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3 years ago

if Michael had four oranges and Timmy had 5 oranges and they combined them together how many oranges do they have in total

Mathematics

2 answers:

bija089 [108]3 years ago

4 0

They have 9 because
$5 + 4 = 9$

Yuri [45]3 years ago

3 0

Hi,

Answer: Timmy and Michael would have 9 Oranges together.

My work: To easily achieve this problem all you have to do is take Michael’s amount of oranges; 4 and Timmy’s amount of oranges; 5 and add them together. (4 +5 =) When you do this you get your answer 9. (4 + 5 = 9)

I Hoped I Helped!

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Does someone knows what equals 12.5 please

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The vertex of the parabola below is at the point (1,3), and the point (2,6) is

Inessa05 [86]

Answer:

B. y=3(x-1)2 + 3

Step-by-step explanation:

Given that

vertex of the parabola is at the point (1,3)

let's verify, if the option B is the correct equation of the parabola.

$y=3(x-1)^2 + 3\\ \\y=3(x^2+1-2x) + 3\\\\y=3x^2+3-6x + 3\\\\y=3x^2-6x + 6....Eq1$

comparing to standard equationof parabola (standard quadratic equation), we get

$a=3, b=-6 and c=6$

to find the vertex we use formula for x- coordinate as $x=-b/2a$

$x=-(-6)/2(3)\\\\x=6/6\\x=1$

to find y put x=1 in the Eq1, we get

$y=3(1)^2-6(1)+6\\\\y=3-6+6\\\\y=3$

vertex =(x,y) = (1, 3)

thus vertex of the parabola from the equation y=3(x-1)2 + 3 is (1,3), thus verified

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3 years ago

Which table is a linear function please help!

SpyIntel [72]

IM TRYING TO ANSWER BUT IT WONT LET ME OMH ITS C

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2 years ago

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

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fredd [130]

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