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alexandr402 [8]
3 years ago
5

Find the roots.1) f(x) = - x^2 + 6x + 272) f(x) = 3x^2 + 17x - 20​

Mathematics
1 answer:
Nookie1986 [14]3 years ago
5 0

Answer:

1)  x1=-3 and x2=9

2) x1=-20/3 and x2=1

Step-by-step explanation:

We can use the quadratic equation to solve every problem related with quadratic term, that is a equation written as ax^2 + bx +c =0

x=\frac{-b+/- \sqrt{b^{2}-4ac } }{2a}

Remember that the +/- sign means that there are two solutions, one using the + sign and other usign the - sign.

For the first polynomial we have a=-1, b = 6, c=27

Using the quadratic equation we have> x1=-3 and x2=9.

For the second polynomial we have a=3, b=17, c=-20.

Using the quadratic equation we have> x1=-20/3 and x2=1.

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What is the value of In e^4?​
alex41 [277]

Answer:

4

Step-by-step explanation:

The easiest way to solve this is just to put it in your calculator.  

But if you want to go deeper, then ln(x)^4 = 4*ln(x)

Since x in this case = e then ln (e)^1 = 1.

ln(e)^4 = 4*ln(e) = 4

5 0
3 years ago
What is the answer to n+ 9 =9
Eva8 [605]
N=0 hope this helpssss
3 0
3 years ago
Read 2 more answers
A fisherman yanks a fish vertically out of the water with an acceleration of 2.5! !! using very light fishing line that has a br
Artyom0805 [142]

Two forces act on the fish, and these are:


1) Its weight mg, downwards and 


2) pressure in the string, T, upwards which is 22 N when the string breaks. 
In the result of these two forces, the fish had an acceleration of 2.5 m/s^2 in the upward way. 


So, T - mg = m*2.5


=> m = T / (2.5 + g)


=> m > 22 / (2.5 + 9.8) ( when the string snaps ) 


=> m > 1.79 kg. 


As a result, the mass of the fish is more than 1.79 kg as the line has been broken.

6 0
3 years ago
The thickness X of aluminum sheets is distributed according to the probability density function f(x) = 450 (x2 - x) if 6 < x
grandymaker [24]

Solution :

Given :

f(x) = \left\{\begin{matrix}\frac{1}{450}(x^2-x) & \text{if  } 6 < x < 12 \\ 0 & \text{otherwise}\end{matrix}\right.

1. Cumulative distribution function

$P(X \leq x) = \int_{- \infty}^x f(x) \ dx$

              $=\int_{- \infty}^6 f(x) dx + \int_{6}^x f(x) dx $

             $=0+\int_6^x \frac{1}{450}(x^2-x) \ dx$

             $=\frac{1}{450} \int_6^x (x^2-x) \ dx$

             $=\frac{1}{450}\left[\frac{x^3}{3}-\frac{x^2}{2}\right]_6^x$

             $=\frac{1}{450}\left[ \left( \frac{x^3}{3} - \frac{x^2}{2}\left) - \left( \frac{6^3}{3} - \frac{6^2}{2} \right) \right]  $

            $=\frac{1}{450}\left[\frac{x^3}{3} - \frac{x^2}{2} - 54 \right]$

2.  Mean $E(x) = \int_{- \infty}^{\infty} \ x \ f(x) \ dx$

                       $=\int_{6}^{12}x . \left( \frac{1}{450} \ (x^2-x)\right)\  dx$

                     $=\frac{1}{450} \int_6^{12} \ (x^3 - x^2) \ dx$

                     $=\frac{1}{450} \left[\frac{x^4}{4} - \frac{x^3}{3} \right]_6^{12} \ dx$

                     $=\frac{1}{450} \left[ \left(\frac{(12)^4}{4} - \frac{(12)^3}{3} \right) -  \left(\frac{(6)^4}{4} - \frac{(6)^3}{3} \right) $

                     $=\frac{1}{450} [4608 - 252]$

                    = 17.2857

5 0
3 years ago
What is the solution to the equation below? log 20 x^3-2 log x=4
guajiro [1.7K]
20 (4)^3-2
20(64)-2
1280-2
1278
4 0
3 years ago
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