What is the area under the curve y=x−x^2and above the x-axis?

Answer:

The range represents the number of users each month for 24 months

Step-by-step explanation:

The range depends on the y axis. Looking at the y axis we see the number of users for each month. Looking at the x axis we see that it's 24 months not 20 months

From the graph, we can interpret the range represents the number of users each month for 24 months. Therefore, option B is the correct answer.

We need to find what can be interpreted from the range of the given graph.

The range is the difference between the smallest and highest numbers in a list or set. To find the range, first put all the numbers in order. Then subtract (take away) the lowest number from the highest.

From the given graph, we can see the x-axis represents the number of months and the y-axis represents the number of users.

So, the range represents the number of users each month for 24 months

20-15.8=4.2

Therefore, option B is the correct answer.

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Answer:

The distance between the origin and the given point is 13.9283 units.  

Step-by-step explanation:

The coordinates of origin are (0,0,0)

We are given a point R(9,7,80

The distance formula:

[tex]\sqrt{(y_1 -x_1)^2 + (y_2 -x_2)^2 + (y_3 -x_3)^2}[/tex], where [tex](x_1, x_2, x_3)[/tex] are coordinates of one point and  [tex](y_1, y_2, y_3)[/tex] are coordinates of other point.

Putting the values as:

[tex]y_1 = 9, y_2 = 7, y_3 = 8\\x_1 =0, x_2 = 0, x_3 = 0[/tex]

We get d = [tex]\sqrt{81 + 64 + 49}[/tex]

d = [tex]\sqrt{194}[/tex]

d = 13.9283

Thus, the distance between the origin and the given point is 13.9283 units.

Answer:

You can make 180 meals.

Step-by-step explanation:

We can look at this problem the following way:

For each appetizers, we can choose 3 salads.

For each salad, we can choose 5 entrees. So we already have 3 salads, each with 5 possible entrees, so there are already 3*5 = 15 possibilities.

For each entree, we can choose 4 desserts. So for each of the 15 possibilites of salad and entrees, there are 4 desserts. So there are 15*4 = 60 possibilities.

There are also 3 appetizers possible for each of the 60 possibilities of salads, entrees and desserts. So there are 60*3 = 180 possibilities.

Answer:

a)Total Labor Cost = $600x + $30y.

b) Total variable cost = $3000.

c) Total labor cost last month = $6000.

d) Unit cost of labor per class = $60.

e) Total variable labor cost = $30 * 150 = $4500, Total Labor Cost = $3000 + $4500 = $7500, and Unit cost of labor per class = $7500/150 = $50.

Step-by-step explanation:

a) It is given that the labor cost includes two components: cost of trainers, which is actually their salaries, and cost of a fitness class taught. Fitness trainer costs $600 and one fitness class costs $30. Assuming there are x number of trainers and y number of classes, therefore the model can be expressed as:

Total Labor Cost = Fitness Trainer Cost * number of trainers + Fitness Class Cost * number of classes.

Total Labor Cost = $600x + $30y.

b) The total variable labor cost will be the cost spent on the number of classes. Since number of classes are 100 and the cost of one class is $30, therefore:

Total variable cost = cost of one class * number of classes.

Total variable cost = $30 * 100.

Total variable cost = $3000.

c) Furthermore, last month, x = 5 and y = 100. Plug these values in the total labor cost equation:

Total labor cost last month = $600(5) + $30(100).

Total labor cost last month = $3000 + $3000.

Total labor cost last month = $6000.

d) The total labor cost is $600. Number of classes are 100. Therefore:

Unit cost of labor per class = Total Labor Cost/Number of classes.

Unit cost of labor per class = $6000/100.

Unit cost of labor per class = $60.

e) If the number of classes are increases by 50%, this means that the number of classes will be 150 instead of 100. Therefore:

Total variable labor cost = $30 * 150 = $4500.

Total Labor Cost = $3000 + $4500 = $7500.

Unit cost of labor per class = $7500/150 = $50!!!

Answer:

System of equations:

[tex]x_1+2x_2+2x_3=6\\2x_1+x_2+x_3=6\\x_1+x_2+3x_3=6[/tex]

Augmented matrix:

[tex]\left[\begin{array}{cccc}1&2&2&6\\2&1&1&6\\1&1&3&6\end{array}\right][/tex]

Reduced Row Echelon matrix:

[tex]\left[\begin{array}{cccc}1&2&2&6\\0&1&1&2\\0&0&1&1\end{array}\right][/tex]

Step-by-step explanation:

Convert the system into an augmented matrix:

[tex]\left[\begin{array}{cccc}1&2&2&6\\2&1&1&6\\1&1&3&6\end{array}\right][/tex]

For notation, R_n is the new nth row and r_n the unchanged one.

1. Operations:

[tex]R_2=-2r_1+r_2\\R_3=-r_1+r_3[/tex]

Resulting matrix:

[tex]\left[\begin{array}{cccc}1&2&2&6\\0&-3&-3&-6\\0&-1&1&0\end{array}\right][/tex]

2. Operations:

[tex]R_2=-\frac{1}{3}r_2[/tex]

Resulting matrix:

[tex]\left[\begin{array}{cccc}1&2&2&6\\0&1&1&2\\0&-1&1&0\end{array}\right][/tex]

3. Operations:

[tex]R_3=r_2+r_3[/tex]

Resulting matrix:

[tex]\left[\begin{array}{cccc}1&2&2&6\\0&1&1&2\\0&0&2&2\end{array}\right][/tex]

4. Operations:

[tex]R_3=\frac{1}{2}r_3[/tex]

Resulting matrix:

[tex]\left[\begin{array}{cccc}1&2&2&6\\0&1&1&2\\0&0&1&1\end{array}\right][/tex]

Answer:

look at the step by step explanation

Step-by-step explanation:

10<x=45

i dunno if this is correct

Answer: The attendance before the increase was 4712.

Step-by-step explanation:

Let the attendance before the increase be 'x'.

Rate of increment = 5%

so, Attendance after increment becomes

[tex]\dfrac{100+5}{100}\times x = 4948\\\\\dfrac{105}{100}\times x=4948\\\\1.05\times x=4948\\\\x=\dfrac{4948}{1.05}\\\\x=4712.38\\\\x=4712[/tex]

Hence, the attendance before the increase was 4712.

Final answer:

To find the original attendance, an equation was formulated where the original attendance (x) increased by 5% equals 4,948. By solving for x, the original attendance before the increase was determined to be approximately 4,712 when rounded to the nearest whole number.

Explanation:

The question asks to find the original attendance before a 5% increase that resulted in a final attendance of 4,948. To calculate the initial attendance, you can set up an equation where the original attendance (which we will call 'x') plus 5% of the original attendance equals the final attendance (4,948).

This can be expressed algebraically as:
x + 0.05x = 4948

Solving for x gives you:
1.05x = 4948
x = 4948 / 1.05
x = 4,712 (rounded to the nearest whole number)

Therefore, the attendance before the increase was approximately 4,712.

Answer:

[tex]52\times \frac{7}{6}[/tex]

[tex]60\frac{2}{3}\text{ pounds}[/tex]

Step-by-step explanation:

Given,

The total pounds eaten by cat in 3 weeks = [tex]3\frac{1}{2}[/tex]= [tex]\frac{7}{2}[/tex]

∵ Each day the pounds of eaten is same,

⇒ Total pounds eaten in each week = [tex]\frac{7}{6}[/tex]

∵ 1 year = 52 weeks,

So, the pounds eaten in 52 weeks or 1 year = 52 × pounds eaten in each week

= 52 × [tex]\frac{7}{6}[/tex]

Which is the required expression,

By solving it,

The number of pounds eaten by cat in a year

[tex]=\frac{364}{6}[/tex]

[tex]=60\frac{2}{3}[/tex]

Answer:

(14a+3, 21+4) = 1

Step-by-step explanation:

We are going to use the Euclidean Algorithm to prove that these two integers have a gcd of 1.

gcd (14a + 3, 21a + 4) = gcd (14a+3, 7a + 1) = gcd (1, 7a+1) = 1

Therefore,

(14a + 3, 21a + 4) = 1

Answer:

$248,000.

Step-by-step explanation:

We have been given that the revenue from manufacturing and selling x units of toaster ovens is given by [tex]R(x)=-0.3x^2+200x-82,000[/tex].

To find the amount of revenue earned from selling 3,000 toaster, we will substitute [tex]x=3,000[/tex] in the given formula as:

[tex]R(3,000)=-0.03(3,000)^2+200(3,000)-82,000[/tex]

[tex]R(3,000)=-0.03*9,000000+600,000-82,000[/tex]

[tex]R(3,000)=-270,000+518,000[/tex]

[tex]R(3,000)=248,000[/tex]

Therefore, the company should expect revenue of $248,000 from selling 3,000 toaster ovens.

same as before here, the bird is up above and from there goes down, so we sum up both amounts.

[tex]\bf \stackrel{mixed}{528\frac{1}{5}}\implies \cfrac{528\cdot 5+1}{5}\implies \stackrel{improper}{\cfrac{2641}{5}}~\hfill \stackrel{mixed}{89\frac{3}{5}}\implies \cfrac{89\cdot 5+3}{5}\implies \stackrel{improper}{\cfrac{448}{5}} \\\\[-0.35em] ~\dotfill\\\\ \cfrac{2641}{5}+\cfrac{448}{5}\implies \stackrel{\textit{using an LCD of 5}}{\cfrac{(1)2641+(1)448}{5}}\implies \cfrac{3089}{5}\implies 617\frac{4}{5}[/tex]

Answer:

[tex]617\frac{4}{5}[/tex]

Step-by-step explanation:

We have been given that a bird flies from its nest [tex]528\frac{1}{5}[/tex] to the bottom of the canyon [tex]-89\frac{3}{5}[/tex].

First of all, we will convert both mixed fractions into improper fraction.

[tex]528\frac{1}{5}\Rightarrow \frac{2640+1}{5}=\frac{2641}{5}[/tex]

[tex]89\frac{3}{5}\Rightarrow \frac{445+3}{5}=\frac{448}{5}[/tex]

To solve our given problem, we will find difference of both elevations as:

[tex]\frac{2641}{5}-(-\frac{448}{5})[/tex]

[tex]\frac{2641}{5}+\frac{448}{5}[/tex]

[tex]\frac{2641+448}{5}[/tex]

[tex]\frac{3089}{5}[/tex]

[tex]617\frac{4}{5}[/tex]

Therefore, the bird flown [tex]617\frac{4}{5}[/tex] units.

Answer:

The number of seniors who scored above 96% is 1.

Step-by-step explanation:

Consider the provided information.

Two percent of all seniors in a class of 50 have scored above 96% on an ext exam.

Now we need to find the number of seniors who scored above 96%

For this we need to find the two percent of 50.

2% of 50 can be calculated as:

[tex]\frac{2}{100}\times50[/tex]

[tex]\frac{100}{100}[/tex]

[tex]1[/tex]

Hence, the number of seniors who scored above 96% is 1.

Answer:

Step-by-step explanation:

Given that t A, B, and C are invertible matrices of the same

size.

To pr that [tex](ABC)^{-1} = C^{-1}B^{-1}A^{-1}.[/tex]

to pr that  [tex]( ABC) (C^{-1}B^{-1}A^{-1})=e[/tex]

LS=[tex]( AB(C (C^{-1})B^{-1}A^{-1})\\=ABIB^{-1}A^{-1})\\=A(BB^{-1})A^{-1})\\=AA^{-1})\\=I[/tex]

Thus proved

To show that the product of invertible matrices A, B, and C (ABC) is invertible and that its inverse is (ABC)-1 = C-1B-1A-1, we use the property that the inverse of a product is the product of inverses in reverse order, and the associativity of matrix multiplication.

A student asked to show that if A, B, and C are invertible matrices of the same size, the product ABC is invertible, and that the inverse of this product is (ABC)-1 = C-1B-1A-1.

To prove this, we can use the properties of matrix multiplication and inverses:

This completes the proof that the product ABC is invertible and its inverse is C-1B-1A-1

The probability of selecting a cookie with either chocolate chips or pecans from a plate of 50 cookies is 36%.

The question asks about the probability of selecting a cookie with either chocolate chips or pecans from a plate of 50 cookies, with some overlapping flavors. Given the information:

Total cookies: 50

Chocolate chip cookies: 10

Pecan cookies: 14

Cookies with both flavors: 6

To calculate the probability of choosing a cookie with either chocolate chips or pecans (or both), we can use the formula:

P(C OR N) = P(C) + P(N) \'u2013 P(C AND N)

where:

P(C) is the probability of choosing a chocolate chip cookie,

P(N) is the probability of choosing a pecan cookie, and

P(C AND N) is the probability of choosing a cookie with both flavors.

Thus:

P(C) = 10/50 = 0.20,

P(N) = 14/50 = 0.28, and

P(C AND N) = 6/50 = 0.12.

Therefore:

P(C OR N) = 0.20 + 0.28 - 0.12 = 0.36 or 36%.

This is the probability that a randomly selected cookie from the plate will have either chocolate chips, pecans, or both.

Sets and set operations are ways of organizing, classifying and obtaining information about objects according to the characteristics they possess, as objects generally have several characteristics, the same object can belong to several sets, an example is the subjects of a school , where students (objects) are classified according to the subject they study (set).

The intersection of sets is a new set consisting of those objects that simultaneously possess the characteristics of each intersected set, the intersection of two subjects will be those students who have both subjects enrolled.

The union of sets is a new set consisting of all the objects belonging to the united sets, the union of two subjects will be all students of both courses.

In this case there are three sets B, C and S of which we are given the following information:

Answer

n(BꓵSꓵC)=5

n(BꓵS)=10 – 5 = 5

n(BꓵC)=12 – 5 = 7

n[(BꓵC)ꓴ(BꓵS)ꓴ(CꓵS)]=21  – 5 – 5 – 7 = 4

n(B)=36 – 5 – 5 – 7 = 19

n(S)=30 – 5 – 5 – 4 = 16

n(C)=34 – 5 – 7 – 4 = 18

Answer: 40.

Step-by-step explanation:

Given : The sweater department ran a sale last week and sold 95% of the sweaters that were on sale.

95% can be written as 0.95  [ by dividing 100 ]

Also, the number of sweaters sold = 38

Let x be the number of sweaters were on sale.

Then , we have the following equation :_

[tex]0.95x=38\\\\\Rightarrow\ x=\dfrac{38}{0.95}=\dfrac{3800}{95}=40[/tex]

Hence, 40 sweaters were on sale.

Answer:

If x = -4, then -x = 4.

Step-by-step explanation:

The procedure for answering this question is straightforward, you just have to substitute the value of x, when you write -x:

If x = -4, then -x = -(-4) = 4 (remember the law of signs)

Therefore: -x = 4.

It actually holds for any real number.

Answer:

8 rental movies will make the costs of the two stores be the same

Step-by-step explanation:

- Mr Cosgrove is comparing movie rentals deals

- Netflix charge a flat rate of $8.50

- Blockbuster charge $4.50 plus a $0.50 per movie

- We need to find how many movie rental make the costs of the two

  stores will be the same

- Assume that the number of rental movies is m

∵ Blockbuster charge $4.50 plus a $0.50 per movie

∴ Blockbuster cost = 4.50 + 0.50 m

∵ The cost of Netflix = 8.50

- Equate the two costs to find the number of the rental movies

∵ 4.50 + 0.50 m = 8.50

- Subtract 4.50 from both sides

∴ 0.50 m = 4

- Divide both sides by 0.50

∴ m = 8 movies

8 rental movies will make the costs of the two stores be the same

Answer:

Claim : The average time spent on a delivery does not exceed 25 minutes.

n = 21

We are given that  p-value for the test was found to be .0284.

Now we are supposed to state the conclusions

a) At α = .025, we fail to reject [tex]H_0[/tex].

p value = 0.0284

α = 0.025

P value > α

So, we accept the null hypothesis i.e. we fail to reject null hypothesis.

b)At α = .05, we fail to reject [tex]H_0[/tex].

p value = 0.0284

α = 0.025

P value < α

So, we reject the null hypothesis

c)At α =.03, we fail to reject [tex]H_0[/tex].

p value = 0.0284

α = 0.025

P value < α

So, we reject the null hypothesis

d)At α =.02, we reject [tex]H_0[/tex].

p value = 0.0284

α = 0.025

P value > α

So, we accept the null hypothesis i.e. we fail to reject null hypothesis.

The probability of selecting a red marble from the jar of marbles is 2/17.

To calculate the probability of selecting a red marble from the jar, we need to determine the total number of marbles in the jar and the number of red marbles.

Total number of marbles = 2 (red) + 3 (white) + 5 (blue) + 7 (green) = 17

Number of red marbles = 2

Now, the probability of selecting a red marble is given by:

Probability = (Number of red marbles) / (Total number of marbles) = 2 / 17

So, the correct probability of selecting a red marble from the jar is 2/17.

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Final answer:

The probability of selecting a red marble from the jar is [tex]\frac{2}{17}[/tex] since there are 2 red marbles out of a total of 17 marbles.

Explanation:

The probability of selecting a red marble from a jar of marbles containing two red marbles, three white marbles, five blue marbles, and seven green marbles can be calculated by dividing the number of red marbles by the total number of marbles in the jar. First, we determine the total number of marbles: 2 (red) + 3 (white) + 5 (blue) + 7 (green) = 17 marbles. Next, we calculate the probability of selecting a red marble by dividing the number of red marbles (2) by the total (17).

Therefore, the probability of selecting a red marble is [tex]\frac{2}{17}[/tex].

Answer: SERIES = 670  Ω

PARALLEL = 91.94 Ω

Step-by-step explanation:

Hi, resistors in series obey the following equation :

R1+ R2 = RT

RT is the equivalent resistance. We have the value of both resistances, so we apply the ecuation:

R1 = 110Ω, and R2 = 560Ω

110Ω+ 560Ω = 670 Ω

When resistors are in parallel, resistors obey the following equation:

1/R1 + 1/R2= 1/RT

so, in our case:

1/ 110Ω +1/560Ω = 1/rt

0.01087Ω = 1/RT

RT= 1/0.01087 Ω= 91.94 Ω

Answer:

240

Step-by-step explanation:

Given:

The cost function for canoes, c(x) = 24,000 + 100x

Revenue function for canoes, R(x) = 200x

here, x is the number of canoes sold

Fixed cost = $24,000

Cost of production = $100

now at break even point, Revenue = cost

thus,

200x = 24,000 + 100x

or

100x = 24,000

or

x = 240 canoes

Hence, at breakeven, the number of canoes sold is 240

Answer:

P = 0.42

Step-by-step explanation:

This probability problem can be solved by building a Venn like diagram for each probability.

I say that we have two sets:

-Set A, that is the probability of receiving this virus through the email.

-Set B, that is the probability of receiving it through the webpage.

The most important information in these kind of problems is the intersection. That is, that he virus enters the system simultaneously by both email and webpage with a probability of 0.17. It means that [tex]A \cap B[/tex] = 0.17.

By email only

The problem states that there is a 40 chance of receiving it through the email. It means that we have the following equation:

[tex]A + (A \cap B) = 0.40[/tex]

[tex]A + 0.17 = 0.40[/tex]

[tex]A = 0.23[/tex]

where A is the probability that the system receives the virus just through the email.

The problem states that there is a 40% chance of receiving it through the email. 23% just through email and 17% by both the email and the webpage.

By webpage only

There is a 35% chance of receiving it through the webpage. With this information, we have the following equation:

[tex]B + (A \cap B) = 0.35[/tex]

[tex]B + 0.17 = 0.35[/tex]

[tex]B = 0.18[/tex]

where B is the probability that the system receives the virus just through the webpage.

The problem states that there is a 35% chance of receiving it through the webpage. 18% just through the webpage and 17% by both the email and the webpage.

What is the probability that the virus does not enter the system at all?

So, we have the following probabilities.

- The virus does not enter the system: P

- The virus enters the system just by email: 23% = 0.23

- The virus enters the system just by webpage: 18% = 0.18

- The virus enters the system both by email and by the webpage: 17% = 0.17.

The sum of the probabilities is 100% = 1. So:

P + 0.23 + 0.18 + 0.17 = 1

P = 1 - 0.58

P = 0.42

There is a probability of 42% that the virus does not enter the system at all.

Answer:

(a) Not a group

(b) Not a group

(c) Abelian group

Step-by-step explanation:

In order for a system <G,*> to be a group, the following must be satisfied

(1) The binary operation is associative, i.e., (a*b)*c = a*(b*c) for all a,b,c in G

(2) There is an identity element, i.e., there is an element e such that a*e = e*a = a for all a in G

(3) For each a in G, there is an inverse, i.e, another element a' in G such that a*a' = a'*a = e (the identity)

If in addition the operation * is commutative (a*b = b*a for every a,b in G), then the group is said to be Abelian

(a)  

The system <G,*> is not a group since there are no identity.  

To see this, suppose there is an element e such that  

a*e = a

then  

a-e = a which implies e=0

It is easy to see that 0 cannot be an identity.

For example  

2*0 = 2-0 = 2

Whereas

0*2 = 0-2 = -2

So 2*0 is not equal to 0*2

(b)

The system <G,*> is not a group either.

If A is a matrix 2x2 and the determinant of A det(A)=0, then the inverse of A does not exist.

(c)

The table of the operation G is showed in the attachment.

It is evident that this system is isomorphic under the identity map, to the cyclic group

[tex]\mathbb{Z}_{5}[/tex]

the system formed by the subset of Z, {0,1,2,3,4} with the operation of addition module 5, which is an Abelian cyclic group

We conclude that the system <G,*> is Abelian.

Attachment: Table for the operation * in (c)

Answer:

The equilibrium point is (3, 3620)

Step-by-step explanation:

We set the supply and the demand equation equal to each other and solve:

[tex]-338q+4634=400q^2+20\\400q^2+338q-4614=0[/tex]

We can solve by factoring:

[tex]2(q-3)(200q+769)=0[/tex]

Setting each factor equal to zero we get:

[tex]q=3\text{ or }q=\displaystyle-\frac{769}{200}[/tex]

Only a positive quantity makes sense, so q=3 is the equilibrium quantity.

To get the equilibrium price we just plug 3 in place of q in any of the functions. Let us use the demand function which is easier to handle:

[tex]D(3)=-338(3)+4634=3620[/tex]

Therefore the equilibrium price is p=3620

In ordered pair form the equilibrium point is (3, 3620)

Answer:

Your column vector is:

[tex]\left[\begin{array}{c}15&10&5&0&-5&-10&-15&-20&-25\end{array}\right][/tex]

Step-by-step explanation:

The first step to solve your problem is knowing what is a column vector:

A column vector is a matrix that only has one column, and multiple rows.

The problem wants the vector to range from 15 to -25.

It means that the biggest value in the vector is 15, and the smallest is -25. Since it is from 15 to -25, the first element of your column vector is 15, and the last element is -25.

With a step size of 5

At each element, you decrease 5. So you have: 15,10,..,-20,-25.

The vector is:

[tex]\left[\begin{array}{c}15&10&5&0&-5&-10&-15&-20&-25\end{array}\right][/tex]

Answer:

false

Step-by-step explanation:

Answer: B

Step-by-step explanation:

twenty thousand is 20,000

10^4 is 10,000

2x10,000 = 20,000

The completed expression would be (2×10,000) + (1×100) + (9×10) which is the correct option (B) 10^4

In math, a number line can be defined as a straight line with numbers arranged at equal segments or intervals throughout. A number line is typically shown horizontally and can be extended indefinitely in any direction.

The numbers on the number line increase as one moves from left to right and decrease on moving from right to left.

Juliana wants to write the numbers twenty thousand, and one hundred ninety in expanded notation.

Given the expression as (2×?)+(1×100)+(9×10)

Here (2×10,000) + (1×100) + (9×10)

Since twenty thousand is 20,000

⇒ 10⁴ is 10,000

⇒ 2×10,000

⇒ 20,000

Hence, the completed expression would be (2×10,000) + (1×100) + (9×10)

Learn more about the number line here:

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Answer with Step-by-step explanation:

We are given that two matrices A and B are square matrices of the same size.

We have to prove that

Tr(C(A+B)=C(Tr(A)+Tr(B))

Where C is constant

We know that tr A=Sum of diagonal elements of A

Therefore,

Tr(A)=Sum of diagonal elements of A

Tr(B)=Sum of diagonal elements of B

C(Tr(A))=[tex]C\cdot[/tex] Sum of diagonal elements of A

C(Tr(B))=[tex]C\cdot[/tex] Sum of diagonal elements of B

[tex]C(A+B)=C\cdot (A+B)[/tex]

Tr(C(A+B)=Sum of diagonal elements of (C(A+B))

Suppose ,A=[tex]\left[\begin{array}{ccc}1&0\\1&1\end{array}\right][/tex]

B=[tex]\left[\begin{array}{ccc}1&1\\1&1\end{array}\right][/tex]

Tr(A)=1+1=2

Tr(B)=1+1=2

C(Tr(A)+Tr(B))=C(2+2)=4C

A+B=[tex]\left[\begin{array}{ccc}1&0\\1&1\end{array}\right]+\left[\begin{array}{ccc}1&1\\1&1\end{array}\right][/tex]

A+B=[tex]\left[\begin{array}{ccc}2&1\\2&2\end{array}\right][/tex]

C(A+B)=[tex]\left[\begin{array}{ccc}2C&C\\2C&2C\end{array}\right][/tex]

Tr(C(A+B))=2C+2C=4C

Hence, Tr(C(A+B)=C(Tr(A)+Tr(B))

Hence, proved.

Answer:

For the given question the correct answer is option 'C'.

Step-by-step explanation:

For a plane triangle we have:

1) Sum of the all the interior angles is 180 degrees. Hence option 'A' is correct.

2) For a right angled triangle from Pythagoras theorem we know that [tex]H^2=a^2+b^2[/tex]

where,

H is the hypotenuse of the triangle

a,b are the sides of the right angled triangle

hence the option 'B' is also correct

3) For any triangle ABC we know that

[tex]\frac{sin(A)}{BC}=\frac{sin(B)}{AC}=\frac{sin(C)}{AB}=constant[/tex]

hence option 'D' is also correct.

Thus only incorrect answer among the given option is 'C'.