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3 years ago

Are 12x^2y^3z and -45zy^3x2 like terms? explain why or why not.

Mathematics

1 answer:

Alex787 [66]3 years ago

4 0

Here in the second term I am considering 2 as power of x .

So rewriting both the terms here:

First term: 12x²y³z

Second term: -45zy³x²

Let us now find out whether they are like terms or not.

"Like terms" are terms whose variables (and their exponents such as the 2 in x²) are the same.

In the given two terms let us find exponents of each variable and compare them for both terms.

z : first and second term both have exponent 1

x: first and second term both have exponent 2

y: first and second term both have exponent 3

Since we have all the exponents equal for both first and second terms variables, so we can say that the two terms are like terms.

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PtichkaEL [24]

(a) $x^{-5}$

(b) $3x^{-7}$

Solution:

To write each of the given expression in the form $ax^n$ :

(a) $\frac{x^3}{x^8}$

Using exponential rule: $\frac{a^x}{a^y}=a^{x-y}$

$$\frac{x^3}{x^8}=x^{3-8}$

$$\frac{x^3}{x^8}=x^{-5}$

(b) $\frac{6x}{2x^8}$

Divide numerator and denominator by the common factor 2, we get

$$\frac{6x}{2x^8}=\frac{3x}{x^8}$

Using exponential rule: $\frac{a^x}{a^y}=a^{x-y}$

$$=3x^{1-8}$

$$\frac{6x}{2x^8} =3x^{-7}$

Divide numerator and denominator by the common factor 7, we get

$$\frac{28x^6}{21x^2}=\frac{4x^6}{3x^2}$

Using exponential rule: $\frac{a^x}{a^y}=a^{x-y}$

$$=\frac{4}{3}x^{6-2}$

$$\frac{28x^6}{21x^2}=\frac{4}{3}x^4$

8 0

3 years ago

Let S be the surface defined by x 2 + 2y 3 + 3z 4 = 6. Let T be the surface defined parametrically by r(u, v) = (1+ln u, 2e v+u−

aleksandrvk [35]

The tangent to $C$ through (1, 1, 1) must be perpendicular to the normal vectors to the surfaces $S$ and $T$ at that point.

Let $f(x,y,z)=x^2+2y^3+3z^4$ . Then $S$ is the level curve $f(x,y,z)=6$ . Recall that the gradient vector is perpendicular to level curves; we have

$\nabla f(x,y,z)=(2x,6y,12z^2)$

so that the gradient of $f$ at (1, 1, 1) is

$\nabla f(1,1,1)=(2,6,12)$

For the surface $T$ , we have

$\begin{cases}1+\ln u=1\\2e^v+u-2=1\\uv+1=1\end{cases}\implies u=1,v=0$

so that $\vec r(1,0)=(1,1,1)$ . We can obtain a vector normal to $T$ by taking the cross product of the partial derivatives of $\vec r(u,v)$ , and evaluating that product for $u=1,v=0$ :

$\dfrac{\partial\vec r}{\partial u}\times\dfrac{\partial\vec r}{\partial v}=\left(u-2ve^v,-1,\dfrac{2e^v}u\right)$

$\left(\dfrac{\partial\vec r}{\partial u}\times\dfrac{\partial\vec r}{\partial v}\right)(1,0)=(1,-1,2)$

Now take the cross product of the two normal vectors to $S$ and $T$ :

$(2,6,12)\times(1,-1,2)=(24,8,-8)$

The direction of vector (24, 8, -8) is the direction of the tangent line to $C$ at (1, 1, 1). We can capture all points on the line containing this vector by scaling it by $t\in\Bbb R$ . Then adding (1, 1, 1) shifts this line to the point of tangency on $C$ . So the tangent line has equation

$\vec\ell(t)=(1,1,1)+t(24,8,-8)=(1+24t,1+8t,1-8t)$

7 0

3 years ago

Please help me answer these 3 questions. Also, please EXPLAIN how you got the answer because sometimes people answer but don't e

nirvana33 [79]

Answer:

3 and 4

Step-by-step explanation:

4. 28/105 is what it is trying to explain

3. -3/12 is what it is trying to explain

7 0

3 years ago

What is the pythagorean theorem? x?

wariber [46]

Answer:

A^2+b^2=c^2

Step-by-step explanation:

Pretty much the two shortest sides of a right triangle squared is equal to the length of the longer side (hypotenuse) squared

6 0

3 years ago

I need help with this

Greeley [361]

Answer:

triangle goes on top

Step-by-step explanation:

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3 years ago