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3 years ago

What is the measure of angle D?

Mathematics

2 answers:

exis [7]3 years ago

6 0

Answer:

Step-by-step explanation:

the answer is 52 because all the angles have to add up to 360 and 128+126+54=308 and 360-308=52

melomori [17]3 years ago

6 0

Answer:

D=52

Step-by-step explanation:

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(Giving brainliest! Pls have at least one sentence to explain why you chose the answer :) )

Kryger [21]

Answer:

Step-by-step explanation:

Hope it helps

4 0

2 years ago

Read 2 more answers

The polynomial x^2+3x-1 is a factor of x^4+3x^3-2x^2-3x+1 true or false?

pashok25 [27]

Answer:

Yes, it is true that $x^2+3x-1$ is a factor of $x^4+3x^3-2x^2-3x+1$ .

Step-by-step explanation:

Let us try to factorize $x^4+3x^3-2x^2-3x+1$

$x^4+3x^3-2x^2-3x+1\\\Rightarrow x^4-2x^2+1-3x+3x^3$

Let us try to make a whole square of the given terms:

$\Rightarrow (x^2)^2-2\times x^2 \times 1+1^2+3x^3-3x\\\Rightarrow (x^2-1)^2+3x^3-3x\\$

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Formula used above:

$a^{2} -2 \times a \times b +b^{2} = (a-b)^2$

In the above equation, we had $a = x, b = 1$ .

--------------

Further solving the above equation, taking $3x$ common out of $3x^3-3x$

$\Rightarrow (x^2-1)^2+3x(x^2-1)\\$

Taking $(x^{2} -1)$ common out of the above term:

$\Rightarrow (x^2-1)((x^2-1)+3x)\\\Rightarrow (x^2-1)(x^2+3x-1)$

So, the two factors are $(x^2-1)\ and\ (x^2+3x-1)$ .

$\therefore$ The statement that $x^2+3x-1$ is a factor of $x^4+3x^3-2x^2-3x+1$ is True.

7 0

2 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

AB=2y+1, BC=y+1,CD=7x-3,DA=3x what is x and y

sattari [20]

I attached the picture associated with this question.

Answer:
x = 2
y = 5

Explanation:
ABCD is a parallelogram. This means that each two opposite sides are equal.
This means that:
1- AB = CD
2y + 1 = 7x - 3 ...........> equation I
2- AD = BC
3x = y + 1
This can be rewritten as:
y = 3x - 1............> equation II

Substitute with equation II in equation I and solve for x as follows:
2y + 1 = 7x - 3 ...........> equation I
2(3x - 1) + 1 = 7x - 3
6x - 2 + 1 = 7x - 3
6x - 1 = 7x - 3
7x - 6x = -1 + 3
x = 2

Substitute with x in equation I to get y as follows:
y = 3x - 1
y = 3(2) - 1
y = 6 - 1
y = 5

Hope this helps :)

3 0

3 years ago

There are 25 students names in a hat. You choose 5 names. Three are boys names and two are girls names.How many of the 25 names

iren [92.7K]

5 is the answer your prop

5 0

3 years ago