Solve each of the following equations using the Principle of Zero Products and match it to its correct pair of solutions on the
right. Each pair on the right will be used once.
(x-9)(3x + 4) = 0
(x-4) (9x + 1) = 0
(x+9)(3x-4)= 0
(x+4)(9x-1)=0
x = -4 or x =
x = -9 or x =
x = 9 or x = -
x = 4 or x=-1

The athletic director can therefore draw the conclusion that the average academic standing of football players is comparable to the average standing of other student athletes.

The following formula is used to determine the test statistic:

SRS, s = 0.54, n = 40 > 30,

x = 3.18

H0: μ = 3.10

H1: μ > 3.10

α = 0.05

The t-distribution table and our alpha level of.05 are then used to determine our crucial values.

Our critical values are 2.093 for a two specimens with 19 levels of freedom and α = 0.05 level of significance.

t = x -  μ / s/√n

t = 3.18 - 3.10/ 54/√20

t = 0.66

As a result, It is clear that sampling error is the only possible explanation for the discrepancy between both the sample mean as well as the hypothesized value.

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The length and width of a rectangle are 36 and 30 yard respectively.

Its the sum of length of the sides used to made the given figure.

We are given that width of a rectangle is 6 yards less than the length. The perimeter is 132 yards.

Therefore, the perimeter = (2 x length)+(2 x width)

The length = 6 + w

Hence, we have;

132 = (2(6 + w))+(2w)

132 = 12 + 2w + 2w

132 = 12 + 4w

120 = 4w

w = 30

Length = 36 yard.

Thus, the length and width of a rectangle are 36 and 30 yard respectively.

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Answer: $18/hour

12 x 1.5 = 18

Answer: C. 2x+3

Step-by-step explanation:

[tex]\displaystyle\\V=\frac{\pi r^2h}{3} \\\\V=\frac{\pi r^2(6)}{3}\\\\V=2\pi r^2\\\\\\8\pi x^2+24\pi x+18\pi \\\\V=2\pi (4x^2+12\pi +9)\\\\V=2\pi ((2x)^2+2(2x)(3)+3^2)\\\\V=2\pi (2x+3)^2\\\\THUS,\\[/tex]

[tex]2\pi r^2=2\pi (2x+3)^2[/tex]

Divide both parts of the equation by 2π:

[tex]r^2=(2x+3)^2[/tex]

Extract the square root of both parts of the equation:

[tex]r=2x+3[/tex]

Answer:

A = 29°

Step-by-step explanation:

set

(3x-4) + (9x+17) + (6x-31) = 180

Solve for x

18x - 18= 180

18x = 198

x = 11

So A will be

A = 3x - 4

A = 3(11) - 4

A = 29°

Answer:

Step-by-step explanation:

The final interval notation is  F ∩ H = [7, 4] &  F ∪ H = [-∞, ∞].

What is the interval notation?

A set of real numbers known as an interval in mathematics contains all real numbers falling inside any two of the set's numbers. For instance, the interval containing 0, 1, and all integers in between is the set of values x satisfying 0 x 1.

We have,

F and H are sets of real numbers defined as follows.

F = { v l v < 4}

H = { v l v ≥ 7}

F is the set of all real numbers that are less than or equal to 4.  H is the set of real numbers greater than 7.

F U H is the union (or combination) of the two sets, so it is the set of all real numbers that are less than equal to 3 or greater than 6:

F ∩ H is the set of real numbers that are in both sets; but since F is less than or equal to 4 and H is greater than 7, there are no numbers that are in both sets.  Hence the answer is the empty set, Ø:

F ∩ H = Ø

F ∩ H = [7, 4]

F ∪ H = [-∞, ∞]

Hence, the final interval notation is  F ∩ H = [7, 4] &  F ∪ H = [-∞, ∞].

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The berries is the bowl is 6 berries fewer than the original amount

The amount of berries in the bowel is solved using addition and subtraction property of equality

36 berries are in a bowl James eats 21 of the berries

= 35 berries - 21 berries

= 15 berries

he put 14 more berries in the bowl

= 15 berries + 14 berries

= 29 berries

How many fewer is solved by

= 35 berries - 29 berries

= 6 berries

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Answer:

Step-by-step explanation:

24

Answer: 24!

Step-by-step explanation:

hope this help

The statement that best describes a sequence of transformations in which the blue figure is the image of the red figure is given as follows:

A. Rotate 90° clockwise about the origin and then dilate with respect to the origin using a scale factor of 1/2.

The horizontal side lengths on the blue figure are given as follows:

8 - 4 = 6 - 2 = 4.

The vertical side lengths on the transformed red figure is of:

4 - 2 = 3 - 1 = 2.

Hence the scale factor of the dilation is given as follows:

2/4 = 1/2.

Which means that options C and D are not correct.

The rules for each rotation are given as follows:

On the second quadrant, of the blue figure, we have that:

On the first quadrant, of the red figure, we have that:

Hence the rotation was clockwise, and option A is correct.

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Answer:

15d10mo

Step-by-step explanation:

Simplify:

(9+6)(mod10)

=15d10mo

The answer is 10.67. It can be calculated by integrating and then applying the limits.

What is the integration of [tex]x^n[/tex]?

The integration of power of a variable is given by the following formula:

[tex]\int x^ndx=\frac{x^{n+1}}{n+1}[/tex]

If limits are there then limits can be applied after integration.

Consider the given function.

[tex]\int^3_1(x^2-4x+9)dx[/tex]

Now, apply integration on each term as follows:

[tex]\int^3_1x^2dx-4\int^3_1xdx+9\int^3_1dx[/tex]

Now, use the following formula to integrate:

[tex]\int x^ndx=\frac{x^{n+1}}{n+1}[/tex]

Now, the integration will look like as follows:

[tex][x^3/3]^3_1-4[x^2/2]^3_1+9[x/1]^3_1[/tex]

Now, apply limits to find the answer as follows:

[tex][3^3/3-1^3/3]-4[3^2/2-1^2/2]+9[3-1]\\[/tex]

[tex][9-\frac{1}{3}]-4[9/2-1/2]+18\\10.67[/tex]

Hence, the answer is 10.67. It can be calculated by integrating and then applying the limits.

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Answer:

-3

Step-by-step explanation:

x - 3y = 4  Write in the slope intercept form.  Subtract x from both sides

-3y = -x + 4  Divide all the way through by -3

y = [tex]\frac{1}{3}[/tex] x - [tex]\frac{4}{3}[/tex]

The slope is the number before the x  ( 1/3 )

The perpendicular slope is the opposite reciprocal or -3

The equation associated with the quotient of twenty and a number, decreased by 4, is equal to zero is 20/x - 4 = 0 and that number will be 5.

Determine the known quantities and designate the unknown quantity as a variable while trying to set up or construct a linear equation to fit a real-world application.

In other words, an equation is a set of variables that are constrained through a situation or case.

Let's say that number is x,

Quotients of 20 and x will be given as 20/x

20/x - 4 = 0

20/x = 4

x = 20/4 = 5

Hence"The equation associated with the quotient of twenty and a number, decreased by 4, is equal to zero is 20/x - 4 = 0 and that number will be 5".

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[tex]\begin{array}{llll} ~\hspace{7em}\textit{negative exponents} \\\\ a^{-n} \implies \cfrac{1}{a^n} ~\hspace{4.5em} a^n\implies \cfrac{1}{a^{-n}} \end{array} ~\hfill \begin{array}{llll} \textit{rational exponents} \\\\ a^{\frac{ n}{ m}} \implies \sqrt[ m]{a^ n} \end{array} \\\\[-0.35em] ~\dotfill[/tex]

[tex]\cfrac{6x^{-1}y^{-1}}{\sqrt[4]{xy^4}}\implies \cfrac{6x^{-1}y^{-1}}{\left( xy^4 \right)^{\frac{1}{4}} }\implies \cfrac{6x^{-1}y^{-1}}{x^{\frac{1}{4}} y^{4\cdot \frac{1}{4}}}\implies \cfrac{6x^{-1}y^{-1}}{x^{\frac{1}{4}} y^1}\implies \cfrac{6x^{-1}y^{-1}x^{-\frac{1}{4}} y^{-1}}{1}[/tex]

[tex]6x^{-1}x^{-\frac{1}{4}}y^{-1} y^{-1}\implies 6x^{-1-\frac{1}{4}} y^{-1-1}\implies {\Large \begin{array}{llll} \underset{\textit{\small a}}{6} \stackrel{\textit{\small ~~ b}}{x^{-\frac{5}{4}}} \stackrel{\textit{\small ~~ c}}{y^{-2}} \end{array}}[/tex]

Answer:

The probability of a head of a die is 1/6 and that of a side of a coin

The probability that the total showing is less than 10 is 5/6.

A mathematical expression is made up of terms (constants and variables) separated by mathematical operators. A mathematical equation is used to equate two expressions. Equation modelling is the process of writing a mathematical verbal expression in the form of a mathematical expression for correct analysis, observations and results of the given problem. The formula to calculate the probability of occurrence of an event 'A' can be written as -  

P(A) = n(A)/n(S)

where -

n(A) = Number of outcomes favorable to event A.

n(S) = Total number of outcomes

Given are two dices rolled.

The total number of favorable outcomes can be calculated as -

[a] = (1,1) to (1,6) = 6

[b] = (2,1) to (2,6) = 6

[c] = (3,1) to (3,6) = 6

[d] = (4,1) to (4,5) = 5

[e] = (5,1) to (5,4) = 4

[f] = (6,1) to (6,3) = 3

n(A) = [a] + [b] + [c] + [d] + [e] + [f] =  6+6+6+5+4+3 = 30

n(S) = 36

So, the probability of occurrence of an event 'A' -

P(A) = n(A)/n(S)

P(A) = 30/36

P(A) = 5/6

Therefore, the probability that the total showing is less than 10 is 5/6.

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Answer:

Step-by-step explanation:

We can use the Side-Side-Side (SSS) Theorem – Three pairs of corresponding sides are congruent in two triangles.

The animal which travel at a rate of 1.5 meters per minutes is ants.

The correct answer option is option D

Based on the given task content, it can be solved using fraction.

Speed refers to the rate of change of distance at a given time period.

Average speed = Total distance / Time taken

Turtle = Total distance / Time taken

= 6/5

= 1.2 meters per minutes

Snail = Total distance / Time taken

= 4/8

= 0.5 meters per minutes

Caterpillar = Total distance / Time taken

= 7/7

= 1 meters per minutes

Ant = Total distance / Time taken

= 12/8

= 1.5 meters per minutes

In conclusion, ant is the animal which has an average speed of 1.5 meters per minutes

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Answer: A transformation in which all distances on the coordinate plane are shortened by multiplying either all x-coordinates (horizontal compression) or all y-coordinates (vertical compression) of a graph by a common factor less than 1.

Step-by-step explanation:

The Maclaurin series of  ln (1 + x³)  is

[tex]= x^3-\dfrac{1}{2}x^6+\frac{1}{3}x^9+\ldots \:[/tex]

The function is given in the question as

h(x) = ln (1 + x³)

Taylor (Maclaurin) series ofln (1 + x³) up to n = 3

A Maclaurin series is given by f(x),

[tex]f\left(x\right)=f\left(a\right)+\frac{f^'\left(a\right)}{1!}\left(x-a\right)+\frac{f^{''}\left(a\right)}{2!}\left(x-a\right)^2+\frac{f^{'''}\left(a\right)}{3!}\left(x-a\right)^3+\ldots[/tex]

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

We need to do to get the desired polynomial is to calculate the derivatives, evaluate them at the given point, and plug the results into the given formula.

f⁽⁰⁾(x) = f(x) = ln (1 + x³)

Evaluate the function at the point: f(0)=0

1st derivative: f(1)(x)=(ln(x3+1))′=3x²/(x³+1)  

1st derivative at the given point: (f(0))′=0

2nd derivative: f(2)(x)=(f(1)(x))′=(3x²/(x³+1) )′=3x(2−x³)/(x⁶+2x³+1) 

Evaluate the 2nd derivative at the given point: (f(0))′′=0

3rd derivative: f(3)(x)=(f(2)(x))′= (3x(2−x³)/(x⁶+2x³+1) )′= 6(x⁶−7x³+1)x9+3x⁶+3x³+1

Evaluate the 3rd derivative at the given point: (f(0))′′′=6

Now, use the calculated values to get a polynomial:

f(x)≈0/0!x⁰+0/1!x¹+0/2!x²+6/3!x³+0/4/!x⁴+0/5!x⁵

Thus, the Maclaurin series of  ln (1 + x³)  is

[tex]= x^3-\dfrac{1}{2}x^6+\frac{1}{3}x^9+\ldots \:[/tex]

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It is a uniform rectilinear movement which is one in which an object moves in a straight line, in one only direction, with a constant speed.

When we spoke of constant speed we mean that the movement retains the same speed, that is; that the object does not move faster, or slower and always at the same speed.

We obtain the data, as specified by the exercise.

Data:

 V = 28 m/s

 T = 10 s

 D = ?

We have that the uniform motion formula is:

     [tex]\large\displaystyle\text{$\begin{gathered}\sf V=\dfrac{d}{t}, \to where \end{gathered}$}[/tex]

We solve the formula, to calculate the distance and we obtain.And we also substitute our data.

[tex]\boxed{\large\displaystyle\text{$\begin{gathered}\sf \bf{d=v*t} \end{gathered}$} }[/tex]

[tex]\boxed{\large\displaystyle\text{$\begin{gathered}\sf \bf{d=28 \ \frac{m}{\not{s}}*10 \not{s} } \end{gathered}$}}[/tex]

[tex]\boxed{\boxed{\large\displaystyle\text{$\begin{gathered}\sf \bf{d=280 \ m } \end{gathered}$}}}[/tex]

Answer:

and the point F lies on the bottom face.

a) Construct a section of a parallelepiped by three given points E, F, G. Explain the construction of each of the segments (22 points).

b) Indicate the name (type) of the resulting polygon and shade its inner part (8 points).

1

Task 3.

An isosceles triangle ABC with base AB is given. From a point S lying outside the plane ABC, a perpendicular is dropped to point C.

a) Draw the linear angle of the dihedral angle SABC. Explain why this particular

They each will pass 12 scales.

What do you mean by algebraic expression?

The concept of algebraic expressions is the use of letters or alphabets to represent numbers without providing their precise values. We learned how to express an unknown value using letters like x, y, and z in the fundamentals of algebra. Here, we refer to these letters as variables. Variables and constants can both be used in an algebraic expression. A coefficient is any value that is added before a variable and then multiplied by it.

It is given that Becky has passed 4 scales, whereas her friend Charlotte has passed 8 of them.

Going forward, Becky has committed to passing 4 scales per week, and Charlotte has committed to passing 2 per week.

Let number of weeks be x

Becky has passed 4x + 4

Charlotte has passed 2x + 8

According to question:

4x + 4 = 2x + 8

4x - 2x = 8 - 4

2x = 4

x = 4/2 = 2

Becky has passed 4x + 4 = 4(2) + 4 = 8 + 4 = 12

Charlotte has passed 2x + 8 = 2(2) + 8 = 4 + 8 = 12

Therefore, They each will pass 12 scales.

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The perimeter of the image of the triangle is approximately equal to 87 units.

In this problem we find the coordinates of the vertices of a triangle: A(x, y) = (- 4, 3), B(x, y) = (4, 3), C(x, y) = (- 4, - 6), of which we must determine the perimeter of its image, which is the result of a scale factor of 3. The perimeter is the sum of the lengths of the triangle. The length of each side of the image of the triangle can be found by Pythagorean theorem and definition of dilation:

l = √[(Δx)² + (Δy)²]

Where:

And the perimeter of the image of the triangle is described below:

p = k · (AB + BC + AC)

Where k is the scale factor.

First, determine the lengths of sides AB, BC and AC:

Side AB

AB = √[[4 - (- 4)]² + (3 - 3)²]

AB = 8

Side BC

BC = √[(- 4 - 4)² + (- 6 - 3)²]

BC = √[(- 8)² + (- 9)²]

BC = √145

Side AC

AC = √[(- 4 - 4)² + (- 6 - 3)²]

AC = 9

Second, determine the perimeter of the image:

p = 3 · (8 + √145 + 9)

p = 3 · (17 + √145)

p = 51 + 3√145

p ≈ 87.125

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