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SOVA2 [1]
3 years ago
9

Which equals ( V20 - 13) (120 + V3)?

Mathematics
1 answer:
Firlakuza [10]3 years ago
3 0

Answer:

17

thats the answer

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Find the shaded area​
REY [17]

The parabola <em>y</em> = <em>x</em> ² and the line <em>x</em> + <em>y</em> = 12 intersect for

<em>x</em> ² = 12 - <em>x</em>

<em>x</em> ² + <em>x</em> - 12 = 0

(<em>x</em> - 3) (<em>x</em> + 4) = 0

===>   <em>x</em> = 3

so you can compute the area by using two integrals,

\displaystyle \int_0^3 x^2\,\mathrm dx + \int_3^{12}(12-x)\,\mathrm dx

Then the area you want is

\displaystyle \frac{x^3}3\bigg|_0^3 + \left(12x-\frac{x^2}2\right)\bigg|_3^{12} = \left(\frac{3^3}3-\frac{0^3}3\right) + \left(12^2-\frac{12^2}2 - 12\times3 + \frac{3^2}2\right) \\\\ = \boxed{\frac{99}2}

Alternatively, you can subtract the area bounded by <em>y</em> = <em>x</em> ², <em>x</em> + <em>y</em> = 12, and the <em>y</em>-axis in the first quadrant from the area of a triangle with height 12 (the <em>y</em>-intercept of the line) and length 12 (the <em>x</em>-intercept).

Such a triangle has area

1/2 × 12 × 12 = 72

and the area you want to cut away from this is given by a single integral,

\displaystyle \int_0^3 ((12-x)-x^2)\,\mathrm dx = \int_0^3(12-x-x^2)\,\mathrm dx

The integral has a value of

\displaystyle \left(12x-\frac{x^2}2-\frac{x^3}3\right)\bigg|_0^3 = 12\times3 - \frac{3^2}2 - \frac{3^3}3 \\\\ = \frac{45}2

and so the area of the shaded region is again 72 - 45/2 = 99/2.

7 0
3 years ago
Calculate the quotient of 2 8 divided by 1 4 .
Lunna [17]

Answer:

2 28 divided by 14 is 2

Step-by-step explanation:

28÷14=2

5 0
3 years ago
After going shopping we had 10% of our money left we spent 180 $ how much money did we have left
Marrrta [24]

Answer:

$18 :)

Step-by-step explanation:

10% of 180 is 18

4 0
3 years ago
The CHS Baseball team was on the field and the batter popped the ball up. The equation b(x)=80t-16t^2+3.5 represents the height
Aleksandr [31]

Answer:

<em>t = 5 seconds</em>

Step-by-step explanation:

We must solve the equation of movement of the ball as a function of time in seconds.

We must find how many seconds it takes until the ball touches the ground. When the ball touches the ground, the height b (t) = 0

Then we equate the equation to zero and solve for t.

80t-16t ^ 2 + 3.5 = 0

Use the quadratic formula.

Where

a = -16\\\\b = 80\\\\c = 3.5

t = \frac{-b \±\sqrt{b^2-4ac}}{2a}\\\\\\t_1 = \frac{-(80)-\sqrt{(80)^2-4(-16)(3.5)}}{2(-16)}\\\\t_1= 5.04\ s\\-------------------------\\t_2 =\frac{-(80)+\sqrt{(80)^2-4(-16)(3.5)}}{2(-16)}\\\\t_2=-0.0434\ s

We take the positive solution

<em>t = 5 seconds</em>

7 0
3 years ago
Find the value 'X' in this given picture
taurus [48]
Hi.
This is just messing around with exterior angles.
I say around 120 degrees, if I forgot what exterior angles were.
But <em>with quick calculation, it is probably 120 degrees</em>
5 0
4 years ago
Read 2 more answers
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