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3 years ago

Spell stun seed backwards and say it

Mathematics

1 answer:

Lisa [10]3 years ago

4 0

Dees nuts because it’s spleens it but yea

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Zack is designing wheels for aconcept car. The diameter of the wheel is18 inches. Zack wants to make spokes inthe wheel that run

Ket [755]

Answer:

The spokes are 9 inches because the radius is half the diameter.

Comment if this is right.

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

F(x) = 2x + 5

Finger [1]

9514 1404 393

Answer:

15. f(x) +g(x) = 7x +2

16. f(x) -g(x) = -3x +8

17. h(x) · g(x) = 35x -21

Step-by-step explanation:

Substitute the function definition for each function expression and simplify.

15. f(x) + g(x) = (2x +5) +(5x -3)

f(x) +g(x) = 7x +2

16. f(x) - g(x) = (2x +5) -(5x -3)

f(x) -g(x) = -3x +8

17. h(x) · g(x) = (7) · (5x -3)

h(x) · g(x) = 35x -21

4 0

3 years ago

What is the value of the expression shown 24-52

Ahat [919]

Answer:

-28

Step-by-step explanation:

It is negative so:

52-24=28

-28

8 0

3 years ago

Linda and Juan went shopping. Linda spent $17 less than Juan.

Mila [183]

Answer:

x - 17

Step-by-step explanation:

x is how much Juan spent. Linda spent 17 less. x - 17 = How much Linda spent

4 0

3 years ago

Read 2 more answers

Please see attachment

Dafna11 [192]

Answer:

a) The value of absolute minimum value = - 0.3536

b) which is attained at $x = \frac{1}{\sqrt{2} }$

Step-by-step explanation:

Step(i):-

Given function

$f(x) = \frac{-x}{2x^{2} +1}$ ...(i)

Differentiating equation (i) with respective to 'x'

$f^{l} = \frac{2x^{2} +1(-1) - (-x) (4x)}{(2x^{2}+1)^{2} }$ ...(ii)

$f^{l}(x) = \frac{2x^{2}-1}{(2x^{2}+1)^{2} }$

Equating Zero

$f^{l}(x) = \frac{2x^{2}-1}{(2x^{2}+1)^{2} } = 0$

$\frac{2x^{2}-1}{(2x^{2}+1)^{2} } = 0$

$2 x^{2}-1 = 0$

$2 x^{2} = 1$

$x^{2} = \frac{1}{2}$

$x = \frac{-1}{\sqrt{2} } , x = \frac{1}{\sqrt{2} }$

Step(ii):-

Again Differentiating equation (ii) with respective to 'x'

$f^{ll}(x) = \frac{(2x^{2} +1)^{2} (4x) - 2(2x^{2} +1) (4x)(2x^{2}-1) }{(2x^{2}+1)^{4} }$

put

$x = \frac{1}{\sqrt{2} }$

$f^{ll} (x) > 0$

The absolute minimum value at $x = \frac{1}{\sqrt{2} }$

Step(iii):-

The value of absolute minimum value

$f(x) = \frac{-x}{2x^{2} +1}$

$f(\frac{1}{\sqrt{2} } ) = \frac{-\frac{1}{\sqrt{2} } }{2(\frac{1}{\sqrt{2} } )^{2} +1}$

on calculation we get

The value of absolute minimum value = - 0.3536

Final answer:-

a) The value of absolute minimum value = - 0.3536

b) which is attained at $x = \frac{1}{\sqrt{2} }$

3 0

3 years ago