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mr_godi [17]
3 years ago
14

Solve the equation. Identify any extraneous solutions.

Mathematics
2 answers:
Brrunno [24]3 years ago
6 0
x=\sqrt{2x+24}\\
x\geq 0 \wedge 2x+24\geq0\\
x\geq 0 \wedge 2x\geq-24\\
x\geq 0 \wedge x\geq-12\\
x\geq0\\\\
x=\sqrt{2x+24}\\
x^2=2x+24\\
x^2-2x-24=0\\
x^2+4x-6x-24=0\\
x(x+4)-6(x+4)=0\\
(x-6)(x+4)=0\\
x=6 \vee x=-4\\\\
-4\not \geq 0

So, <span>6 is a solution to the original equation. The value –4 is an extraneous solution.</span>

Mice21 [21]3 years ago
5 0
x=\sqrt{(2x+24)}\\\\The\ domain:\\2x+24\geq0\to x\geq-12\\x\geq0\\D:x\geq0\\\\x=\sqrt{2x+24}\ \ \ \ |square\ both\ sides\\\\x^2=(\sqrt{2x+24)})^2\\\\x^2=2x+24\ \ \ |-2x-24\\\\x^2-2x-24=0\\\\x^2+4x-6x-24=0\\\\x(x+4)-6(x+4)=0\\\\(x+4)(x-6)=0\iff x+4=0\ \vee\ x-6=0\\\\x=-4\notin D\ \vee\ x=6\in D

Answer: 6 is a solution to the original equation. The value –4 is an extraneous solution.

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Evaluate: (8y+5) -2z; if y =3 and z =9
iVinArrow [24]

Answer:

11

Step-by-step explanation:

(8y + 5) - 2z

y = 3, z = 9

Let's plug in the given values.

(8(3) + 5) - 2(9)

First let's multiply.

(24 + 5) - 18

Now simplify within the parentheses.

29 - 18

Subtract.

11

This is your answer.

Hope this helps!

5 0
3 years ago
Please tell me if I was right.
Gennadij [26K]
Im pretty sure you are i have done this question before and im pretty sure thats what i put and i got it right. plus just go with your gut

8 0
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2 years ago
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Use the exponential decay​ model, Upper A equals Upper A 0 e Superscript kt​, to solve the following. The​ half-life of a certai
Akimi4 [234]

Answer:

It will take 7 years ( approx )

Step-by-step explanation:

Given equation that shows the amount of the substance after t years,

A=A_0 e^{kt}

Where,

A_0 = Initial amount of the substance,

If the half life of the substance is 19 years,

Then if t = 19, amount of the substance = \frac{A_0}{2},

i.e.

\frac{A_0}{2}=A_0 e^{19k}

\frac{1}{2} = e^{19k}

0.5 = e^{19k}

Taking ln both sides,

\ln(0.5) = \ln(e^{19k})

\ln(0.5) = 19k

\implies k = \frac{\ln(0.5)}{19}\approx -0.03648

Now, if the substance to decay to 78​% of its original​ amount,

Then A=78\% \text{ of }A_0 =\frac{78A_0}{100}=0.78 A_0

0.78 A_0=A_0 e^{-0.03648t}

0.78 = e^{-0.03648t}

Again taking ln both sides,

\ln(0.78) = -0.03648t

-0.24846=-0.03648t

\implies t = \frac{0.24846}{0.03648}=6.81085\approx 7

Hence, approximately the substance would be 78% of its initial value after 7 years.

5 0
3 years ago
[4 -4 0 0]+[4 -2 2 -4]
jok3333 [9.3K]

Answer:

[8 -6 2 -4].

Step-by-step explanation:

[4 -4 0 0]+[4 -2 2 -4]

= [4+4 -4+-2 0+2 0+-4]

= [8 -6 2 -4]

6 0
3 years ago
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