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2 years ago

Find variable values for two variables in an equation

Mathematics

1 answer:

never [62]2 years ago

5 0

Simplifying

2x + 3y = 12

Solving

2x + 3y = 12

Solving for variable 'x'.

Move all terms containing x to the left, all other terms to the right.

Add '-3y' to each side of the equation.

2x + 3y + -3y = 12 + -3y

Combine like terms: 3y + -3y = 0

2x + 0 = 12 + -3y

2x = 12 + -3y

Divide each side by '2'.

x = 6 + -1.5y

Simplifying

x = 6 + -1.5y

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Determine the length of the missing side of the right triangle? ? inches 15 inches 8 inches

Ksenya-84 [330]

Answer:

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3 years ago

Is 7.8 rational or irrational

ivanzaharov [21]

Answer:

Rational!

Step-by-step explanation:

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2 years ago

On the beach are 2 towels. On each towel<br> are 4 buckets. How many buckets are there<br> in alle

Ahat [919]

Dude, are u serious?

The answer would be 2*4,

Which it's 8

the answer is 8

Hope it helps!!!!

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3 years ago

Read 2 more answers

Find the simplified product:<br> sqrt 2x^2 * sqrt18^5

makvit [3.9K]

Answer:

1944x^2.

Step-by-step explanation:

sqrt 2x^2 * sqrt18^5

= sqrt2x^2 * (sqrt(9^5*2^5))

= sqrt 2 x^2 * (sqrt9^5 * sqrt 2^5)

= sqrt2 x^2 * (243* sqrt2^5)

= 2^0.5 (243 * 2^2.5) x^2

= 2^3 * 243 x^2

= 1944x^2.

5 0

2 years ago

Determine the open t-intervals on which the curve is concave downward or concave upward. (Enter your answer using interval notat

Rashid [163]

Solution :

We have been given a parametric curve :

x = sin t , y = cos t , 0 < t < π

In order to determine concavity of the given parametric curve, we need to evaluate its second derivative first.

Therefore,

$\frac{dx}{dt} = \cos t, \ \frac{dy}{dt}= - \sin t$

$\therefore \frac{dy}{dx}= \frac{- \sin t}{\cos t}$

$=- \tan t$

Taking double derivatives of the above equation:

$\frac{d^2y}{dx^2}= - \frac{d}{dx}(\tan t)$

$= - \sec^2 t \frac{dt}{dx}$

$= - \sec^2 t \left(\frac{1}{\cos t}\right)$

$= - \sec^3 t$

For the concave up, we have

$\frac{d^2y}{dx^2} > 0$

$\Rightarrow - \sec^3 t > 0$

∴ $t \ \epsilon \left( \frac{\pi }{2}, \pi \right)$

For the concave down, we have

$\frac{d^2y}{dx^2} < 0$

$\Rightarrow - \sec^3 t < 0$

$t \ \epsilon \left( 0,\frac{\pi }{2} \right)$

8 0

2 years ago