MATHEMATICS COLLEGE

In the multiplication below, each of A, B andC represents a different digit. What is ABC?
A B C
X
3
В В В

Answers

Answer 1
Answer:

The value of ABC is 340.

The multiplication problem can be written as follows:

ABC

x 3

-----

BBBB

The product of the two numbers is a 4-digit number, so the product of the units digits must be 4. Since 3 x 4 = 12, and the product of the units digits is 4, the units digit of the product is 4, and the carryover is 1.

The product of the tens digits must be a multiple of 10, since the product of the units digits is 4 and the carryover is 1. The only possibility is 3 x B = 1B, where B is an even digit. Since B is a single digit, the only even digit that works is 4. Therefore, B = 4, and the carryover is 1.

The product of the hundreds digits must be a multiple of 100, since the product of the tens digits is a multiple of 10 and the carryover is 1. The only possibility is 3 x A + 1 = 1A, where A is a digit greater than or equal to 3. Since A is a single digit, the only digit greater than or equal to 3 that works is 3. Therefore, A = 3, and the product of the hundreds digits is 10.

The product of the thousands digits must be 0, since the product of the hundreds digits is 10 and the carryover is 0. Therefore, C = 0.

Therefore, the value of ABC is 340.

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Answer 2
Answer:

Answer:

ABC = 148, 3*148 = 444

Step-by-step explanation:

We know that 111 = 3*  37, so all numbers of the form BBB has the factor 37.

So we need a multiple of 37 such thant when multiplied, we get three digits the same as the middle digit.

Try 4*37 = 148, 148*3 = 444, bingo, we got the right combination.

So ABC is 148.

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If A= B and AB= 3x-5 BC= 5x-6 and AC = 2x-9 find the value of X

Answers

Therefore, the value of x is -4 if A= B and AB= 3x-5 BC= 5x-6 and AC = 2x-9.

What is equation?

An equation is a mathematical statement that shows the equality of two expressions. It is composed of two sides, separated by an equals sign (=), indicating that the two sides are equivalent in value. An equation may contain variables, which are unknown values represented by letters, as well as constants, which are known values. Equations are used in many areas of mathematics and science to model and solve problems. For example, the equation y = mx + b is a linear equation that describes the relationship between the variables x and y in a straight line, where m is the slope of the line and b is the y-intercept. Equations can be solved by manipulating the variables and using mathematical operations to isolate the unknown value.

Here,

Since A = B, we know that AB = B². So, we can rewrite the equation AB = 3x - 5 as B² = 3x - 5.

Similarly, we can rewrite BC = 5x - 6 as B² = 5x - 6, and AC = 2x - 9 as A² - B² = (2x - 9) - (B^2).

Since we know that A = B, we can substitute B for A in the last equation to get:

B² - B² = (2x - 9) - (B²)

Simplifying this equation, we get:

0 = 2x - 9 - B²

Now we can substitute the equation B² = 3x - 5 into the above equation to get:

0 = 2x - 9 - (3x - 5)

Simplifying this equation, we get:

0 = -x - 4

Solving for x, we get:

x = -4

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What is the solution set for - 4x + 10 = 5(x + 11)? HELPPPPPPP​

Answers

Answer:

-5

Step-by-step explanation:

-4x +10 =5(x +11)

-4x +10 =5x +55

-4x - 5x =55 - 10

-9x =45

x=-45 :9

x=-5

Solve for x. Round to the nearest tenth. triangle 67, 29, x​

Answers

Final answer:

To solve for x in a triangle with side lengths of 67, 29, and x, we can use the Law of Cosines. The value of x is approximately 47.6.

Explanation:

To solve for x in the given triangle with side lengths 67, 29, and x, we can use the Law of Cosines. The Law of Cosines states that for any triangle with side lengths a, b, and c and angle C opposite side c, the following equation holds true: c^2 = a^2 + b^2 - 2ab*cos(C). In this case, we can substitute the given values into the equation and solve for x. Let's calculate it:

x^2 = 67^2 + 29^2 - 2*67*29*cos(C)

x^2 = 4489 + 841 - 3886cos(C)

Solving for x, we find that x is approximately 47.6.

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Answer:

84°

Step-by-step explanation:

Because every triangle has a combined side length of 180°

67+29=96°

180-96=84°

[10 points] Let v1, v2 and v3 be three linearly independent vectors in R 3 . (a) Find the rank of the matrix A = (v1 − v2) (v2 − v3) (v3 − v1) . (b) Find the rank of the matrix B = (v1 + v2) (v2 + v3) (v3 + v1) .

Answers

Answer:

The solution and complete explanation for the above question and mentioned conditions is given below in the attached document.I hope my explanation will help you in understanding this particular question.

Step-by-step explanation:

Three friends — let’s call them X, Y , and Z — like to play pool (pocket billiards). There are some pool games that involve three players, but these people instead like to play 9-ball, which is a game between two players with the property that a tie cannot occur (there’s always a winner and a loser in any given round). Since it’s not possible for all three of these friends to play at the same time, they use a simple rule to decide who plays in the next round: loser sits down. For example, suppose that, in round 1, X and Y play; then if X wins, Y sits down and the next game is between X and Z. Question: in the long run, which two players square off against each other most often? Least often? So far what I’ve described is completely realistic, but now we need to make a (strong) simplifying assumption. In practice people get tired and/or discouraged, so the probability that (say) X beats Y in any single round is probably not constant in time, but let’s pretend it is, to get a kind of baseline analysis: let 0 < pXY < 1 be the probability that X beats Y in any given game, and define 0 < pXZ < 1 and 0 < pY Z < 1 correspondingly. Consider the stochastic process P that keeps track of

Answers

Answer:

Step-by-step explanation:

(a) If the state space is taken as S = \{(XY),(XZ),(YZ)\} , the probability of transitioning from one state, say (XY) to another state, say (XZ) will be the same as the probability of Y losing out to X, because if X and Y were playing and Y loses to X, then X and Z will play in the next match. This probability is constant with time, as mentioned in the question. Hence, the probabilities of moving from one state to another are constant over time. Hence, the Markov chain is time-homogeneous.

(b) The state transition matrix will be:

P=\begin{vmatrix} 0 & p_(XY) & (1-p_(XY))\n p_(XZ)& 0& (1-p_(XZ))\n p_(YZ)&(1-p_(YZ)) & 0\end{vmatrix},

where as stated in part (b) above, the rows of the matrix state the probability of transitioning from one of the states S = \{(XY),(XZ),(YZ)\} (in that order) at time n and the columns of the matrix state the probability of transitioning to one of the states S = \{(XY),(XZ),(YZ)\} (in the same order) at time n+1.

Consider the entries in the matrix. For example, if players X and Y are playing at time n (row 1), then X beats Y with probability p_(XY), then since Y is the loser, he sits out and X plays with Z (column 2) at the next time step. Hence, P(1, 2) = p_(XY). P(1, 1) = 0 because if X and Y are playing, one of them will be a loser and thus X and Y both together will not play at the next time step. P(1, 3) = 1 - p_(XY), because if X and Y are playing, and Y beats X, the probability of which is1 - p_(XY), then Y and Z play each other at the next time step. Similarly,P(2, 1) = p_(XZ), because if X and Z are playing and X beats Z with probabilityp_(XZ), then X plays Y at the next time step.

(c) At equilibrium,

vP = v,

i.e., the steady state distribution v of the Markov Chain is such that after applying the transition probabilities (i.e., multiplying by the matrix P), we get back the same steady state distribution v. The Eigenvalues of the matrix P are found below:

:det(P-\lambda I)=0\Rightarrow \begin{vmatrix} 0-\lambda & 0.6 & 0.4\n 0.975& 0-\lambda& 0.025\n 0.95& 0.05& 0-\lambda\end{vmatrix}=0

\Rightarrow -\lambda ^3+0.9663\lambda +0.0337=0\n\Rightarrow (\lambda -1)(\lambda ^2+\lambda +0.0337)=0

The solutions are

\lambda =1,-0.0349,-0.9651. These are the eigenvalues of P.

The sum of all the rows of the matrixP-\lambda I is equal to 0 when \lambda =1.Hence, one of the eigenvectors is :

\overline{x} = \begin{bmatrix} 1\n 1\n 1 \end{bmatrix}.

The other eigenvectors can be found using Gaussian elimination:

\overline{x} = \begin{bmatrix} 1\n -0.9862\n -0.9333 \end{bmatrix}, \overline{x} = \begin{bmatrix} -0.0017\n -0.6666\n 1 \end{bmatrix}

Hence, we can write:

P = V * D * V^(-1), where

V = \begin{bmatrix} 1 & 1 & -0.0017\n 1 & -0.9862 & -0.6666 \n 1 & -0.9333 & 1 \end{bmatrix}

and

D = \begin{bmatrix} 1 & 0 & 0\n 0 & -0.9651 & 0 \n 0 & 0 & -0.0349 \end{bmatrix}

After n time steps, the distribution of states is:

v = v_0P^n\Rightarrow v = v_0(VDV^(-1))^n=v_0(VDV^(-1)VDV^(-1)...VDV^(-1))=v_0(VD^nV^(-1)).

Let n be very large, say n = 1000 (steady state) and let v0 = [0.333 0.333 0.333] be the initial state. then,

D^n \approx \begin{bmatrix} 1 & 0 & 0\n 0& 0 &0 \n 0 & 0 & 0 \end{bmatrix}.

Hence,

v=v_0(VD^nV^(-1))=v_0(V\begin{bmatrix} 1 & 0 & 0\n 0 &0 &0 \n 0& 0 & 0 \end{bmatrix}V^(-1))=[0.491, 0.305, 0.204].

Now, it can be verified that

vP = [0.491, 0.305,0.204]P=[0.491, 0.305,0.204].

Josie bought 750 bags of peanuts for $375 he intends to sell each bag for $.15 more than you paid how much should he charge for each bag

Answers

Step-by-step explanation:

Jose bought 750 bags for $375.00

Each bag cost $375/750= $0.50

He  intends to sells them for $0.15 more

He should charge : $0.5 +$0.15=$0.65
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