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4 years ago

What do i do here to work this out?

Mathematics

1 answer:

den301095 [7]4 years ago

5 0

In order to combine fractions, they need to have the same denominator.

To figure out what value to make the denominator, find the least common denominator of 4 and 5:

4: 4, 8, 12, 16, 20

5: 5, 10, 15, 20

The LCD is 20, so multiply 4/5 by 4/4 and -1/4 by 5/5 to make the denominator = 20

$(\frac{4}{4} )\frac{4}{5} -(\frac{5}{5} )\frac{1}{4} =$

$\frac{16}{20} -\frac{5}{20} =$ Now combine the fractions

$\frac{16}{20}- \frac{5}{20} =\frac{11}{20}$

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vladimir1956 [14]

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Step-by-step explanation:

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3 years ago

If y=x^2+2x+3 find the falue of Y when x=3

yarga [219]

Y=18 is the correct answer

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4 years ago

Read 2 more answers

11. Write 850 as the product of its prime factors. A. 17 × 50 B. 5 × 10 × 17 C. 10 × 85 D. 2 × 5 × 5 × 17

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4 0

3 years ago

A above ground garden is rectangular and has a length of 15 feet and a height of 2 feet. It will hold 375 cubic feet of soil. Wh

Fittoniya [83]

Answer: $12.5\ ft$

Step-by-step explanation:

Given

Length L=15 feet

width b=2 feet

Volume V=375 cubic feet

Suppose h is the height

Volume is given by $V=l\times b\times h$

$\Rightarrow 375=15\times 2\times h\\\Rightarrow h=\frac{375}{30}=12.5\ ft$

4 0

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

3 0

3 years ago