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3 years ago

Find how many six-digit numbers can be formed from the digits 2, 3, 4, 5, 6 and 7 (with repetitions), if:

Mathematics

1 answer:

Goshia [24]3 years ago

4 0

Answer:

case 1 = 2592

case 2 = 729

case 1 + case 2 = 2916

(this is not a direct adition, because case 1 and case 2 have some shared elements)

Step-by-step explanation:

Case 1)

6 digits numbers that can be divided by 25.

For the first four positions, we can use any of the 6 given numbers.

For the last two positions, we have that the only numbers that can be divided by 25 are numbers that end in 25, 50, 75 or 100.

The only two that we can create with the numbers given are 25 and 75.

So for the fifth position we have 2 options, 2 or 7,

and for the last position we have only one option, 5.

Then the total number of combinations is:

C = 6*6*6*6*2*1 = 2592

case 2)

The even numbers are 2,4 and 6

the odd numbers are 3, 5 and 7.

For the even positions we can only use odd numbers, we have 3 even positions and 3 odd numbers, so the combinations are:

3*3*3

For the odd positions we can only use even numbers, we have 3 even numbers, so the number of combinations is:

3*3*3

we can put those two togheter and get that the total number of combinations is:

C = 3*3*3*3*3*3 = 3^6 = 729

If we want to calculate the combinations togheter, we need to discard the cases where we use 2 in the fourth position and 5 in the sixt position (because those numbers are already counted in case 1) so we have 2 numbers for the fifth position and 2 numbers for the sixt position

Then the number of combinations is

C = 3*3*3*3*2*2 = 324

Case 1 + case 2 = 324 + 2592 = 2916

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Amy is pulling a wagon with a force of 30 pounds up a hill at an angle of 25°. Give the force exerted on the wagon as a vector a

Fofino [41]

Answer:

Vector (ordered pair - rectangular form)

$\vec F = (27.189,12.679)\,[lbf]$

Vector (ordered pair - polar form)

$\vec F = (30\,lbf, 25^{\circ})$

Sum of vectorial components (linear combination)

$\vec F = 27.189\cdot \hat{i} + 12.679\cdot \hat{j}\,[N]$

Step-by-step explanation:

From statement we know that force exerted on the wagon has a magnitude of 30 pounds-force and an angle of 25° above the horizontal, which corresponds to the +x semiaxis, whereas the vertical is represented by the +y semiaxis.

The force ( $\vec F$ ), in pounds-force, can be modelled in two forms:

Vector (ordered pair - rectangular form)

$\vec F = \left(\|\vec F\|\cdot \cos \theta, \|\vec F\|\cdot \sin \theta\right)$ (1)

Vector (ordered pair - polar form)

$\vec F = \left(\|\vec F\|, \theta\right)$

Sum of vectorial components (linear combination)

$\vec {F} = \left(\|\vec F\|\cdot \cos \theta\right)\cdot \hat{i} + \left(\|\vec F\|\cdot \sin \theta \right)\cdot \hat{j}$ (2)

Where:

$\|\vec F\|$ - Norm of the vector force, in newtons.

$\theta$ - Direction of the vector force with regard to the horizontal, in sexagesimal degrees.

$\hat{i}$ , $\hat{j}$ - Orthogonal axes, no unit.

If we know that $\|\vec F\| = 30\,lbf$ and $\theta = 25^{\circ}$ , then the force exerted on the wagon is:

Vector (ordered pair - rectangular form)

$\vec F= \left(30\cdot \cos 25^{\circ}, 30\cdot \sin 25^{\circ}\right)\,[lbf]$

$\vec F = (27.189,12.679)\,[lbf]$

Vector (ordered pair - polar form)

$\vec F = (30\,lbf, 25^{\circ})$

Sum of vectorial components (linear combination)

$\vec F = (30\cdot \cos 25^{\circ})\cdot \hat{i} + (30\cdot \sin 25^{\circ})\cdot \hat{j}\,[N]$

$\vec F = 27.189\cdot \hat{i} + 12.679\cdot \hat{j}\,[N]$

8 0

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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