After a little manipulation, the given diff'l equation will look like this:

e^y * dy = (2x + 1) * dx.
x^2
Integrating both sides, we get e^y = 2------- + x + c, or e^y = x^2 + x + c
2

Now let x=0 and y = 1, o find c:

e^1 = 0^2 + 0 + c. So, c = e, and the solution is e^y = x^2 + x + e.

4 0

3 years ago

I NEED HELP !!! 30 points!!!! and brainliest :) please give sensible answer :(

Allushta [10]

Answer:

Step-by-step explanation:

9) PQR Is an isosceles triangle

=> ∠PRQ = (180° - x)/2

PRS is an isosceles right triangle

=> ∠PRS = 45°

Have: ∠PRS + ∠PRQ = 115°

=> $\frac{180-x}{2}+45=115 \\$

=> 180 - x = (115 - 45).2 = 140

<=> x = 180 - 140 = 40

10) ABD is an isosceles right triangle => ∠ABD = 45°

BCD is an equilateral triangle => ∠CBD = 60°

have: x = ∠ABD + ∠CBD = 45° + 60° = 105°

11) have: x = y (2)

PQT is an isosceles triangle => ∠PQT = 180 - 70.2 = 40

QTS is an isosceles triangle => ∠TQS = 180 -2x

QRS is an isosceles triangle => ∠RSQ = y

have: 40 + 180 - 2x + y = 180 => 2x - y = 40 (1)

(1)(2) => $\left \{ {{y=x} \atop {2x-y=40}} \right.\\\\=> \left \{ {{x=40} \atop {y=40}} \right.$

=> x + y = 80

12) EFJ Is an equilateral triangle => ∠FJE = 60

∠FJE is the outer angle of the triangle FHJ but FHJ is an isosceles triangle

=> 60 = 2.∠JHF => ∠JHF = 30°

∠JHF is the outer angle of the triangle FHG

=> 30° = 2x

<=> x = 15°

7 0

2 years ago

Solve 64^x = 16^x−1. 1. x = −2 2 x = −1 3. x = - 1 /4 4. x = negative 1 over 3

DochEvi [55]

$64^x = 16^{x-1} \\\\ (4^3)^x=(4^2)^{x-1} \\\\ 4^{3x}=4^{2x-2}\\\\ 3x=2x-2 \\\\ \boxed{x=-2}$

8 0

3 years ago

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