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Nikolay [14]
4 years ago
12

Complete the statement with the choice that best

Mathematics
2 answers:
DerKrebs [107]4 years ago
7 0

Answer:

Factors of the constant term

chubhunter [2.5K]4 years ago
5 0

Answer:

Factors of the constant term

Step-by-step explanation:

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What is 5\18 equal to
Evgesh-ka [11]
The answer is 3.3 hope this helps you  .Skateboards
6 0
4 years ago
How do i solve x squared = 4x -5
ICE Princess25 [194]

Answer:

x = 2 ±i

Step-by-step explanation:

x^2 = 4x-5

Subtract 4x from each side

x^2 - 4x = 4x-5 -4x

x^2 - 4x= -5

Complete the square

Take the coefficient of the x term, divide by 2 and square it

-4/2 =2   2^2 =4

Add 4 to each side

x^2 -4x+4 = -5 +4

The left side is (x-coefficient of the x term/2)^2

(x-2)^2 = -1

Take the square root of each side

sqrt((x-2)^2) = sqrt(-1)

x-2 = ±i

Add 2 to each side

x-2+2 = 2 ±i

x = 2 ±i

6 0
3 years ago
Chloe has 3 red pencils, 5 blue pencils, and 1 yellow pencil in her backpack . She chooses one pencil at random, puts it back, a
Gnesinka [82]

Answer:

then she only has eight pencils in the bag.

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
At the beginning of an experiment, a scientist has 300 grams of radioactive goo. After 150 minutes, her sample has decayed to 37
monitta

Answer:

Half-life of the goo is 49.5 minutes

G(t)= 300e^{-0.014t}

191.7 grams of goo will remain after 32 minutes

Step-by-step explanation:

Let M_0\,,\,M_f denotes initial and final mass.

M_0=300\,\,grams\,,\,M_f=37.5\,\,grams

According to exponential decay,

\ln \left ( \frac{M_f}{M_0} \right )=-kt

Here, t denotes time and k denotes decay constant.

\ln \left ( \frac{M_f}{M_0} \right )=-kt\\\ln \left ( \frac{37.5}{300} \right )=-k(150)\\-2.079=-k(150)\\k=\frac{2.079}{150}=0.014

So, half-life of the goo in minutes is calculated as follows:

\ln \left ( \frac{50}{100} \right )=-kt\\\ln \left ( \frac{50}{100} \right )=-(0.014)t\\t=\frac{-0.693}{-0.014}=49.5\,\,minutes

Half-life of the goo is 49.5 minutes

\ln \left ( \frac{M_f}{M_0} \right )=-kt\Rightarrow M_f=M_0e^{-kt}

So,

G(t)= M_f=M_0e^{-kt}

Put M_0=300\,\,grams\,,\,k=0.014

G(t)= 300e^{-0.014t}

Put t = 32 minutes

G(32)= 300e^{-0.014(32)}=300e^{-0.448}=191.7\,\,grams

7 0
3 years ago
Find a third-degree polynomial equation with the rational coefficients that has roots -3 and 1+i
jolli1 [7]

Answer:

x³ + x² − 4x + 6 = 0

Step-by-step explanation:

Imaginary roots come in conjugate pairs.  So if 1+i is a root, then 1−i is also a root.

(x − (-3)) (x − (1+i) (x − (1−i)) = 0

(x + 3) (x² − (1+i) x − (1−i) x + (1+i) (1−i)) = 0

(x + 3) (x² − x − ix − x + ix + 1 − i²) = 0

(x + 3) (x² − 2x + 2) = 0

x (x² − 2x + 2) + 3 (x² − 2x + 2) = 0

x³ − 2x² + 2x + 3x² − 6x + 6 = 0

x³ + x² − 4x + 6 = 0

3 0
3 years ago
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