Can someone help me with this question an my other work?

Answer:

[tex]A = \left[\begin{array}{cccc}1&-1&-3&-5\\4&2&0&-2\\7&5&3&1\end{array}\right][/tex]

Step-by-step explanation:

A = (aij)

i representa a linha e j a coluna.

Tipo 3x4

Isto implica que a matriz tem 3 linhas e 4 colunas.

aij = 3i – 2j.

Primeira linha:

[tex]a_{1,1} = 3(1) - 2(1) = 1[/tex]

[tex]a_{1,2} = 3(1) - 2(2) = -1[/tex]

[tex]a_{1,3} = 3(1) - 2(3) = -3[/tex]

[tex]a_{1,4} = 3(1) - 2(4) = -5[/tex]

Segunda linha:

[tex]a_{2,1} = 3(2) - 2(1) = 4[/tex]

[tex]a_{2,2} = 3(2) - 2(2) = 2[/tex]

[tex]a_{2,3} = 3(2) - 2(3) = 0[/tex]

[tex]a_{2,4} = 3(2) - 2(4) = -2[/tex]

Terceira linha:

[tex]a_{3,1} = 3(3) - 2(1) = 7[/tex]

[tex]a_{3,2} = 3(3) - 2(2) = 5[/tex]

[tex]a_{3,3} = 3(3) - 2(3) = 3[/tex]

[tex]a_{3,4} = 3(3) - 2(4) = 1[/tex]

Matriz:

A matriz é dada por:

[tex]A = \left[\begin{array}{cccc}1&-1&-3&-5\\4&2&0&-2\\7&5&3&1\end{array}\right][/tex]

Answer:

[tex]{ \tt{rate \: of \: change \: in \: A = 9}}[/tex]

Rate of change in function A is two times than that in function B

Answer:

2.56% chance of being selected

Step-by-step explanation:

Given

[tex]n = 39[/tex] --- you and 38 others

Required

Chance of you being selected

To do this, we simply calculate the probability using:

[tex]Pr(x) = \frac{n(x)}{n}[/tex]

Where:

[tex]n(x)= 1[/tex] --- i.e you are just 1 person

So:

[tex]Pr(x) = \frac{1}{39}[/tex]

[tex]Pr(x) = 0.0256[/tex]

Express as percentage

[tex]Pr(x) = 2.56\%[/tex]

Answer:

188 degrees

Step-by-step explanation:

The measure of the arc is the center angle, that is double of the circumference one

94 * 2 = 188 degrees

Answer:

387287i32

Step-by-step explanation:

i did it

Answer:

1) 5.2, and 4.94117647

2) Steph curry

Step-by-step explanation:

Answer:

Atomic No.

Element

Atomic Wt.

Europium

To take a systematic sample, first choose any number between one and five and write it here: k-23

Using the atomic number to count off elements, let the k value you just chose represent the first element in your

sample. Then, count off every fifth element thereafter. For instance, if you chose k-2, your sample will include

Helium (2), Nitrogen (7), Magnesium (12), and so forth. List each atomic number and atomic weight, starting with k

and adding 5 to the atomic number cach time.

Atomic No. Element

Atomic Wt.

3. Lithium 6.94

63

Isla6

8

v४

16

oxygen

Erbium 167.26

13 nluminum 24.98 23 Tantalum

180.45

18 Argon

39.95

28

platinum

195.DK

23 Vanadium 50.94 83 Bismuth 208.48

Niin

28

86

158.69

Radium 226

33

Arsenic 74.92

93 Neptunium 237

38 Strontium

98

87.62

Californium 251

43 Technutium 98

103 Lawrencium 262

UR

cocaina [12.41

108

Hassium

277

Iodine 126.9

113

No element

o

658 Cerium 140:11 118 Unumactiun

63

294

Find the mean atomic weight of this sample and enter it here: x xytomate : 137

AP statistics

Step-by-step explanation:

Step-by-step explanation:

[tex](\sec A - \csc A)(1 + \cot A + \tan A)[/tex]

[tex]=(\sec A - \csc A)\left(1 + \dfrac{\cos A}{\sin A} + \dfrac{\sin A}{\cos A} \right)[/tex]

[tex]=(\sec A - \csc A)\left(1 + \dfrac{\cos^2 A + \sin^2 A}{\sin A\cos A} \right)[/tex]

[tex]=(\sec A - \csc A)\left(\dfrac{1 + \sin A \cos A}{\sin A \cos A} \right)[/tex]

[tex]=\left(\dfrac{\frac{1}{\cos A} - \frac{1}{\sin A}+\sin A - \cos A}{\sin A\cos A}\right)[/tex]

[tex]=\dfrac{\sin A - \sin A \cos^2A - \cos A + \cos A\sin^2A}{(\sin A\cos A)^2}[/tex]

[tex]=\dfrac{\sin A(1 - \cos^2A) - \cos A (1 - \sin^2 A)}{(\sin A\cos A)^2}[/tex]

[tex]=\dfrac{\sin^3A - \cos^3A}{\sin^2A\cos^2A}[/tex]

[tex]=\dfrac{\sin A}{\cos^2A} - \dfrac{\cos A}{\sin^2A}[/tex]

[tex]=\left(\dfrac{1}{\cos A}\right)\left(\dfrac{\sin A}{1}\right) - \left(\dfrac{1}{\sin^2A}\right) \left(\dfrac{\cos A}{1}\right)[/tex]

[tex]=\sec^2A\csc A - \csc^2A\sec A[/tex]

Answer:

Factoring a perfect square trinomial.

Step-by-step explanation:

The left side was able to be simplified via factoring.

Answer:

reflection

Step-by-step explanation:

B is the mirror image of A across a line between the two images

Given:

The numbers are [tex]\dfrac{3}{5},2,0,1,-0.45, 1.44[/tex].

To find:

All the values that cannot be probabilities.

Solution:

We know that,

[tex]\text{Probability}=\dfrac{\text{Favorable outcomes}}{\text{Total outcomes}}[/tex]

The minimum value of favorable outcomes is 0 and the maximum value is equal to the total outcomes. So, the value of probability lies between 0 and 1, inclusive. It other words, the probability lies in the interval [0,1].

[tex]0\leq \text{Probability}\leq 1[/tex]

From the given values only [tex]\dfrac{3}{5}, 0, 1[/tex] lie in the interval [0,1]. So, these values can be probabilities.

The values [tex]2,-0.45, 1.44[/tex] does not lie in the interval [0,1]. So, these values cannot be probabilities.

Therefore, the correct values are [tex]2,-0.45, 1.44[/tex].

Answer:

[tex]f(g(2)) = 2[/tex]

Step-by-step explanation:

Given

[tex]x \to f(x)[/tex]

[tex]-6 \to 1[/tex]

[tex]-3 \to 2[/tex]

[tex]g(x) = f^{-1}(x)[/tex] --- inverse

Required

[tex]f(g(2))[/tex]

For two functions f(x) and g(x) where f(x) and g(x)are inverse;

[tex]f(g(x)) = x[/tex]

So, by comparison:

[tex]f(g(2)) = 2[/tex]

Answer:

The range is from 1.62 to 1.98.

Step-by-step explanation:

We have to solve for the percentage of the particular value if the range of the answer should be +/- 10% of the particular value.

The value given is 1.8, we thus want to find 10% of that: 1.8 * 10/100 = 0.18

Then, add this value to the original value of 1.8: 1.8+0.18 = 1.98

Furthermore, subtract .18 from from the original value of 1.8: 1.8-0.18 = 1.62

The range will be between these two numbers, so the range is from 1.62 to 1.98.

Answer:

the wording (punctuation) of the question can lead to different interpretations....

I assume that the question was >17 & even which is "5/19",

BUT... it can also be read as two questions

first >17 which is "10/19"

and second an even number which is "9/19"

BUT !!! I think that the question answer is 5/19

Step-by-step explanation:

Even Number  = 18/38 = 9/19

greater 17 = 20/38 = 10/19

Even & greater 17 = 10/38 = 5/19

Answer:

///////

Step-by-step explanation:

Answer:

usai964s46s694s4o6s64694s946649s469 opps

Answer:

[tex](x,y)=(7,5)[/tex]

Step-by-step explanation:

Megan's equation will be:

[tex]20x+35y=315[/tex]

[tex]x+y=12[/tex]

Substitute [tex]x=12-y[/tex] in the first equation:

[tex]20(12-y)+35y=315[/tex]

[tex]15y=75[/tex]

[tex]y=75/15[/tex]

[tex]y=5[/tex]

Find x:

[tex]x=12-5[/tex]

[tex]x=7[/tex]

Where x and y represent 30-second and 60-second ads sold, we find that Meghan's sales were:

[tex](x,y)=(7,5)[/tex]

hope this helps....

Answer:

[tex]\sqrt{x} -\frac{16}{\sqrt{x} }[/tex]

Step-by-step explanation:

Answer:

hey there hope this answer helps you out

Step-by-step explanation:

we have two functions f(x) and g(x) such that

[tex]f(x) = \frac{1}{x} [/tex]

and

[tex]g(x) = {x}^{2} + 6x[/tex]

solving for f ° g, we'll substitute the x in f(x) by the value of g(x)

[tex]fog = \frac{1}{g(x)} [/tex]

[tex]fog \: = \frac{1}{ {x}^{2} + 6x } [/tex]

taking x common in denominator

[tex]fog = \frac{1}{x(x + 6)} [/tex]

for a function to exist it should not have 0 in its denominator

checking the values of x for which the denominator of f ° g becomes 0 :-

so the function doesn't exist at values x = 0, -6

True

Step-by-step explanation:

[tex]P(t) = I^2(t)R[/tex]

Taking the derivative of P(t) with respect to time,

[tex]\dfrac{dP(t)}{dt} = 2I(t)\dfrac{dI(t)}{dt}R[/tex]

[tex]\:\:\:\:\:\:\:\:= 2(1)(-0.5)(500) = -500[/tex]

Answer:

n(t) = 11e^0.019t

Step-by-step explanation:

The estimated population in 2010 = 11,000,000 = initial population

The growth rate = 1.9% per year = 1.9/100 =

The exponential growth model follows the general form :

n(t) = ae^rt

a = Initial population ; r = growth rate ; t = period

Hence, we have ;

n(t) = 11e^0.019t in millions

gjgb go ftc in the class is not a short time to be be free and ear is not a short time to

Answer:

at least 99

Step-by-step explanation:

each number starting from 2 can be moved 11 times per thing.

Answer:

18

Step-by-step explanation:

1/3 of 27 is 9. 9 times 2 is 18.

Answer:

[tex]\boxed {\boxed {\sf \$15.06 \ billion }}[/tex]

Step-by-step explanation:

We are given the function f(x). This is the value of diamond production for the year in billions.

[tex]f(x)= 0.21x+13.8[/tex]

We want to find the value of the diamond production in 2006.

We know that x is the number of years after 2000. The year 2006 was 6 years after 2000 because 2006 - 2000 = 6. Therefore, x (years after 2000) is equal to 6. We can substitute 6 in for x.

[tex]f(6)= 0.21(6)+ 13.8[/tex]

Solve according to PEMDAS (Parentheses, Exponents, Multiplication, Division, Addition, Subtraction).

Multiply 0.21 and 6.

[tex]f(6)= 1.26+13.8[/tex]

Add 1.26 and 13.8

[tex]f(6)=15.06[/tex]

Remember the answer is in billions of dollars.

[tex]f(6)= \$ 15.06 \ billion[/tex]

Diamond production in 2006 is expected to be worth 15.06 billion dollars.

Answer:

Step-by-step explanation:

I always advise my students to make a table of information for these story problems because trying to keep track of the information otherwise is a nightmare. The table will look like this:

           d      =      r      *      t

m

c

m is motorcycle and c is car.

First thing we are told is that the ratio of m's speed to c's speed is 5:4; that means that we can divide 5/4 to find out how many times faster m is going than c.

5/4 = 1.25 so we have a couple of values to put into the table right away, along with the fact that they are both traveling the same distance of 160 miles.

             d      =     r      *      t

m        160     =   1.25r

c         160     =     r

The last thing we have to fill in is the time. If m travels a half hour less than c, c is driving a half hour more than m, right? Filling that in:

            d      =      r      *      t

m       160     =   1.25r   *     t

c        160     =      r       *  t + .5

Now we have our 2 equations. Looking at the top row of the table gives us the formula we need to solve this problem. It tells us, in other words, what we are going to be doing with these columns of numbers. Distance equals the rate times the time. For the motorcycle, the equation is:

160 = (1.25r)t  and that seems pretty useless since we still have 2 unknowns in there and you can only have 1 unknown in 1 equation. Let's see what the equation for the car is.

160 = (t + .5)r  Same problem.

Let's go back to the equation for the motorcycle and since we are looking for the rates of each, let's solve that equation for time in terms of rate (solve it for t):

[tex]t=\frac{160}{1.25r}[/tex] and sub that into the car's equation in place of t:

[tex]160=r(\frac{160}{1.25r})+.5r[/tex] and simplify. The r's to the left of the plus sign cancel out leaving us with:

[tex]160=(\frac{160}{1.25})+.5r[/tex] and divide those numbers inside the parenthesis to get:

160 = 128 + .5r and subtract 128 from both sides to get:

32 = .5r and finally divide by .5 to get

r = 64 miles/hour

The car goes 64 mph and the motorcycle goes 1.25 times that so,

m = 1.25(64) and

m = 80 mph

Answer:

answer is C

Step-by-step explanation:

General equation of a line is expressed as shown:

y = mx+c where;

m is the slope or gradient of the line

c is the intercept of the line

Given the equation of the line graph as y =2.5x

Comparing the given equation with the general equation, it is seen that m = 2.5 and c = 0 (since there is no value for the intercept)

Based on the explanation, the y-intercept of the graph is therefore 0

Answer:

B

Step-by-step explanation:

To find the x-intercept, substitute in

0 for y and solve for x

To find the y-intercept, substitute in 0 for x and solve for y

x-intercept(s): None

y-intercept(s): (0,1)

Given:

The graph that shows the ennoblement for college R between 1950 and 2000.

To find:

The two decades that has the greatest change in enrollment.

Solution:

From the given graph, it is clear that the change in the enrollment is:

From 1950 to 1960 is [tex]4-3.5=0.5[/tex] thousand.

From 1960 to 1970 is [tex]5-4.5=1.5[/tex] thousand.

From 1970 to 1980 is [tex]5.5-5=0.5[/tex] thousand.

From 1980 to 1990 is [tex]6.5-5.5=1[/tex] thousand.

From 1990 to 2000 is [tex]7-6.5=0.5[/tex] thousand.

The two decades 1960-1970 and 1980-1990 have the greatest change in enrollment.

Answer:1980 to 1990

Step-by-step explanation: