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vova2212 [387]
3 years ago
8

A probable passes through (3,0) (-2,3) and (-1,4) what function does this represent

Mathematics
1 answer:
xenn [34]3 years ago
6 0
Thought you'd want to know:  If you're talking about parabolas, it's parabolas, not probables.  ;)

The standard equation of a a quadratic is  y = ax^2 + bx + c.  We need to find the values of the coefficients a, b and c.

Taking the first point:  When x=3, y=0, so write 0 = a(3)^2 + b(3) + c, or
                                                                           0 = 9a + 3b + 1c

Do the same for points (-2,3) and (-1,4).

You will have obtained three linear equations in a, b and c:

3= a(-2)^2 + b(-2) + c, or 3 = 4a - 2b + 1c, also

4 = a(-1)^2 + b(-1) + 1c, or 1a - 1b + 1c.

I used matrix operations to solve this system.  The results are:

a= -2/5, b= 1/5, c= 21/5

and so the function f(x) is f(x) = (-2/5)x^2 + (1/5)x + 21/5.
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can company makes a cylindrical can that has a radius of 6 cm and a height of 10 cm. One of the company's clients needs a cylind
shutvik [7]

the new radius be to meet the client's need is 4.9 cm .

<u>Step-by-step explanation:</u>

Here we have , can company makes a cylindrical can that has a radius of 6 cm and a height of 10 cm. One of the company's clients needs a cylindrical can that has the same volume but is 15 cm tall. We need to find What must the new radius be to meet the client's need . Let's find out:

Let we have two cylinders of volume V_1 , V_2 with parameters as follows :

r_1=6cm\\h_1=10cm\\r_2=?\\h_2=15cm

We know that volume of cylinder is \pi r^2h , According to question volume of both cylinder is equal i.e

⇒ V_1=V_2

⇒ \pi (r_1)^2h_1= \pi (r_2)^2h_2

⇒ (r_1)^2h_1= (r_2)^2h_2

⇒ \frac{(r_1)^2h_1}{h_2}= (r_2)^2

⇒ (r_2) =\sqrt{ \frac{(r_1)^2h_1}{h_2}}                   Putting all values

⇒ (r_2) =\sqrt{ \frac{(6)^2(10)}{15}}

⇒ (r_2) =\sqrt{ \frac{36(10)}{15}}

⇒ (r_2) =\sqrt{ \frac{360}{15}}

⇒ (r_2) =\sqrt{24}

⇒ (r_2) =4.9cm

Therefore , the new radius be to meet the client's need is 4.9 cm .

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3 years ago
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