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3 years ago

Write an algebraic expressions for 11 fewer than a number f ?

Mathematics

1 answer:

meriva3 years ago

6 0

F-11 because its 11 fewer than f which means you subtract 11 from f

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Jonah drew two squares with the same dimension. He then added 2 inches to the length of one square to make it a rectangle. He al

Lubov Fominskaja [6]

Answer:

Perimeter of rectangle added 2 to length:

perimeter of rectangle= length + width + length + width

P = ( l + 2 ) + w + ( l + 2 ) + w

P = 2 ( l + 2 ) + 2 w

P = 4 + 2 l + 2 w

Perimeter of rectangle added 2 to width:

P = l + w + l + w

p = l + ( w + 2) + l + ( w + 2)

p = 2 l + 2 ( w + 2)

p = 2 l + 2 w + 4

5 0

4 years ago

I am a number less than 3000. When you divide me by 32, my remainder is 30. When you divide me by 58, my remainder 44. What numb

yKpoI14uk [10]

Let Q1 and Q2 be the quotients from dividing 32 and 58 respectively.Then 32 Q1 + 30 = 58 Q2 + 4432 Q1 = 58 Q2 + 1416 Q1 = 29 Q2 + 716 Q1 = (16 + 13)Q2 + 716 Q1 = 16 Q2 + 13 Q2 + 716(Q1 - Q2) = 13 Q2 + 7Because Q1 and Q2 are integers, we have to find Q2 whereby 13Q2 + 7 is divisible by 16.After some try-and-error, we get Q2 = 13. That is 13*13 + 7 = 176Therefore, Q1 - Q2 = 11 --> Q1 - 13 = 11 --> Q1 = 2432*24 + 30 = 58*13 + 44 = 798.I am 798

3 0

4 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 $f(0,0)=0, \ \f(0 \pm1)=1, \ \ f(0, \pm \sqrt3/2)=9/16, \ \ f(1,0 )= 2 \text{ and}\ \ f(-1,0)= -2$