ [Split] }}[/tex]

[tex]\boldsymbol{\sf{-2n=-4 }}[/tex]

[tex]\bf{Divide \ both \ sides \ by \ -2. }[/tex]

[tex]\boldsymbol{\sf{n=\dfrac{-4}{-2} }}[/tex]

[tex]\bf{Two \ negatives \ make \ a \ positive. }[/tex]

[tex]\boldsymbol{\sf{n=\dfrac{4}{2} \ \ \longmapsto \ \ [Split] }}[/tex]

[tex]\blue{\boxed{\boldsymbol{\sf{Answer \ \ \longmapsto \ \underline{n=2}}}}}[/tex]

Answer:

CotO

Step-by-step explanation:

The process is int the picture

Answer:

11.7647%

Hope this helps!

Answer:

Step-by-step explanation:

Miles per hour: [tex]\frac{10miles}{4hrs} = 2.5\frac{miles}{hour}[/tex]

Feet per hour:

      [tex]2.5\frac{miles}{hours} (\frac{5280ft}{1mile} )[/tex]

     [tex]=13,200\frac{ft\\}{hr}[/tex]

The lateral area of a cone is 147ft².

The total surface area of the cylinder is 867ft².

The total surface area is 1014ft²

The lateral area of a cone is defined as the area covered by the curved surface of the cone.

Lateral area of a cone = πr x √(r² + h²)

Where:

3.14 x 6 x √(36 + 25) = 147ft²

Total surface area of the cylinder : πr(r + 2h),

(3.14 x 6) x (6 + 40) = 867ft²

Total surface area = 867 + 147=  1014ft²

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Answer:

You can find either x or y by applying the Theorem which states that sum of opposite angles of a cyclic quadrilateral is 180°

Step-by-step explanation:

The above is a cyclic quadrilateral.

A Cyclic quadrilateral is having all the vertices of a quadrilateral ( four sided figure) inside a circle, that is a quadrilateral inscribed in a circle.

To find either x or y, we should note:

The sum of opposite angles of a cyclic quadrilateral is 180° or Angles in opposite segment are supplementary.

Therefore,

2x + 2x + 4= 180

4x = 180 - 4

4x = 176

x = 44°

To find y

3y + 8 + y = 180°

4y = 180 - 8

4y = 172

y = 43°

Therefore, you can find either x or y by applying the Theorem which states that sum of opposite angles of a cyclic quadrilateral is 180°

Answer: 9 cm² : 25 cm²

Step-by-step explanation:

    The area of a square can be found with x² where x is the length of one side.

    x² : x²

    3² : 5²

    9 : 25

           9 cm² : 25 cm²

Answer:

$0.34 per ounce

Step-by-step explanation:

A 5 pound box can be changed to ounces. There are 16 ounces in a pound. So we have 5 groups of 16.

5 × 16 is 80 ounces.

There are 80 ounces in 5lbs.

To get a price per ounces, divide.

dollars/ounces

= 27.20/80

= 0.34 dollars/oz.

This is 34cents per ounce.

The second-order equation as its equivalent system of first-order equations is

[tex]\left[\begin{array}{ccc}u(1)\\\\v(1)\end{array}\right][/tex] = [tex]\left[\begin{array}{ccc}7.5\\\\9\end{array}\right][/tex]

An equivalent system that has the identical answer is known as an equivalent structure. Given a gadget of two equations, we can produce an equal system by way of replacing one equation by means of the sum of the 2 equations, or by way of changing an equation by means of a couple of of itself.

Systems of linear equations are equivalent if and handiest in the event that they have an equal set of solutions. In other phrases, two systems are equal if and only if each answer of one in all of them is likewise a solution of the opposite.

In the structures sciences, a machine equivalent system is the conduct of a parameter or thing of a machine in a way just like a parameter or component of a distinctive system. Similarity means that mathematically the parameters and additives will be indistinguishable from each different.

Taking v = u, we have:

u" + 4u' + 6u = 4sin(3t)

--> v' + 4v + 6u = 4sin(3t)

So the system of equations is:

u' = 0u + 1v

v' = -6u - 4v + 4sin(3t)

So we can write it as:

u(1) = 7.5

v(1) = u'(1) = 9

So the initial condition matrix is:

[tex]\left[\begin{array}{ccc}u(1)\\\\v(1)\end{array}\right][/tex] = [tex]\left[\begin{array}{ccc}7.5\\\\9\end{array}\right][/tex]

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