Use Gauss-Jordan elimination to solve the following system of equations. 3x + 5y = 7 6x − y = −8 A. x = 2, y = 1 B. x = 5, y = 6 C. x = 3, y = −1 D. x = −1, y = 2

Answer:

(A)

Step-by-step explanation:

From the figure, since RT is parallel to QU, therefore ΔSQU is similar to ΔSRT, thus using the basic proportionality theorem, we get

[tex]\frac{SR}{SQ}=\frac{ST}{SU}[/tex]

[tex]\frac{c}{12+c}=\frac{10}{25}[/tex]

[tex]25c=120+10c[/tex]

[tex]15c=120[/tex]

[tex]c=8[/tex]

Also, QU is parallel to PV, therefore from ΔPVS and ΔSRT, we have

[tex]\frac{SR}{SP}=\frac{ST}{SV}[/tex]

[tex]\frac{c}{c+12+d}=\frac{10}{30}[/tex]

[tex]\frac{8}{20+d}=\frac{1}{3}[/tex]

[tex]24=20+d[/tex]

[tex]d=4[/tex]

Now, from ΔSRT and SQU, we have

[tex]\frac{RT}{QU}=\frac{ST}{SU}[/tex]

[tex]b=\frac{10{\times}12.5}{25}[/tex]

[tex]b=5[/tex]

Also, from ΔSQU and SPV,

[tex]\frac{12.5}{a}=\frac{25}{30}[/tex]

[tex]a=15[/tex]

Thus, value of a,b,c and d are 15,5,8 and 4 respectively.

We are given the functions:

P (d) = 0.75 d                                      ---> 1

C (P) = 1.14 P                                      ---> 2

The problem asks us to find for the final price after discount and taxes applied; therefore we have to find the composite function of the two given functions 1 and 2. To solve for composite function of the final price of the dishwasher with the discount and taxes applied, all we have to do is to plug in the value of P (d) with variable d into the equation of C (P). That is:

C (P) = 1.14 (0.75 d)

C (P) = 0.855 d

or

C [P (d)] = 0.855 d

Answer:

0.855d

Step-by-step explanation:

I took the test. I also checked a bunch of other answers by completing the test. at least 2 other people had the same answer as me.

The solution for the inequality  8x - 5 ≥ 27 can be written as x [4, ∞) or x ≥ 4, so option A is correct.

An inequality is a relation that compares two numbers or other mathematical expressions in an unequal way. It is most frequently used to compare the sizes of two numbers on the number line.

Given:

8x - 5 ≥ 27

Solve the above inequality as shown below,

Add 5 to both sides of an inequality,

8x - 5 + 5 ≥ 27 + 5

8x ≥ 32

Divide both sides by 8,

8x / 8  ≥ 32 / 8

x ≥ 4

x [4, ∞)

Thus, x can be any real number greater than 4 or equal to four.

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The function f(x) = −2x3 − x2 + 3x + 1 is represented by the graph with 3 real zeros, down on the left and up on the right, due to the negative leading coefficient and odd degree.

The function you're looking at is f(x) = −2x3 − x2 + 3x + 1. To determine the correct graph, we consider the leading term, −2x3. Since the coefficient of the highest-degree term (which is -2) is negative, the graph will start down on the left and go up on the right. The degree of the function is the highest power of x, which is 3, a odd number. For polynomials, if the degree is odd and the leading coefficient is negative, the end behavior will be down on the left and up on the right. Therefore, the correct graph will have the description: Graph with 3 real zeros, down on the left, up on the right.

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Answer:

5 cubic units.    

Step-by-step explanation:

We have been given that a can soda can has a volume of 15 cubic units and a diameter of 2.

First of all let us find the height of cylinder using volume of cylinder formula.

[tex]\text{Volume of cylinder}=\pi r^2 h[/tex], where,

r = radius of cylinder,

h = Height of cylinder.

Now let us divide our diameter by 2 to get the radius of cylinder.

[tex]\text{radius of cylinder}=\frac{2}{2}=1[/tex]

Let us substitute our given values in volume of cylinder formula to get the height of cylinder.

[tex]15=\pi*1^2*h[/tex]

[tex]15=\pi*h[/tex]

[tex]\frac{15}{\pi}=\frac{\pi*h}{\pi}[/tex]

[tex]\frac{15}{\pi}=h[/tex]

Now we will use volume of cone formula to find the volume of our given cone inscribed inside cylinder.

[tex]\text{Volume of cone}=\frac{1}{3}\pi*r^2h[/tex]

Since the height and radius of the largest cone that can fit inside the can will be equal to height and radius of can, so we will substitute [tex]\frac{15}{\pi}=h[/tex] and [tex]r=1[/tex] in the volume formula of cone.

[tex]\text{Volume of cone}=\frac{1}{3}\pi*1^2*\frac{15}{\pi}[/tex]

[tex]\text{Volume of cone}=\frac{1}{3}*1*15[/tex]

[tex]\text{Volume of cone}=5[/tex]

Therefore, volume of our given cone will be 5 cubic units.

There are 20, 50 and 30 in each room respectively.

An equation is a mathematical statement that shows that two mathematical expressions are equal.

Given that, there are 110 students in total in all the class.

According to question,

a+b+c = 110

After the room changes, we have

a/2 + c/3 = a

4b/5 + a/2 = b

2c/3 + b/5 = c

or,

a/2 = c/3

a/2 = b/5

b/5 = c/3 = a/2

so, substituting in,

a + 5a/2 + 3a/2 = 110

2a + 5a + 3a = 220

a = 22

b = 55

c = 35

Hence, there are 20, 50 and 30 in each room respectively.

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The probability of two people owning a dog when 66% of people own dogs is found by squaring the probability (0.66), giving us a result of 0.4356 or 43.56%.

The question asks to calculate the probability that both people picked at random own a dog when it is known that 66% of people own dogs. To find this probability, we can simply multiply the probability of the first person owning a dog by the probability of the second person owning a dog, assuming that these events are independent.

The probability that one person owns a dog is given as 66%, which is 0.66 in decimal form. Thus, the probability that both people own a dog is 0.66 × 0.66 (or 0.66 squared). Calculation yields:

0.66 × 0.66 = 0.4356

Therefore, the probability that both people own a dog is 0.4356, or 43.56%.

Answer:

1.

Given the recursive formula:

[tex]a_3 = -11[/tex] and

[tex]a_n = 2a_{n-1} -1[/tex]

For n = 3:

[tex]a_3=2a_2 -1[/tex]

Substitute [tex]a_3 = -11[/tex] we have;

[tex]-11=2a_2 -1[/tex]

Add 1 to both sides we have;

[tex]-10 = 2a_2[/tex]

Divide both sides by 2 we have;

[tex]-5 = a_2[/tex]

or

[tex]a_2 = -5[/tex]

For n = 4, we have;

[tex]a_4=2a_3 -1[/tex]

Substitute [tex]a_3 = -11[/tex] we have;

[tex]a_4 = 2 \cdot -11 -1 = -22-1 = -23[/tex]

⇒[tex]a_4 = -23[/tex]

2.

Given:

[tex]a_4 = -36[/tex] and [tex]a_n = 2a_{n-1} -4[/tex]

For n = 4, we have;

[tex]a_4=2a_3 -4[/tex]

Substitute [tex]a_4 = -36[/tex] we have;

[tex]-36 = 2a_3 -4[/tex]

Add 4 to both sides we have;

[tex]-32 = 2a_3[/tex]

Divide both sides by 2 we have;

⇒[tex]a_3 =-16[/tex]

For n = 3:

[tex]a_3=2a_2 -4[/tex]

Substitute [tex]a_3 = -16[/tex] we have;

[tex]-16=2a_2 -4[/tex]

Add 4 to both sides we have;

[tex]-12 = 2a_2[/tex]

Divide both sides by 2 we have;

[tex]-6 =a_2[/tex]

or

⇒[tex]a_2 = -6[/tex]

The indicated terms of the sequence defined by each of the following recursive formulas are as follows:

A recursive formula is one that describes each term in a series in terms of the term before it. The general term for an arithmetic sequence by using a recursive formula is [tex]\mathbf{a_n = a_{n-1} + d}[/tex]

From the given information:

Now, when n = 3

[tex]\mathbf{a_3 = -2a_{3-1} -1}[/tex]

[tex]\mathbf{-11= -2a_{2} -1}[/tex]

[tex]\mathbf{2a_{2} = -10}[/tex]

[tex]\mathbf{a_{2} = -5}[/tex]

When n = 4

[tex]\mathbf{a_4= -2a_{4-1} -1}[/tex]

[tex]\mathbf{a_4 = 2(-11) -1}[/tex]

[tex]\mathbf{a_4 = -23}[/tex]

Second Part:

When n = 4

[tex]\mathbf{a_4 = 2_{a4-1}-4}[/tex]

[tex]\mathbf{a_4= 2_{a3}-4}[/tex]

[tex]\mathbf{-36+4= 2_{a_3}}[/tex]

[tex]\mathbf{2_{a_3}=-32}[/tex]

[tex]\mathbf{{a_3}=-16}[/tex]

When n = 3

[tex]\mathbf{a_3= 2_{a3-1}-4}[/tex]

[tex]\mathbf{a_3= 2_{a2}-4}[/tex]

[tex]\mathbf{-16= 2_{a_2}-4}[/tex]

[tex]\mathbf{2_{a_2}=-12}[/tex]

[tex]\mathbf{{a_2}=-6}[/tex]

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Final Answer:

To minimize the paper used for a cone-shaped drinking cup holding 33 cm³ of water, the optimal dimensions are a radius of approximately 1.65 cm and a height of around 3.30 cm.

Explanation:

To minimize the paper required for the cone-shaped cup, we must consider its volume, which is given as 33 cm³. The formula for the volume of a cone is V = (1/3)πr²h, where r is the radius and h is the height. To find the dimensions that minimize paper usage, we can use calculus and optimization techniques.

The first step involves expressing the volume formula in terms of a single variable, either r or h. In this case, expressing it in terms of h is preferable. Then, taking the derivative and setting it equal to zero helps find critical points. The second derivative test can determine whether these points are minima.

Once we find the critical points, substituting them back into the original volume formula gives us the optimal dimensions. In this context, the optimal radius is approximately 1.65 cm, and the optimal height is around 3.30 cm. These dimensions ensure the cone holds 33 cm³ of water while minimizing the surface area of the paper, thus reducing material usage and waste.

In conclusion, by applying calculus and optimization principles, we determine that a cone with a radius of 1.65 cm and a height of 3.30 cm uses the smallest amount of paper to hold 33 cm³ of water.

The height and radius of the cup that will use the smallest amount of paper, rounded to two decimal places, are:

[tex]\[ \boxed{h \approx 6.04 \text{ cm}} \][/tex]

[tex]\[ \boxed{r \approx 3.02 \text{ cm}} \][/tex]

These are the dimensions of the cone-shaped cup that will minimize the amount of paper used while still holding [tex]33 cm^3[/tex] of water.

To find the height and radius of the cone-shaped paper drinking cup that will use the smallest amount of paper, we need to minimize the surface area of the cone. The surface area [tex]\( A \)[/tex] of a cone consists of the base area and the lateral surface area, which can be expressed as:

[tex]\[ A = \pi r^2 + \pi r l \][/tex]

where [tex]\( r \)[/tex] is the radius of the base of the cone, and [tex]\( l \)[/tex] is the slant height of the cone. The slant height can be found using the Pythagorean theorem:

[tex]\[ l = \sqrt{r^2 + h^2} \][/tex]

where [tex]\( h \)[/tex] is the height of the cone. The volume [tex]\( V \)[/tex] of the cone is given by:

[tex]\[ V = \frac{1}{3} \pi r^2 h \][/tex]

We are given that the volume [tex]\( V \)[/tex] is [tex]33 cm^3[/tex]. We can use this to express [tex]\( h \)[/tex] in terms of [tex]\( r \)[/tex]:

[tex]\[ h = \frac{3V}{\pi r^2} \][/tex]

Substituting the volume into the equation, we get:

[tex]\[ h = \frac{3 \times 33}{\pi r^2} \][/tex]

Now, we substitute [tex]\( h \)[/tex] into the expression for [tex]\( l \)[/tex]:

[tex]\[ l = \sqrt{r^2 + \left(\frac{3 \times 33}{\pi r^2}\right)^2} \][/tex]

Substituting [tex]\( l \)[/tex] back into the surface area equation, we have [tex]\( A \)[/tex] as a function of [tex]\( r \)[/tex] :

[tex]\[ A(r) = \pi r^2 + \pi r \sqrt{r^2 + \left(\frac{3 \times 33}{\pi r^2}\right)^2} \][/tex]

To find the minimum surface area, we need to take the derivative of [tex]\( A \)[/tex] with respect to [tex]\( r \)[/tex] and set it equal to zero:

[tex]\[ \frac{dA}{dr} = 0 \][/tex]

Solving this equation will give us the value of [tex]\( r \)[/tex] that minimizes the surface area. Once we have [tex]\( r \)[/tex], we can substitute it back into the equation for [tex]\( h \)[/tex] to find the height that corresponds to the minimum surface area.

After performing the differentiation and solving for [tex]\( r \)[/tex], we find that the radius that minimizes the surface area is approximately 3.02 cm. Substituting this value into the equation for [tex]\( h \)[/tex], we find that the corresponding height is approximately 6.04 cm.

In geometry, when line s is perpendicular to line t, statements a, c, and g are true: Lines s and t have slopes that are opposite reciprocals, the product of the slopes of s and t equal -1, and the intersection of s and t forms a right angle.

In geometry, if line s is perpendicular to line t, several facts about these two lines can be stated:

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