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EastWind [94]
3 years ago
8

(x2 + 1) dy /dx + 3x(y − 1) = 0, y(0) = 3 how do i go about solving this mess

Mathematics
1 answer:
monitta3 years ago
7 0
Consider, pls, this option.

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Help anyone can help me do this question,I will mark brainlest.​
vovikov84 [41]

Answer:

for a. you first shade the intersectoin between a and b the shade the intersectio

Step-by-step explanation:

5 0
3 years ago
The expression 4x^2represents<br> (4)(x)(x)<br> (4x)(4x)<br> (4)(x)(2)
Daniel [21]

The expression 4x^2represents 4.x.x

5 0
3 years ago
If y varies directly as x and x=2 when y=−4, find y when x=−6.
kakasveta [241]

Answer:

Equation: x × -2 = y

Solution: y = 12

Step-by-step explanation:

if (x = 2) is (y = -4) then (x = 1) is (y = -2) so if (x = -6) is (y = 12) if we follow the equation [x × -2 = y] which was found using the first set of variables and dividing each by 2.

4 0
3 years ago
Solve for x and y <br> x+y=1<br> 2y-x=8
lara31 [8.8K]

Answer:

(- 2, 3 )

Step-by-step explanation:

Given the 2 equations

x + y = 1 → (1)

2y - x = 8 → (2)

Adding the 2 equations term by term will eliminate the x- term

(x - x) + (y + 2y) = (1 + 8), that is

3y = 9 ( divide both sides by 3 )

y = 3

Substitute y = 3 into either of the 2 equations and solve for y

Substituting y = 3 in (1)

x + 3 = 1 ( subtract 3 from both sides )

x = - 2

Solution is (- 2, 3 )

7 0
3 years ago
There is a parallelogram ABCD with diagonals AC and BD. The diagonals AC and BD intersects each other at point E. Side AB is con
grandymaker [24]

Answer:

SAS theorem

Step-by-step explanation:

Given

\square ABCD

\[ \lvert \[ \lvert AB =\[ \lvert \[ \lvert CD

\angle BAC = \angle  DCA

Required

Which theorem shows △ABE ≅ △CDE.

From the question, we understand that:

AC and BD intersects at E.

This implies that:

\[ \lvert \[ \lvert AE = \[ \lvert \[ \lvert EC

and

\[ \lvert \[ \lvert BE = \[ \lvert \[ \lvert ED

So, the congruent sides and angles of △ABE and △CDE are:

\[ \lvert \[ \lvert AB =\[ \lvert \[ \lvert CD ---- S

\angle BAC = \angle  DCA ---- A

\[ \lvert \[ \lvert BE = \[ \lvert \[ \lvert ED or \[ \lvert \[ \lvert AE = \[ \lvert \[ \lvert EC  --- S

<em>Hence, the theorem that compares both triangles is the SAS theorem</em>

4 0
2 years ago
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