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Bezzdna [24]
3 years ago
8

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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Look at pic and answer pls
PtichkaEL [24]

(a)  x^{-5}

(b)  3x^{-7}

(c)  $\frac{4}{3}x^4

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}

(c)  \frac{28x^6}{21x^2}

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:

i dont know what im doing

4 0
3 years ago
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