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alexgriva [62]
3 years ago
6

Or yk dont answer my question thats coo :/

Mathematics
1 answer:
san4es73 [151]3 years ago
6 0

Answer:

whats the question i might be able to help

Step-by-step explanation:

:)))))))

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Sistemas por el método de reducción por suma o resta.<br> (3x+y=4) y (2x-4y=5)
Cloud [144]

The solution to the system of equation is (3/2, -1/2)

<h3>System of equations</h3>

Given the following system of equations as shown;

(3x+y=4 .... * 4

2x-4y=5 ..... * 1

_______________

12x+4y=16

2x-4y=5

Add both equations

12x+2x = 16 + 5

14x = 21

x = 3/2

Substitute x = 3/2 into any of the equation to have;

2x-4y=5

2(3/2) - 4y = 5

3 - 4y = 5

-4y = 5 - 3

-4y = 2

y = -1/2

Hence the solution to the system of equation is (3/2, -1/2)

Learn more on system of equation here: brainly.com/question/25976025

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5 0
1 year ago
Pls help !!!!!! i need the answer for this pleasee &lt;33
Serjik [45]
Isishsiahsuxhzgfzgsj
6 0
2 years ago
Read 2 more answers
Which of the following expressions will help him determine the length of segment BE? BE = AD BE = CD BE equals CD times AD over
Sati [7]

Answer:

BE equals CD times AB over AC

Step-by-step explanation:

took the test

6 0
2 years ago
(Prove) The angle subtended by an arc at the center is double the angle subtended by it at any
Pavlova-9 [17]

Answer:

The angle subtended by an arc at the center is double the angle subtended by it at any  point on the remaining part of the circle.

Step-by-step explanation:

Let us consider the image attached.

Center of circle be O.

Arc AB subtends the angle \angle APB on the circle and \angle AOB on the center of the circle.

To prove:

\angle AOB = 2 \times \angle APB

Proof:

In \triangle PAO: AO and PO are radius of the circles so AO = PO

And angles opposite to equal sides of a triangle are also equal in a triangle.

So, \angle PAO = \angle OPA

Using external angle property, that external angle is equal to sum of two opposite internal angles of a triangle.

\angle AOQ = \angle PAO + \angle OPA=2 \times \angle APO .... (1)

Similarly,

In \triangle PBO: BO and PO are radius of the circles so BO = PO

And angles opposite to equal sides of a triangle are also equal in a triangle.

So, \angle PBO = \angle OPB

Using external angle property, that external angle is equal to sum of two opposite internal angles of a triangle.

\angle BOQ = \angle PBO + \angle OPB=2 \times \angle BPO .... (2)

Now, we can see that:

\angle AOB = \angle AOQ+\angle BOQ

Using equations (1) and (2):

\angle AOB = 2\angle APO+2\angle BPO\\\angle AOB = 2(\angle APO+\angle BPO)\\\bold{\angle AOB = 2(\angle APB)}

Hence, proved.

4 0
3 years ago
Point P (-5,2) is translated using the rule (x+3,y-1) what is the x coordinate of P​
Licemer1 [7]

Answer:

(-2, 1)

Step-by-step explanation:

(-5 + 3),(2 -1)

4 0
3 years ago
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