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Over [174]
2 years ago
12

If you can answer this and show your work ill give brainly

Mathematics
2 answers:
Arlecino [84]2 years ago
8 0

Answer:

First answer is 127 and second answer is 15

marusya05 [52]2 years ago
7 0

1501.30435

I may be wrong.

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Why is “h” negative in math
tresset_1 [31]

Answer:

Just remember that when it comes to vertex form and parabola transformations a negative h means shift right and positive h left. We anumberd to the number lgraphbitself itself having negative integers on left and positive on right and I can see why this throws people off.

Step-by-step explanation:

can i get brainlist brainly deleted all my answers for no reason

5 0
3 years ago
Read 2 more answers
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
What is the least common multiple of 18
Leno4ka [110]
2! is the lcm of eighteen
6 0
4 years ago
Read 2 more answers
Does Anyone know how to do thsi
Likurg_2 [28]
473
______
4)692
4
____
29
28
____
12
12
____
0



6 0
3 years ago
Read 2 more answers
£6000 is divided between Adam, Ben and<br> Chris in the ratio 1:3:4. How much does Ben receive?
Gala2k [10]

Answer:

Chris gets £3000

Step-by-step explanation:

A : B : C

1 : 3 : 4 = 8 (add all the parts together)

£6000 ÷ 8 = 750

so the multiplier is x 750

1 : 3 : 4

↩ x 750

750 : 2250 : 3000

Chris gets £3000

3 0
3 years ago
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