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IRINA_888 [86]
3 years ago
6

HELP ASAP Find the Difference. SHOW YOUR WORK 3x + 5 - (2x + 2)

Mathematics
1 answer:
REY [17]3 years ago
3 0

Answer:

3x + 5 - (2x + 2)

3x + 5 -2x-2

x+3

Step-by-step explanation:

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What is partial derivative of z=(2x+3y)^10 with respect to x,y?
maw [93]
Not sure if you mean to ask for the first order partial derivatives, one wrt x and the other wrt y, or the second order partial derivative, first wrt x then wrt y. I'll assume the former.

\dfrac\partial{\partial x}(2x+3y)^{10}=10(2x+3y)^9\times2=20(2x+3y)^9

\dfrac\partial{\partial y}(2x+3y)^{10}=10(2x+3y)^9\times3=30(2x+3y)^9

Or, if you actually did want the second order derivative,

\dfrac{\partial^2}{\partial y\partial x}(2x+3y)^{10}=\dfrac\partial{\partial y}\left[20(2x+3y)^9\right]=180(2x+3y)^8\times3=540(2x+3y)^8

and in case you meant the other way around, no need to compute that, as z_{xy}=z_{yx} by Schwarz' theorem (the partial derivatives are guaranteed to be continuous because z is a polynomial).
3 0
3 years ago
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I need help hurry plz
Scorpion4ik [409]

Answer:

Hey there!

The solution of -14.5<x, shows that x is greater than and not equal to -14.5.

The third graph from the top shows this.

Hope this helps :)

7 0
3 years ago
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5
a_sh-v [17]

Answer:

I could be wrong, but I believe that the answer is D, none. This is because there is only proof of one congruent side and angle.

Step-by-step explanation:

4 0
2 years ago
Out of a group of 30 volunteers, 17 people are chosen to participate in a survey about the number of phone calls they make each
mel-nik [20]

Answer:

i dont know

Step-by-step explanation:

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6 0
3 years ago
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Find an upper limit for the zeroes 2x^4 -7x^3 + 4x^2 + 7x - 6 = 0
erik [133]

<u>Answer-</u>

2 is the upper limit for the zeros.

<u>Solution-</u>

The given function f(x) is,

2x^4 -7x^3 + 4x^2 + 7x - 6 = 0

For calculating the zeros,

\Rightarrow f(x)=0

\Rightarrow 2x^4 -7x^3 + 4x^2 + 7x - 6 = 0

\Rightarrow 2x^4-4x^3-3x^3+6x^2-2x^2+ 4x+3x-6=0

\Rightarrow 2x^3(x-2)-3x^2(x-2)-2x(x-2)+3(x-2)=0

\Rightarrow (x-2)(2x^3-3x^2-2x+3)=0

\Rightarrow (x-2)(x^2(2x-3)-1(2x-3))=0

\Rightarrow (x-2)(x^2-1)(2x-3)=0

\Rightarrow (x-2)(x+1)(x-1)(2x-3)=0

\Rightarrow x-2=0,\ x+1=0,\ x-1=0,\ 2x-3=0

\Rightarrow x=2,\ x=-1,\ x=1,\ x=\frac{3}{2}

From all the 4 roots, it can be obtained that 2 is the greatest zero.

7 0
3 years ago
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