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kipiarov [429]
3 years ago
11

What are the zeros of f(x) = x2 + x - 12?​

Mathematics
1 answer:
lukranit [14]3 years ago
6 0

Answer:

Factoring

(x +4) (x-3)

x = -4

x =3

Step-by-step explanation:

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rjkz [21]

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Assume that X has a normal distribution, and find the indicated probability.
GrogVix [38]

Answer: 0.6827

Step-by-step explanation:

I believe it is 0.6827

6 0
3 years ago
Find the value of k and x2<br> x^2+ 13x + k = 0, x1=-9
Anarel [89]

Given:

The quadratic equation is:

x^2+13x+k=0

x_1=-9

To find:

The value of k and x_1.

Solution:

We have,

x^2+13x+k=0                ...(i)

Putting x=-9, we get

(-9)^2+13(-9)+k=0

81-117+k=0

-36+k=0

k=36

Putting k=36 in (i), we get

x^2+13x+36=0

Splitting the middle term, we get

x^2+9x+4x+36=0

x(x+9)+4(x+9)=0

(x+9)(x+4)=0

x=-9,-4

Here, x_1=-9 and x_2=-4.

Therefore, the required values are k=36 and x_2=-4.

3 0
2 years ago
Between 0 degrees Celsius and 30 degrees Celsius, the volume V (in cubic centimeters) of 1 kg of water at a temperature T is giv
Ira Lisetskai [31]

Answer:

T = 3.967 C

Step-by-step explanation:

Density = mass / volume

Use the mass = 1kg and volume as the equation given V, we will come up with the following equation

D = 1 / 999.87−0.06426T+0.0085043T^2−0.0000679T^3

   = (999.87−0.06426T+0.0085043T^2−0.0000679T^3)^-1

Find the first derivative of D with respect to temperature T

dD/dT = \dfrac{70000000\left(291T^2-24298T+91800\right)}{\left(679T^3-85043T^2+642600T-9998700000\right)^2}

Let dD/dT = 0 to find the critical value we will get

\dfrac{70000000\left(291T^2-24298T+91800\right)}  = 0

Using formula of quadratic, we get the roots:

T =  79.53 and T = 3.967

Since the temperature is only between 0 and 30, pick T = 3.967

Find 2nd derivative to check whether the equation will have maximum value:

-\dfrac{140000000\left(395178T^4-65993368T^3+3286558821T^2+2886200857800T-121415215620000\right)}{\left(679T^3-85043T^2+642600T-9998700000\right)^3}

Substituting the value with T=3.967,

d2D/dT2 = -1.54 x 10^(-8)    a negative value. Hence It is a maximum value

Substitute T =3.967 into equation V, we get V = 0.001 i.e. the volume when the the density is the highest is at 0.001 m3 with density of

D = 1/0.001 = 1000 kg/m3

Therefore T = 3.967 C

3 0
3 years ago
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