True or False? An equiangular triangle can never be scalene. A. True B. False

-114=-5x-2(4x+18) ????????

SVETLANKA909090 [29]

Primeiro multiplica o -2 com tudo que tem dentro do parenteses (chuveirinho) -114 = -5x -2(4x+18)

-2 . 4x = -8x + -2 . 18 = -36 = -8x-36

Agora soma os números com incógnitas (no caso x)

-114 = - 5x - 8x -36

Joga a incógnita do outro, tornando positiva

-114 = -13x - 36

13x -114 = -36

e o 114 do outro lado, ficando positiva também

13x -114 = -36

13x = - 36 + 114

13x = 78

x = 78 /13

x= 6

ESPERO TER AJUDADO :)

8 0

3 years ago

In 2017 the state University’s men’s basketball team finished the season ranked 34 out

Alex Ar [27]

Answer:

sorry I don't know the answer

8 0

3 years ago

PLEZ NEED HELP ASAP 24.455−18.246

Leviafan [203]

Answer:

The answer is 6.209

6 0

3 years ago

Read 2 more answers

PLEASE HELP! GIVING 20 POINTs

MakcuM [25]

Answer:

Box A-10

Box B-15

Box C-20

i think

5 0

3 years ago

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Separable differential equation y’ln^2y+ysqrtx=0 y(0)=e

Maksim231197 [3]

By applying the theory of separable ordinary differential equations we conclude that the solution of the differential equation $\frac{dy}{dx} \cdot (\ln y)^{2} + y\cdot \sqrt{x} = 0$ with y(0) = e is $y = e^{\sqrt [3]{-2\cdot x^{\frac{3}{2} }+1}}$ .

<h3>How to solve separable differential equation</h3>

In this question we must separate each variable on each side of the equivalence, integrate each side of the expression and find an explicit expression (y = f(x)) if possible.

$\frac{dy}{dx} \cdot (\ln y)^{2} + y\cdot \sqrt{x} = 0$

$(\ln y)^{2}\,dy = -y \cdot \sqrt{x}\, dx$

$-\frac{(\ln y)^{2}}{y}\, dy = \sqrt{x} \,dx$

$-\int {\frac{(\ln y)^{2}}{y} } \, dy = \int {\sqrt{x}} \, dx$

If u = ㏑ y and du = dy/y, then:

$-\int {u^{2}\,du } = \int {x^{\frac{1}{2} }} \, dx$

$-\frac{1}{3}\cdot u^{3} = \frac{2\cdot x^{\frac{3}{2} }}{3} + C$

$u^{3} = -2\cdot x^{\frac{3}{2} } + C$

$(\ln y)^{3} = - 2\cdot x^{\frac{3}{2} } + C$

$C = (\ln e)^{3}$

$C = 1$

And finally we get the explicit expression:

$\ln y = \sqrt [3]{-2\cdot x^{\frac{3}{2} }+ 1}$

$y = e^{\sqrt [3]{-2\cdot x^{\frac{3}{2} }+1}}$

By applying the theory of separable ordinary differential equations we conclude that the solution of the differential equation $\frac{dy}{dx} \cdot (\ln y)^{2} + y\cdot \sqrt{x} = 0$ with y(0) = e is $y = e^{\sqrt [3]{-2\cdot x^{\frac{3}{2} }+1}}$ .

To learn more on ordinary differential equations: brainly.com/question/14620493

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6 0

2 years ago