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kykrilka [37]
3 years ago
13

(as a expression)

Mathematics
2 answers:
motikmotik3 years ago
4 0

Answer:

4 or 5

Step-by-step explanation:

shutvik [7]3 years ago
4 0
28-18+2x
That’s the equation hope it helps
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What is the intersection of x+2y=-4 and 4x=3x+12? And what do i do to find this out?
Andrews [41]
Clean up the equation with x first and x=12.
Substitute this value in to the other equation to solve for y.
You should get a solution of (12,-8)
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3 years ago
What is the simplified form of <br><img src="https://tex.z-dn.net/?f=%20%5Cfrac%7Bx%20%2B%206%7D%7B3%7D%20%20-%20%20%5Cfrac%7Bx%
solmaris [256]

Answer:

4/3

Step-by-step explanation:

x + 6 - (x + 2) / 3

= x - x + 6 - 2 / 3  

=  4/3

6 0
3 years ago
Read 2 more answers
1. In which quadrant does the point (-9,7) lie?<br> A. I<br> B. II<br> C. III<br> D. IV
astra-53 [7]
Point (-9,7) lies in quadrant II
6 0
3 years ago
Let w(s,t)=f(u(s,t),v(s,t)) where u(1,0)=−6,∂u∂s(1,0)=5,∂u∂1(1,0)=7 v(1,0)=−8,∂v∂s(1,0)=−8,∂v∂t(1,0)=6 ∂f∂u(−6,−8)=−1,∂f∂v(−6,−8
Blababa [14]
w(s,t)=f(u(s,t),v(s,t))

From the given set of conditions, it's likely that you are asked to find the values of \dfrac{\partial w}{\partial s} and \dfrac{\partial w}{\partial t} at the point (s,t)=(1,0).

By the chain rule, the partial derivative with respect to s is

\dfrac{\partial w}{\partial s}=\dfrac{\partial f}{\partial u}\dfrac{\partial u}{\partial s}+\dfrac{\partial f}{\partial v}\dfrac{\partial v}{\partial s}

and so at the point (1,0), we have

\dfrac{\partial w}{\partial s}\bigg|_{(s,t)=(1,0)}=\dfrac{\partial f}{\partial &#10;u}\bigg|_{(u,v)=(-6,-8)}\dfrac{\partial u}{\partial s}\bigg|_{(s,t)=(1,0)}+\dfrac{\partial f}{\partial &#10;v}\bigg|_{(u,v)=(-6,-8)}\dfrac{\partial v}{\partial s}\bigg|_{(s,t)=(1,0)}
\dfrac{\partial w}{\partial s}\bigg|_{(s,t)=(1,0)}=(-1)(5)+(2)(-8)=-21

Similarly, the partial derivative with respect to t would be found via

\dfrac{\partial w}{\partial t}\bigg|_{(s,t)=(1,0)}=\dfrac{\partial f}{\partial &#10;u}\bigg|_{(u,v)=(-6,-8)}\dfrac{\partial u}{\partial t}\bigg|_{(s,t)=(1,0)}+\dfrac{\partial f}{\partial &#10;v}\bigg|_{(u,v)=(-6,-8)}\dfrac{\partial v}{\partial t}\bigg|_{(s,t)=(1,0)}
\dfrac{\partial w}{\partial t}\bigg|_{(s,t)=(1,0)}=(-1)(7)+(2)(6)=5
6 0
4 years ago
Melissa and her friends are stretching rubber bands for an activity in science class. Melissa stretched her elastic to 10 inches
Butoxors [25]
Melissa, Juan, Sara, and Marsha.

Hope this helps! :)
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3 years ago
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