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2 years ago

9

Write a quadratic equation with -4 and 3/2 as it’s roots. Write the equation in the form ax^2 + bx + c = 0, where a, b, and c ar

e integers.

1 answer:

jonny [76]2 years ago

5 0

9514 1404 393

Answer:

2x² +5x -12 = 0

Step-by-step explanation:

When p and q are roots of a quadratic, its factored form can be written as ...

(x -p)(x -q) = 0

Here, the roots are given as -4 and 3/2, so the factored form would be ...

(x -(-4))(x -(3/2) = 0

Multiplying by 2 gives us ...

(x +4)(2x -3) = 0

Expanding the product, we find the desired quadratic is ...

2x² +5x -12 = 0

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Step-by-step explanation:

that's the answer

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The lowest temperature in January was less than−6°. Which could be the lowest temperature in January?

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Step-by-step explanation:

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3 years ago

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Solve the following system of equations for all three variables.

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Answer:

Step-by-step explanation:

-2x + 3y – 4z = 8 ----------------(I)

5x – 3y + 5z = -8 -----------------(II)

7x – 3y + 3z = 8 ------------------(III)

Add equation (I) & (II) and thus y will be eliminated

(I) -2x + 3y – 4z = 8

(II) <u>5x – 3y + 5z = -8</u> {Add}

3x + z = 0 ------------------------(A)

Multiply equation (II) by (-1) and then add with equation (III). Thus y will be eliminated.

(II) * (-1) -5x + 3y - 5z = +8

<u>7x – 3y + 3z = 8</u> {Add}

2x -2z = 16 ---------------(B)

Multiply equation (A) by 2 and then add. Thus z will be eliminated and we will get the value of x

(A) * 2 6x + 2z = 0

(B) <u>2x - 2z = 16</u> {Add}

8x = 16

Divide both sides by 8

x = 16/8

x = 2

Plugin x = 2 in equation (A)

3x + z = 0

3*2 + z = 0

6 + z = 0

z = -6

Plug in x = 2 and z = - 6 in equation (I)

-2x +3y - 4z = 8

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3 years ago

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The half-life of a radioactive element is 130 days, but your sample will not be useful to you after 80% of the radioactive nu

gtnhenbr [62]

Answer:

We can use the sample about 42 days.

Step-by-step explanation:

Decay Equation:

$\frac{dN}{dt}\propto -N$

$\Rightarrow \frac{dN}{dt} =-\lambda N$

$\Rightarrow \frac{dN}{N} =-\lambda dt$

Integrating both sides

$\int \frac{dN}{N} =\int\lambda dt$

$\Rightarrow ln|N|=-\lambda t+c$

When t=0, N= $N_0$ = initial amount

$\Rightarrow ln|N_0|=-\lambda .0+c$

$\Rightarrow c= ln|N_0|$

$\therefore ln|N|=-\lambda t+ln|N_0|$

$\Rightarrow ln|N|-ln|N_0|=-\lambda t$

$\Rightarrow ln|\frac{N}{N_0}|=-\lambda t$ .......(1)

$\frac{N}{N_0}=e^{-\lambda t}$ .........(2)

Logarithm:

$ln|\frac mn|= ln|m|-ln|n|$
$ln|ab|=ln|a|+ln|b|$

$ln|e^a|=a$

$ln|a|=b \Rightarrow a=e^b$
$ln|1|=0$

130 days is the half-life of the given radioactive element.

For half life,

$N=\frac12 N_0$ , $t=t_\frac12=130$ days.

we plug all values in equation (1)

$ln|\frac{\frac12N_0}{N_0}|=-\lambda \times 130$

$\rightarrow ln|\frac{\frac12}{1}|=-\lambda \times 130$

$\rightarrow ln|1|-ln|2|-ln|1|=-\lambda \times 130$

$\rightarrow -ln|2|=-\lambda \times 130$

$\rightarrow \lambda= \frac{-ln|2|}{-130}$

$\rightarrow \lambda= \frac{ln|2|}{130}$

We need to find the time when the sample remains 80% of its original.

$N=\frac{80}{100}N_0$

$\therefore ln|{\frac{\frac {80}{100}N_0}{N_0}|=-\frac{ln2}{130}t$

$\Rightarrow ln|{{\frac {80}{100}|=-\frac{ln2}{130}t$

$\Rightarrow ln|{{ {80}|-ln|{100}|=-\frac{ln2}{130}t$

$\Rightarrow t=\frac{ln|80|-ln|100|}{-\frac{ln|2|}{130}}$

$\Rightarrow t=\frac{(ln|80|-ln|100|)\times 130}{-{ln|2|}}$

$\Rightarrow t\approx 42$

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2/3 <7/8

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3 years ago