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JulijaS [17]
3 years ago
13

Options - positive - negative - 0 -undefined

Mathematics
1 answer:
Brilliant_brown [7]3 years ago
6 0
The slope of the line is 0 since in the graph, the y intercept stays at a certain point throughout
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John works as a carpenterfor a small construction company earning $13.50per hour. in the busy season. he works 50 hours per week
mylen [45]
A) $675
b) $202.50

Multiply $13.50 by the total number hours worked each week during each season 
7 0
3 years ago
50 PTSS HELP ME I NEED IT ASAPPP PLEASE
USPshnik [31]

Answer: It has congruent segments and angles, and both parallel and perpendicular lines.

Step-by-step explanation:

It may have special shapes depending on what the seat and back look like.

If it is a square seat and back then it has some special shapes for sure, like a rectangle.

Hope this helps!

5 0
3 years ago
Read 2 more answers
Which of these could change a taxpayer's adjusted gross
KiRa [710]

Answer:

Student~loan~interest~pay

Step-by-step explanation:

That~should~be~the~answer~to~your~question.

5 0
3 years ago
Find the absolute maximum and minimum values of f on the set D. f(x,y)=2x^3+y^4, D={(x,y) | x^2+y^2<=1}.
castortr0y [4]

Answer:

absolute maximum is f(1, 0) = 2 and the absolute minimum is f(−1, 0) = −2.

Step-by-step explanation:

We compute,

$ f_x = 6x^2, f_y=4y^3 $

Hence, $ f_x = f_y = 0 $  if and only if (x,y) = (0,0)

This is unique critical point of D. The boundary equation is given by

$ x^2+y^2=1$

Hence, the top half of the boundary is,

$ T = \{ x, \sqrt{1-x^2} : -1 \leq x \leq 1\}

On T we have, $ f(x, \sqrt{1-x^2} = 2x^3 +(1-x^2)^2 = x^4 +2x^3-2x^2+1  \text{ for}\ -1 \leq x \leq 1$

We compute

$ \frac{d}{dx}(f(x, \sqrt{1-x^2}))= 4x^3+6x^2-4x = 2x(2x^2+3x-2)=2x(2x-1)(x+2)=0$

0 if and  only if x=0, x= 1/2 or x = -2.

We disregard  $ x = -2 \notin [-1,1]$

Hence, the critical points on T are (0,1) and $(\frac{1}{2}, \sqrt{1-(\frac{1}{2})^2}=\frac{\sqrt3}{2})$

On the bottom half, B, we have

$ f(x, \sqrt{1-x^2})= f(x,-\sqrt{1-x^2})$

Therefore, the critical points on B are (0,-1) and $( 1/2, -\sqrt3/2)  

It remains to  evaluate f(x, y) at the points $ (0,0), (0 \pm1), (1/2, \pm \sqrt3/2) \text{ and}\  (\pm1, 0)$ .

We should consider  latter two points, $(\pm1,0)$, since they are the boundary points for the T and also  B. We compute $ f(0,0)=0, \ \f(0 \pm1)=1, \ \ f(0, \pm \sqrt3/2)=9/16, \ \ f(1,0 )= 2 \text{ and}\ \ f(-1,0)= -2 $

We conclude that the  absolute maximum = f(1, 0) = 2

And the absolute minimum = f(−1, 0) = −2.

6 0
3 years ago
Evaluate each expression if x=-1, y=3, z=-4.<br> 3|z-x|+|2-y|
denpristay [2]
3|-4-(-1)|+|2-3|=10
Final answer: 10
5 0
3 years ago
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