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Gennadij [26K]
3 years ago
5

Help please hurry please

Mathematics
1 answer:
DIA [1.3K]3 years ago
5 0
I have no idea sorry btw I would skip it if I were you because it is only worth 1 point
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Which graph represents y=3sqrt x+6-3?
Lubov Fominskaja [6]

Answer:

graph for y=3sqrt x+6-3

Step-by-step explanation:

hope this helps :]

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3 years ago
If r = 25 cm and s = 7 cm, what is the length of t?
Alina [70]
The length would be 15
8 0
4 years ago
P minus p over 6 = p over three plus 2
mel-nik [20]
The sum of opposites equals zero, which makes the equation zero over 6 equals p over 3 + 2 then you simplify the equation, multiply both sides by 3 which you will get 0=p+6 now we move variables to the left side and change its sign which is -p=6 so your answer will be p=-6 hope that helps you understand 
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3 years ago
Which of the following are trinomials?
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6 0
2 years ago
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
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