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2 years ago

Hi There , Please I need your help and I hope you can help me !

1 answer:

7 0

Kinetic Energy = (1/2)mv².

where m = mass in kg= 2.5 * 10⁵kg = 250000 kg, v = velocity in m/s = 3*10² m/s = 300 m/s

Kinetic Energy = (1/2)*250000*300² = 0.5 * 250000 * 300 * 300

= 1.125 × 10¹⁰ Joules.

2) Relationship between aⁿ and a⁻ⁿ

a⁻ⁿ = 1 / aⁿ

or

aⁿ = 1 / a⁻ⁿ

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8_murik_8 [283]

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2 years ago

SOMEONE HELP ASAP slope of graph

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Answer:

Step-by-step explanation:

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3 years ago

At a particular restaurant, each onion ring has 65 calories and each slider has 250 calories. A combination meal with onion ring

andre [41]

Answer:

The number of onion rings is 3 in the combination meal and the number of sliders is 2 in the combination meal.

Step-by-step explanation:

Given:

At a particular restaurant, each onion ring has 65 calories and each slider has 250 calories. A combination meal with onion rings and sliders has a total of 5 onion rings and sliders altogether and contains 695 calories.

Now, to find the number of onion rings in the combination meal and the number of sliders in the combination meal.

Let the number of onion rings in the combination meal be $x.$

And let the number of sliders in the combination meal be $y.$

So, the total number of onion rings and sliders are:

$x+y=5\\\\y=5-x\ \ \ .....(1)$

Now, the total number of calories each onion rings and sliders has:

$x(65)+y(250)=695$

Substituting the value of $y$ from equation (1) we get:

$x(65)+(5-x)(250)=695\\\\65x+1250-250x=695\\\\1250-185x=695$

Subtracting both sides by 1250 we get:

$-185x=-555$

Dividing both sides by -185 we get:

$x=3.$

The number of onion rings in the combination meal = 3.

Now, substituting the value of $x$ in equation (1) we get:

$y=5-x\\\\y=5-3\\\\y=2.$

The number of sliders in the combination meal = 2.

Hence, the number of onion rings is 3 in the combination meal and the number of sliders is 2 in the combination meal.

8 0

3 years ago

What is the solution of x=2+ square root x-2?

alex41 [277]

So I'm going to assume that this question is asking for non extraneous solutions, or solutions that are found in the equation and are valid solutions when plugged back into the equation. So firstly, subtract 2 on both sides of the equation:

$x-2=\sqrt{x-2}$

Next, square both sides:

$(x-2)^2=x-2\\(x-2)(x-2)=x-2\\x^2-4x+4=x-2$

Next, subtract x and add 2 to both sides of the equation:

$x^2-5x+6=0$

Now we are going to be factoring by grouping to find the solution(s). Firstly, what two terms have a product of 6x^2 and a sum of -5x? That would be -3x and -2x. Replace -5x with -2x - 3x:

$x^2-2x-3x+6=0$