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3 years ago

What is the first term of the expression? −6xyz + 9xy + 12x + 18y TIA!

Mathematics

2 answers:

tatyana61 [14]3 years ago

4 0

For this case we have that by definition:

The terms of a polynomial expression are those that are composed of coefficients and variables separated by signs of addition and subtraction.

We then have the following expression:

$-6xyz + 9xy + 12x + 18y$

According to the definition, we can conclude that the first term is given by:

$-6xyz$

Answer:

The first term of the expression is:

$-6xyz$

Mkey [24]3 years ago

4 0

Answer:

Answer:

The first term of the expression is:

-6xyz

Step-by-step explanation:

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The answer to the equation is x < 2.

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Answer:

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2 years ago

If f(x)=|x+7|, find f(−12). −19 −5 5 19

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Answer:

Step-by-step explanation:

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3 years ago

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3y=11-2x, 3x=y-11 linear equation solve using the algebraic method check solution

tamaranim1 [39]

Answer:

(x, y) = (- 2, 5)

Step-by-step explanation:

given the 2 equations

3y = 11 - 2x → (1)

3x = y - 11 → (2)

Rearrange (2) expressing y in terms of x

add 11 to both sides

y = 3x + 11 → (3)

Substitute y = 3x + 11 into (1)

3(3x + 11) = 11 - 2x

9x + 33 = 11 - 2x ( add 2x to both sides )

11x + 33 = 11 ( subtract 33 from both sides )

11x = - 22 ( divide both sides by 11 )

x = - 2

Substitute x = - 2 in (3) for corresponding value of y

y = (3 × - 2) + 11 = - 6 + 11 = 5

As a check

substitute x = - 2, y = 5 into (1) and (2) and if the left side equals the right side then these values are the solution.

(1) : left side = (3 × 5) = 15

right side = 11 - (2 × - 2) = 11 + 4 = 15 ⇒ left = right

(2) : left side = (3 × - 2 ) = - 6

right side = 5 - 11 = - 6 ⇒ left = right

solution = (- 2, 5 )

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3 years ago

2 points) Sometimes a change of variable can be used to convert a differential equation y′=f(t,y) into a separable equation. One

Stells [14]

$y'=(t+y)^2-1$

Substitute $u=t+y$ , so that $u'=y'$ , and

$u'=u^2-1$

which is separable as

$\dfrac{u'}{u^2-1}=1$

Integrate both sides with respect to $t$ . For the integral on the left, first split into partial fractions:

$\dfrac{u'}2\left(\frac1{u-1}-\frac1{u+1}\right)=1$

$\displaystyle\int\frac{u'}2\left(\frac1{u-1}-\frac1{u+1}\right)\,\mathrm dt=\int\mathrm dt$

$\dfrac12(\ln|u-1|-\ln|u+1|)=t+C$

Solve for $u$ :

$\dfrac12\ln\left|\dfrac{u-1}{u+1}\right|=t+C$

$\ln\left|1-\dfrac2{u+1}\right|=2t+C$

$1-\dfrac2{u+1}=e^{2t+C}=Ce^{2t}$

$\dfrac2{u+1}=1-Ce^{2t}$

$\dfrac{u+1}2=\dfrac1{1-Ce^{2t}}$

$u=\dfrac2{1-Ce^{2t}}-1$

Replace $u$ and solve for $y$ :

$t+y=\dfrac2{1-Ce^{2t}}-1$

$y=\dfrac2{1-Ce^{2t}}-1-t$

Now use the given initial condition to solve for $C$ :

$y(3)=4\implies4=\dfrac2{1-Ce^6}-1-3\implies C=\dfrac3{4e^6}$

so that the particular solution is

$y=\dfrac2{1-\frac34e^{2t-6}}-1-t=\boxed{\dfrac8{4-3e^{2t-6}}-1-t}$

3 0

3 years ago