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3 years ago

How do i solve 2 1/3 - 1 3/4

Mathematics

1 answer:

Ber [7]3 years ago

8 0

Answer:

7/12

Step-by-step explanation:

The first step is converting the mixed fractions to improper fractions (meaning there will be more in the numerator than denominator). So:

2 & 1/3 is equal to 6/3 + 1/3 = 7/3, since we take the whole number and multiply it by the denominator, then put it over the denominator. Another way of looking at this is what divided by the denominator (3) will get us 2, which, in this case, is 3. Then 1 & 3/4 is 4/4 + 3/4 = 7/4.

The next step is: when subtracting fractions, the denominators must always be the same. So you must set the denominators of 7/3 & 7/4 to be the same, by finding the least common denominator. In this case, it is 12, since 4 & 3 can both be multiplied to get 12.

$\frac{7}{3} * \frac{4}{4} = \frac{28}{12}$

and

$\frac{7}{4} * \frac{3}{3} = \frac{21}{12}$

Then, you take the difference between these two fractions:

$\frac{28}{12} - \frac{21}{12} = \frac{7}{12}$

So, your final answer is 7/12.

Hopefully this helped you understand the topic better, sorry if my explanation is a little confusing. Good luck in your studies & have a wonderful day!

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3 years ago

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A rocket is fired with an initial vertical velocity of 40 m/s from a launch pad 45 m high, and its height is given by the formul

N76 [4]

Answer:

1) 125 meters

2) For 9 seconds.

Step-by-step explanation:

The rocket's height is given by the formula $-5t^2+40t+45$ .

Notice that this is a quadratic.

Part 1)

Since this is a quadratic, the highest height the rocket goes will be the vertex of our quadratic. Remember that the vertex of a quadratic in standard form is:

$(-\frac{b}{2a}, f(-\frac{b}{2a}))$

So, let's identify our coefficients. The standard quadratic form is:

$ax^2+bx+c$

Therefore, in this case, our a is -5, b is 40, and c is 45.

So, let's find the x-coordinate of our vertex. Substitute 40 for b and -5 for a. This yields:

$x=\frac{-(40)}{2(-5)}$

Multiply and reduce. So:

$x=-40/-10=4$

Now, substitute 4 for t to find the height. So:

$-5(4)^2+40(4)+45$

Evaluate:

$=-5(16)+40(4)+45$

Multiply and add:

$=-80+160+45=125\text{ meters}$

Therefore, the highest the rocket goes up is 125 meters.

Part 2)

To find out for how much time the rocket is in the air, we can think about after how many seconds after launch will the rocket land.

If the rocket lands, the height h will be 0. So, we can set our expression equal to 0 and solve for t:

$-5t^2+40t+45=0$

First, let's divide everything by -5. This yields:

$t^2-8t-9=0$

We can factor:

$(t+1)(t-9)=0$

Zero Product Property:

$t+1=0\text{ or } t-9=0$

Solve for t:

$t=-1\text{ or } t=9$

In this case, since t is our time in seconds, -1 seconds does not make sense. So, we can remove -1 from our solution set.

Therefore, 9 seconds after launch, the rocket will touch the ground.

Therefore, the rocket was in the air for 9 seconds.

And we're done!

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4 years ago

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ElenaW [278]

Answer:

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Step-by-step explanation:

using the y=mx+b equation, where m=slope and b=y-intercept, we can place our known slope and y-intercept to form our answer, $y=-\frac{3}{2}x-2$ :-) hope this helped!

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3 years ago

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Answer:

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