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3 years ago

Can some answer pls.

Mathematics

2 answers:

Alja [10]3 years ago

6 0

The answer is E (7,-2) if u look at the graph u can find the answer easily. And remember if u need help ask ur teachers to help you

kirill [66]3 years ago

5 0

Answer:

E. (7,-2)

Step-by-step explanation:

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find the cost of a car using the installment plan if the down payment is $1,500 and the monthly payments are $385 for 4 years, s

Blizzard [7]

Well, if this is assuming there is no tax on the vehicle and it is fully paid off by the end of the payments, an equation can be set up like the following. 385x+1500=C, x being months and C being total cost. 385(12*4)+1500= C, 385(48)+1500=C, 18480+1500=C, 19980=C.

8 0

3 years ago

The radius of a circle is 16 cm. Find its area in terms of \piπ

Kitty [74]

Answer256pi

Step-by-step explanation:

Area=pi*r^2

=pi*16^2

256pi

3 0

3 years ago

Let M be the set of all nxn matrices. Define a relation on won M by A B there exists an invertible matrix P such that A = P BP S

Sophie [7]

Answer:

Recall that a relation is an equivalence relation if and only if is symmetric, reflexive and transitive. In order to simplify the notation we will use A↔B when A is in relation with B.

Reflexive: We need to prove that A↔A. Let us write J for the identity matrix and recall that J is invertible. Notice that $A=J^{-1}AJ$ . Thus, A↔A.

Symmetric: We need to prove that A↔B implies B↔A. As A↔B there exists an invertible matrix P such that $A=P^{-1}BP$ . In this equality we can perform a right multiplication by $P^{-1}$ and obtain $AP^{-1} =P^{-1}B$ . Then, in the obtained equality we perform a left multiplication by P and get $PAP^{-1} =B$ . If we write $Q=P^{-1}$ and $Q^{-1} = P$ we have $B = Q^{-1}AQ$ . Thus, B↔A.

Transitive: We need to prove that A↔B and B↔C implies A↔C. From the fact A↔B we have $A=P^{-1}BP$ and from B↔C we have $B=Q^{-1}CQ$ . Now, if we substitute the last equality into the first one we get

$A=P^{-1}Q^{-1}CQP = (P^{-1}Q^{-1})C(QP)$ .

Recall that if P and Q are invertible, then QP is invertible and $(QP)^{-1}=P^{-1}Q^{-1}$ . So, if we denote R=QP we obtained that

$A=R^{-1}CR$ . Hence, A↔C.

Therefore, the relation is an equivalence relation.

4 0

3 years ago

Brainliest for the first solution

mr Goodwill [35]

Q1 Solution:

x = 3 or x = -1

Step-by-step explanation:

x²-2x-3 = 0

In order to solve the quadratic equation by factoring, we have to determine two numbers whose sum is -2 and their product -3. By trial and error the two numbers are found to be; -3 and 1. The next step is to split the middle term by substituting it with the above two numbers found;

x²+x-3x-3 = 0

x(x+1)-3(x+1) = 0

(x-3)(x+1) = 0

Finally we apply the zero Product Property :

If ab = 0 then a = 0 or b = 0

This implies;

x-3= 0 or x+1 = 0

x = 3 or x = -1 are the solutions to x²-2x-3 = 0

Q2 Solution:

x = -1/2 or x = 3

Step-by-step explanation:

2x²-5x-3 =0

In order to solve the quadratic equation by factoring, we have to determine two numbers whose sum is -5 and their product 2(-3)=-6. By trial and error the two numbers are found to be; -6 and 1. The next step is to split the middle term by substituting it with the above two numbers found;

2x²-6x+x-3 = 0

2x(x-3)+1(x-3) = 0

(2x+1)(x-3) = 0

2x+1 = 0 or x-3 = 0

2x = -1 or x = 3

x = -1/2 or x = 3 are the solutions of the given quadratic equation.

Q3 Soution:

x = 4 or x = 3

Step-by-step explanation:

x²-7x = -12

x²-7x+12 = 0

In order to solve the quadratic equation by factoring, we have to determine two numbers whose sum is -7 and their product 12. By trial and error the two numbers are found to be; -4 and -3. The next step is to split the middle term by substituting it with the above two numbers found;

x²-4x-3x+12 = 0

x(x-4)-3(x-4) = 0

(x-4)(x-3) = 0

x-4 = 0 or x-3 = 0

x = 4 or x = 3 are the solutions of the given quadratic equation.

Q4:

x = -2/3 or x = 6

Step-by-step explanation:

3x² = 16x+12

3x²-16x-12 = 0

In order to solve the quadratic equation by factoring, we have to determine two numbers whose sum is -16 and their product 3(-12)= -36. By trial and error the two numbers are found to be; -18 and 2. The next step is to split the middle term by substituting it with the above two numbers found;

3x²-18x+2x-12 = 0

3x(x-6)+2(x-6) = 0

(3x+2)(x-6) = 0

3x+2 = 0 or x-6 =0

3x = -2 or x = 6

x = -2/3 or x = 6 are the solutions of the given quadratic equation.

Q5:

x = 6 or x = -4

Step-by-step explanation:

x²-2x-24 = 0

In order to solve the quadratic equation by factoring, we have to determine two numbers whose sum is -2 and their product -24. By trial and error the two numbers are found to be; -6 and 4. The next step is to split the middle term by substituting it with the above two numbers found;

x²-6x+4x-24 = 0

x(x-6)+4(x-6) = 0

(x-6)(x+4) = 0

x-6 = 0 or x+4 = 0

x = 6 or x = -4 are the solutions to the given quadratic equation.

Q6:

x = 4/3 or x = -1

Step-by-step explanation:

3x² = x+4

3x²-x-4 = 0

In order to solve the quadratic equation by factoring, we have to determine two numbers whose sum is -1 and their product -12. By trial and error the two numbers are found to be; -4 and 3. The next step is to split the middle term by substituting it with the above two numbers found;

3x²-4x+3x-4 = 0

x(3x-4)+1(3x-4) =0

(3x-4)(x+1) = 0

3x-4 =0 or x+1 =0

3x = 4 or x = -1

x = 4/3 or x = -1 are the solutions to the given quadratic equation.

5 0

3 years ago

Read 2 more answers

Write the resultant of the two vectors as an ordered pair. –6 5 and 6 –5

RideAnS [48]

Answer: