What is a 2-digit special number?.

Answer:

Not sure what you are asking, but 6 to the third power = 216 and 6 times 3 = 18. Hopefully one of those is the answer you are looking for.

b) h = [tex]tan \theta[/tex]  x 17 is the required equation.

c) The height of the streetlight is 20.4 ft.

The height of the person is 6 ft.

The distance at which the person is standing from the streetlight = 12 ft.

The distance at which the person's shadow appears = 5ft.

a) Hence, we can make the following diagram.

The right triangle below gives a visual representation of the problem. It shows the known quantities and a variable 'h' to indicate the height of the streetlight.

b) Let's now find an equation trigonometric function involving the unknown quantity 'h'.

from the below diagram, we can say that,

[tex]tan \theta[/tex] = h/base

base= 12+5=17

hence we have, [tex]tan \theta[/tex] = h/17

h = [tex]tan \theta[/tex]  x 17 is the required equation.

c) Now we have to find the height of the streetlight.

firstly let's find the value of [tex]\theta[/tex],

[tex]\theta[/tex] = tan 6/5.

[tex]tan^{-1}[/tex](6/5) = 50.2°

hence, tan 50.2 = h/17

                   1.20 = h/17

                         h= 1.20 x 17

                         h= 20.4 ft.

Hence, the height of the streetlight is 20.4 ft.

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Answer:B

Step-by-step explanation: Reducing 2x-2+3x results in 5x-2 because 2x+3x is 5x

Answer:

B)

Step-by-step explanation:

The answer is option B)

2x - 2 + 3x = 5x - 2

Equation of the tangent line to the curve  [tex]xe^{y} + ye^{x} = 5[/tex] at5 the point (0,5) is y = - ([tex]e^{5}[/tex] + 5)x + 5.

Given curve is,

[tex]xe^{y} + ye^{x} = 5[/tex]

We will differentiate both sides with respect to x,

[tex]\frac{d}{dx}[/tex][tex][xe^{y} + ye^{x} ][/tex] = [tex]e^{y} + xe^{y}\frac{dy}{dx} + e^{x}\frac{dy}{dx} + ye^{x}[/tex] = 0

⇒ [tex]\frac{dy}{dx}[/tex] [tex][xe^{y} + e^{x} ] = - [e^{y} + ye^{x} ][/tex]

⇒ [tex]\frac{dy}{dx}[/tex] = [tex]- \frac{e^{y}+ye^{x} }{xe^{y}+e^{x} }[/tex]

We have to find the equation at point (0,5).

Therefore substituting values of x and y,

that is x = 0 and y = 5.

[tex]\frac{dy}{dx} = - \frac{e^{5}+(5)e^{0} }{(0)e^{5} + e^{0} }[/tex]

    = [tex]- \frac{e^{5} + 5}{1}[/tex]

[tex]\frac{dy}{dx}[/tex]  = - [[tex]e^{5}[/tex] + 5] = gradient

Therefore the equation of tangent line is

(y - y1) = [m(x - x1)]

where m is the gradient

⇒ y - 5 = - [([tex]e^{5}[/tex]+5)(x - 0)]

⇒  y = - ([tex]e^{5}[/tex] + 5)x + 5

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Answer:

there fore kx = -13x^2-73x+5

then k = -13x^2-73x+5/x

Step-by-step explanation:

given that

5x+1x-2+4x^2+kx+9x^2-5+61x+6x-3 =

= you can write it as

4x^2+9x^2+kx+6x+61x+6x-5

= 13x^2+kx+73x-5

then find kx

-kx = 13x^2+73x-5

there fore

kx = -13x^2-73x+5

then k = -13x^2-73x+5/x

There are 2,540,160 strings of SALESPERSONS with no consecutive S's.

Explanation:

The number of r-combinations of a set of n distinct objects is C(n,r) = [tex]\frac{n!}{r!(n-r)!}[/tex]

Multiplication principle : If one event can occur in m ways AND a second event can occur in n ways, then the number of ways that the two events can occur in sequence is then mn.

We want to place the 4S's in the string of the other letters such that there are not consecutive. Since there are 9 positions including spaces between two consecutive letters in the 8-letter string and two positions on either side of the string, we need to place 4 of the 9 positions to contain a single S each.

By the definition of combination, there are ⁹C₄ ways to select 4 of the 9 positions to place the S's (as the order of the positions is not important since a different order of the S's leads to the same string).

By the multiplication principle, there are then ⁹C₄. [tex]\frac{8!}{2}[/tex] (divided by 2 since , in remaining 8 characters 2E are identical) possible strings of SALESPERSONS with no consecutive S's.

⁹C₄ . [tex]\frac{8!}{2}[/tex] = [tex]\frac{9!}{4!(9-4)!}[/tex] [tex]\frac{8!}{2}[/tex]

           = [tex]\frac{9!}{4!.5!} . 20,160[/tex]

           = 126.20,160

           = 2,540,160

Thus, there are 2,540,160 strings of SALESPERSONS with no consecutive S's.

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The value of the side EC for the given triangle is 28.75 .

A line parallel to one side of a triangle splits the other two sides of the triangle proportionally if the line also crosses the other two sides.

In the given triangle,

Side AD = 12, BD = 20 and AC = 56,

Also, the side BC is parallel to side DE.

Let the side EC is x,

Then side AE = 56-x

Since, lines BC and DE are parallel

Therefore, by triangle proportionality theorem,

AD/BD = AE/EC

12/20 = 56-x/x

3/5 = 56-x/x

3x = 230 - 5x

8x = 230

x = 230/8

x = 28.75

The required side EC is 28.75.

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No, every linear equation is not a quadratic equation.

It is defined as the relation between two variables, if we plot the graph of the linear equation we will get a straight line.

If in the linear equation, one variable is present, then the equation is known as the linear equation in one variable.

Any equation of the form ax²+bx+c=0  where x is variable and a, b, and c are any real numbers where a ≠ 0 is called a quadratic equation.

Every linear equation is also a quadratic equation. A polynomial equation is a linear equation or a quadratic equation.

Thus, not every linear equation is not a quadratic equation.

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The required set of 5 positive integers is 3, 3, 4, 4, 16.

The mean is the mathematical average of a group of integers.

Given, to find a set of 5 positive integers such that their mode is 3 and 4, the median is 4, and the mean is 6.

Since the median is 4, the middle number or the number at the 3rd position will be 4.

Since the mode of the data set is 3 and 4, the frequency of 3 and 4 must be 2.

Now, the numbers found so far are 3,3,4,4.

Let the fifth number be x.

The mean of the data set is given by:

mean = (3 + 3 + 4 + 4 + x)/5

Given that mean is 6, therefore,

(3 + 3 + 4 + 4 + x)/5 = 6

14 + x = 30

x = 30 -14

x = 16

Hence, the required set of 5 positive integers is 3, 3, 4, 4, 16.

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In multiple regression, a large value of the test statistic f indicates that most of the variation in the response is unexplained by the regression equation and a small value of f indicates that most of the variation in the response is explained by the regression equation. False.

Multiple regression refers to a statistical technique utilized to evaluate the relationship between a single dependent variable and several independent variables. The main goal of the analysis is to use the independent variables with known values to predict the value of the single dependent value. Test statistic f refers to a statistical measure used to test the significance of regression coefficients in linear regression models. In testing the validity of a multiple regression model, a large value of the F-test statistic indicates that most of the variation in y is explained by the regression equation, hence most of the variation in the response is explained by the regression equation.

Note: The question is incomplete as it is missing options which are a) True. b) False.

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[tex]\qquad \textit{Amount for Exponential Decay} \\\\ A=P(1 - r)^t\qquad \begin{cases} A=\textit{current amount}\dotfill & \$10000\\ P=\textit{initial amount}\dotfill &15000\\ r=rate\to 4\%\to \frac{4}{100}\dotfill &0.04\\ t=years\\ \end{cases}[/tex]

[tex]10000=15000(1 - 0.04)^{t} \implies \cfrac{10000}{15000}=(1 - 0.04)^{t}\implies \cfrac{2}{3}=0.96^t \\\\\\ \log\left( \cfrac{2}{3} \right)=\log(0.96^t)\implies \log\left( \cfrac{2}{3} \right)=t\log(0.96) \\\\\\ \cfrac{\log\left( \frac{2}{3} \right)}{\log(0.96)}=t\implies 9.93\approx t\qquad \textit{about 9 years and 11 months}[/tex]

A number c divided by 8 is greater than or equal to 26:

c/8 ≥ 26

The statement of pre-image is dilated is the resulting image congruent to the origina is false.

A dilation is a change that creates a picture that is a similar shape as the first, however is an alternate size. • An expansion that makes a bigger picture is called a development. • An expansion that makes a more modest picture is known as a decrease.

As we all know that dilation is used to resize the object. It make the objects larger or smaller.

It produces an image that is the same shape as the original, but is a different size.

This means, the original figure and the transformed figure would not be congruent.

Therefore, the statement 'When an object is dilated, the resulting image is congruent to the pre-image.' is False

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The point estimate of the population mean is  11.51.

A population is the entire set of people in a group, whether that group is a country or a collection of people who share a certain trait.

For the data given ;

11.17, 11.32, 11.97, 11.54, 11.77, 11.29, 11.45, 11.64

Let X0 be the population mean.

Then,

X0 = ∑X / n

∑X =  summation of all data

n = total number of observations.

X0 = 11.17 + 11.32 + 11.97 + 11.54 + 11.77 + 11.29 + 11.45 + 11.64 / 8

X0 = 92.15/8

X0 = 11.51

Thus, a point estimate of the population mean is  11.51.

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The estimated probability of the occupation of firefighters is 0.62

A recent Harris Poll survey of 1010 U.S. adults selected at random showed that 627 consider the occupation of firefighter to have very great prestige. i.e. The sample size of U.S. adults : n= 1010

The number of U.S. adults consider the occupation of firefighter to have very great prestige: x= 627

Now we see, the probability that a U.S adult selected at random thinks the occupation of firefighters has very great prestige will be:

p = x/n

 = 627/ 1010 = 0.62 [ To the nearest hundredth]

        Therefore, the estimated probability that a U.S adult selected at random thinks the occupation of firefighters has very great prestige = 0.62

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The percent of the parcels weighed between 14 ounces and 35 ounces will be 0.818. (b) The percentage of the parcels weighing more than 49 ounces will be 0.14 %.

(a)The percent of the parcels weighed between 14 ounces and 35 ounces will be 0.818.

The given data in the problem is;

The probability of parcels weighng 35 ounces or less, P₁=0.841

The probability of parcels weighing 14 ounces or less, P₂=0.023

The probability of parcels weighing between 14 and 35 ounces is found as;

P=P₁-P₂

P=0.841-01.023

P=0.818

The probability in the percentage will be 0.818 x 100 = 81.8

(b) The percent of the parcels weighing more than 49 ounces will be 0.14 %.

The probability of parcels weighing more than 49 ounces will be;

P₃ = 1 - 0.99835

P₃ = 0.00135

The probability in the percentage will be 0.00135 x 100 = 0.135% = 0.14%

Hence the percentage of the parcels weighed between 14 ounces and 35 ounces will be 0.818.

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Disclaimer

The question given by you is incomplete, a similar question has been solved above

The question is

Use the Empirical Rule to answer each question.

During 1 week an overnight delivery company found that the weights of its parcels were normally distributed, with a mean of 28 ounces and a standard deviation of 7 ounces.

(a) What percent of the parcels weighed between 14 ounces and 35 ounces? (Round your answer to one decimal place.)

(b) What percent of the parcels weighed more than 49 ounces? (Round your answer to two decimal places.)

for every one (1) you have, you have 3 times the first

Ok,

for example for every 1 penny you have, you have 3dollars,

you counted and you have 4 pennies, in order to find out how many dollars you have, you have to times 4 by 3 to get 12

altogether you would have 4 pennies: 12 dollars.

The equation of a line that is perpendicular and passing through a point is  y = -1/m + (y1 + x1/m).

Let y = mx + b be the perpendicular line and (x1,y1) be the point.

slope of y = mx + b is:

= m

we know that,

perpendicular slope = -1/m

substitute m and (x1,y1)

y1 = -1/m(x1) + b

y1 + x1/m = b

substitute b value in y = -1/m + b.

y = -1/m + (y1+x1/m)

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Correct option is:

Use a computer program to select the locations who get coupons and those who don't.

What is Volunteer role?

Volunteer work involves people giving freely of their time, labour, and skill; they frequently complete tasks that organizations would otherwise have to pay for. Governors are one example of a voluntary position in education, Trustees.

Given: Lucky's Market wanted to see if they could increase the number of sales per day before they close their locations nation wide.

To test this, they randomly chose half of their 80 locations and send 50% off coupons to their customers only at those locations, while the other half of locations did not receive the 50% off coupons.

They also collected information the amount of money spend on these extra customers.

So, the correct option is, use a computer program to select the locations who get coupons and those that don't.

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The HCF and LCM of 14 and 28 are 14 and 28 respectively.

Highest Common Factor:

The full form of HCF in maths is the Highest Common Factor. HCF of two or more numbers is the greatest factor that divides the numbers. For example, 2 is the HCF of 4 and 6.

Lowest common multiple:

LCM stands for least common multiple. The least common multiple of two numbers is the smallest number which is a multiple of both of them.

Here we have to find the HCF and LCM of 14 and 28.

The two numbers are 14 and 28.

We can write the number as:

14 = 2 × 7

28 = 2² × 7

The HCF is the common number in both the number, which is 2 and 7.

So HCF of two number = is 2 × 7

= 14

The LCM is the largest common multiple of the given numbers.

So LCM of two number = is 2² × 7

= 28

Therefore the HCF and LCM are 14 and 28 respectively.

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The probability of rolling a number greater than 2 is 2/3

Given that a six sided die with sides labelled with numbers 1,2,3,4,5 and 6 is rolled once.

Since die is said to be equally likely for each number, die is unbiased

P(1) = P(2)=...=P(6) = 1/6

For a number greater than 2, we must not include 2 and hence numbers can be 3 or 4 or 5 or 6

Probability of rolling a number >2 = Probability of rolling 3 or 4 or 5 or 6

= P(3)+P(4)+P(5)+P(6)

= 1/6+1/6+1/6+1/6

=4/6

=2/3

Therefore, the probability of rolling a number greater than 2 is 2/3

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c is a constant of the function y = ax² + bx + c is -3

An equation is an expression that shows the relationship between two or more numbers and variables.

Given that:

y = ax² + bx + c

At point (5, 32)

32 = a(5²) + 5b + c   (1)

At point (6,45)

45 = a(6²) + 6b + c   (2)

At point (4, 21)

21 = a(4²) + 4b + c   (3)

Hence:

a = 1, b = 2 and c = -3

The value of c which is a constant of the function y = ax² + bx + c, is -3.

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there are 41479 to distribute eight different toys among four children if the first child gets at least two toys.

It is given that,

Total number of toys: 8

Total number of children: 4

The total number of ways to distribute eight different toys among four children is 4⁸.

If the first child gets no toys then the total number of ways to distribute eight different toys among the four children is 3⁸, and if the first child gets one toy then the total number of ways to distribute eight different toys among the four children is,

⁸C1 * 3⁷

The number of ways there to distribute eight different toys among four children if the first child gets at least two toys is calculated as follows.

Number of ways

=4⁸−3⁸−(⁸C1 ∗ 3⁷)

=65536 − 6561 −17496

=41479

Therefore, the required number of ways is 41479

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√(3/4) and √(5/4) are the irrational numbers

Irrational numbers are numbers that cannot be expressed in the form a/b. They are mostly in the form of square roots when evaluated e.g √7, 2/√5, 3√11, etc.

The opposite of irrational numbers is rational numbers. Rational numbers are numbers that can be written in the form of a/b, where a and b are integers, and b is not equal to 0. e.g 1/3, 2/3, √4 = 4/1, √(1/9) = 1/3, etc.

Given: √(1/4), √(2/4), √(3/4), √(4/4), and √(5/4)

For √(1/4):

√(1/4) = 1/2

This is a rational number

For √(2/4):

√(2/4) = (√2) / 2

This is an  irrational number

For √(3/4):

√(3/4) = (√3) / 2

This is an irrational number

For √(4/4):

√(4/4) = 2/2 = 1

This is a rational number

For √(5/4):

√(5/4) = (√5) / 2

This is an  irrational number

Therefore, the irrational numbers are √(3/4) and √(5/4). So select them

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0 is the minimum (and maximum) of x−y for x, y ∈ (0, 1).

What is real numbers?

A real number in mathematics is a number that can be used to represent a continuous one-dimensional quantity, such as a length of time, temperature, or distance. In this context, continuous denotes that values may vary by arbitrary little amounts. An endless decimal expansion allows for a nearly universal representation of every real number.

The given equation is,

[tex]log_xa+log_ya=4log_x_ya[/tex]

Using the log rule: [tex]log_ab = \frac{ln b}{lna}[/tex]

⇒ [tex]\frac{lna}{lnx}+\frac{lna}{lny} = \frac{4lna}{lnxy}[/tex]

Multiply both sides by, [tex]\frac{lnxy}{lna}[/tex]

⇒  [tex]\frac{lnxy}{lnx}+\frac{lnxy}{lny} = 4[/tex],  a ≠ e

Using log rule: [tex]ln(ab) = lna + lnb[/tex]

⇒

[tex]\frac{lnx+lny}{lnx} +\frac{lnx+lny}{lny} = 4\\ \frac{lnx}{lnx}+\frac{lny}{lnx}+ \frac{lnx}{lny}+\frac{lny}{lny} = 4\\1+\frac{lny}{lnx}+ \frac{lnx}{lny} +1=4\\\frac{lny}{lnx}+\frac{lnx}{lny}=2[/tex]

[tex]\frac{(lnx)^2+(lny)^2}{lnx(lny)} =2\\(lnx)^2+(lny)^2 = 2 lnx ln y\\(lnx)^2 - 2lnxlny+(lny)^2=0\\(lnx - lny)^2 = 0\\lnx - ln y = 0\\x = y[/tex]

Hence, 0 is the minimum (and maximum) of x−y for x, y ∈ (0, 1).

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The probability that a person selected at random to have the test until (s)he passes the test is, 0.03125

Given, a test occurred 5 times in a class.

we have to find that how many times a person selected at random to have the test until (s)he passes it.

the probability of getting pass in the exam be, 1/2

and similarly the probability of getting fail in the exam also be, 1/2

so, the probability that a person selected at random to have the test until (s)he passes the test is,

(1/2)^5

= 1/32

= 0.03125

Hence, the probability that a person selected at random to have the test until (s)he passes the test is, 0.03125

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The examples of ASA congruence in its only form, ASA is calculated via way of means of: ASA = Total Wait Time for Answered Calls/Total # of Answered Calls.

The concept at the back of ASA is to get a top level view of trendy performance.any angles and the aspect covered among the angles of 1 triangle are equal to the corresponding angles and aspect covered among the angles of the second one triangle, then the 2 triangles are stated to be congruent via way of means of ASA rule.

Examples on ASA Congruence RuleExample 1: For Δ ABC, display that AB = AC and Δ ABC is isosceles if the bisector AD of ∠ A is perpendicular to aspect BC. Also, this situation satisfies the homes of the isosceles triangles, thus, Δ ABC is an isosceles triangle. Example 2: Prove that △ACF≅△AEB, if ∠C=∠E and AC=AE.

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andi $8,$8,0xFF is the instructions that clears all the bits in register $8 except for the low order byte, which is unchanged.

The byte with the least amount of the value would be the low-order byte. For instance, a 16-bit int with the value 5,243 might be represented in hex as 0x147B. The low-order byte is 0x7B, whereas the high-order byte is 0x14. A word's bits are numbered 0 through 15, with bit 0 being the least important. The word's low byte, which contains bit 0, and high byte, which contains bit 15, are referred to as such. Byte ordering comes in two flavors. Big-endian format places the byte with the most important bit in the first memory location, followed by bytes whose importance decreases.

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The equation which describes the problem statement would be 250x = 750.

What is the equation in one variable?

A linear equation in one variable is an equation that has only one solution and is expressed in the form ax+b = 0, where a and b are two integers and x is a variable. 2x+3=8, for example, is a linear equation with a single variable.

The rent for staying one night is $250.

and if a couple stays in the standard room for x night and their total rent was $750 then this can be expressed as

             250x = 750

Hence, the equation which describes the problem statement would be 250x = 750.

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Answer:

slope =-  [tex]\frac{1}{3}[/tex]

Step-by-step explanation:

calculate the slope m using the slope formula

m = [tex]\frac{y_{2}-y_{1} }{x_{2}-x_{1} }[/tex]

with (x₁, y₁ ) = (2, - 3 ) and (x₂, y₂ ) = (5, - 4 )

m = [tex]\frac{-4-(-3)}{5-2}[/tex] = [tex]\frac{-4+3}{3}[/tex] = [tex]\frac{-1}{3}[/tex] = - [tex]\frac{1}{3}[/tex]

Answer:

points are ;

(5,-4).......X2, Y2

now

slope(m)=y2 -Y1 / X2-X1

= -4+3 / 5-2

-1 / 3