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3 years ago

15

Find the ordered triple of these equations. 2x - 3y - 4z = -21 -4x + 2y - 3z = -14 -3x - 4y + 2z = -10

1 answer:

sineoko [7]3 years ago

4 0

I did my best hope this helps

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I dont understand how to simplify an expression with exponents

Mashutka [201]

Answer:

3^4

Step-by-step explanation:

Simplifying an expression with expression with exponents:

If they are dividing, we keep the base and subtract the exponents. For example:

$\frac{a^x}{a^y}=a^{x-y}$

In this question:

$\frac{3^2}{3^{-2}}=3^{2-(-2)}=3^{2+2}=3^4$

8 0

1 year ago

Solve the Differential equation (x^2 + y^2) dx + (x^2 - xy) dy = 0

natita [175]

Answer:

$\frac{y}{x}-2ln(\frac{y}{x}+1)=lnx+C$

Step-by-step explanation:

Given differential equation,

$(x^2 + y^2) dx + (x^2 - xy) dy = 0$

$\implies \frac{dy}{dx}=-\frac{x^2 + y^2}{x^2 - xy}----(1)$

Let y = vx

Differentiating with respect to x,

$\frac{dy}{dx}=v+x\frac{dv}{dx}$

From equation (1),

$v+x\frac{dv}{dx}=-\frac{x^2 + (vx)^2}{x^2 - x(vx)}$

$v+x\frac{dv}{dx}=-\frac{x^2 + v^2x^2}{x^2 - vx^2}$

$v+x\frac{dv}{dx}=-\frac{1 + v^2}{1 - v}$

$v+x\frac{dv}{dx}=\frac{1 + v^2}{v-1}$

$x\frac{dv}{dx}=\frac{1 + v^2}{v-1}-v$

$x\frac{dv}{dx}=\frac{1 + v^2-v^2+v}{v-1}$

$x\frac{dv}{dx}=\frac{v+1}{v-1}$

$\frac{v-1}{v+1}dv=\frac{1}{x}dx$

Integrating both sides,

$\int{\frac{v-1}{v+1}}dv=\int{\frac{1}{x}}dx$

$\int{\frac{v-1+1-1}{v+1}}dv=lnx + C$

$\int{1-\frac{2}{v+1}}dv=lnx + C$

$v-2ln(v+1)=lnx+C$

Now, y = vx ⇒ v = y/x

$\implies \frac{y}{x}-2ln(\frac{y}{x}+1)=lnx+C$

5 0

3 years ago

Someone please tell me and maybe ill pay you

rjkz [21]

4x = 6x -10
2x = 10
x = 5

answer
x = 5

6 0

3 years ago

What is the measure of angle RQS in the figure?

Deffense [45]

Answer:

108°

I hope it's helps you

4 0

3 years ago

Read 2 more answers

What is -6 to the negative third power

Montano1993 [528]

(-6)^-3 is the same as
$\frac{1}{(-6)^{3}}$
Which is
$\frac{1}{216}$

5 0

3 years ago