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sergij07 [2.7K]

2 years ago

7

Solve for x not y please I am stuck

2 answers:

Dvinal [7]2 years ago

8 0

Answer:

x=2, x=1

Step-by-step explanation:

Katarina [22]2 years ago

4 0

let's solve :

${4}^{2x} - 20(4 {}^{x} ) + 64 = 0$

let's take

$2 {}^{2x} = y$

now,

${(2{}^{2x} )} {}^{2} - 20(2) {}^{2x} + 64 = 0$

$y {}^{2} - 20y + 64 = 0$

$y {}^{2} - 16y - 4y + 64 = 0$

$y(y - 16) - 4( - 16) = 0$

$(y - 16) (y - 4) = 0$

$y = 16 \: \: or \: \: y = 4$

if y = 16, then :

$2 {}^{2x} = 16$

${2}^{2x} = 2 {}^{4}$

$2x = 4$

$x = 2$

and if y = 4 :

${2}^{2x} = 4$

${2}^{2x} = {2}^{2}$

$2x = 2$

[tex]x = 1[/tex

hence, value of x = 1 or x = 2

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d1i1m1o1n [39]

Step-by-step explanation:

$x^{2} - 4x - 7 = 0$

First, let's move the $7$ to the right-hand side so we can determine what constant we'll need on the left-hand side to complete the square:

$x^{2} - 4x = 7$

From here, since the coefficient of the $x$ term is $-4$ , we know the square will be $(x - 2)$ (since $-2$ it's half of $-4$ ).

To complete this square, we will need to add $(-2)^{2}$ to both sides of the equation:

$x^{2} - 4x + (-2)^2 = 7 + ^{-2}$

$x^{2} - 4x + 4 = 7 + 4$

$(x - 2)^{2} = 11$

Now we can take the square root of both sides to figure out the solutions to $x$ :

$x - 2 = \pm \sqrt{11}$

$x = 2 \pm \sqrt{11}$

7 0

3 years ago

9.014 which digit is in the hundredths place

jekas [21]

Answer:

Step-by-step explanation:

1 is in the hundredths

4 0

3 years ago

Read 2 more answers

Yall help “How old am i if 200 is reduced by 2 times my age is 16?”

Nikitich [7]

Answer:

Two times of your age = 16 × 2 = 32

So, 200 reduced = 32 - 200 =<u> - 163 </u><u>years</u>

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5 0

3 years ago

What is the answer to 2m-6=48

Marrrta [24]

2m-6=48
+6 +6
__________
2m= 42
÷2 ÷2
_________
m=21

m=21 is the answer

5 0

3 years ago

A given field mouse population satisfies the differential equation dp dt = 0.5p − 410 where p is the number of mice and t is the

ohaa [14]

Answer:

a) $t = 2 *ln(\frac{82}{5}) =5.595$

b) $t = 2 *ln(-\frac{820}{p_0 -820})$

c) $p_0 = 820-\frac{820}{e^6}$

Step-by-step explanation:

For this case we have the following differential equation:

$\frac{dp}{dt}=\frac{1}{2} (p-820)$

And if we rewrite the expression we got:

$\frac{dp}{p-820}= \frac{1}{2} dt$

If we integrate both sides we have:

$ln|P-820|= \frac{1}{2}t +c$

Using exponential on both sides we got:

$P= 820 + P_o e^{1/2t}$

Part a

For this case we know that p(0) = 770 so we have this:

$770 = 820 + P_o e^0$

$P_o = -50$

So then our model would be given by:

$P(t) = -50e^{1/2t} +820$

And if we want to find at which time the population would be extinct we have:

$0=-50 e^{1/2 t} +820$

$\frac{820}{50} = e^{1/2 t}$

Using natural log on both sides we got:

$ln(\frac{82}{5}) = \frac{1}{2}t$

And solving for t we got:

$t = 2 *ln(\frac{82}{5}) =5.595$

Part b

For this case we know that p(0) = p0 so we have this:

$p_0 = 820 + P_o e^0$

$P_o = p_0 -820$

So then our model would be given by:

$P(t) = (p_o -820)e^{1/2t} +820$

And if we want to find at which time the population would be extinct we have:

$0=(p_o -820)e^{1/2 t} +820$

$-\frac{820}{p_0 -820} = e^{1/2 t}$

Using natural log on both sides we got:

$ln(-\frac{820}{p_0 -820}) = \frac{1}{2}t$

And solving for t we got:

$t = 2 *ln(-\frac{820}{p_0 -820})$

Part c

For this case we want to find the initial population if we know that the population become extinct in 1 year = 12 months. Using the equation founded on part b we got:

$12 = 2 *ln(\frac{820}{820-p_0})$

$6 = ln (\frac{820}{820-p_0})$

Using exponentials we got:

$e^6 = \frac{820}{820-p_0}$

$(820-p_0) e^6 = 820$

$820-p_0 = \frac{820}{e^6}$

$p_0 = 820-\frac{820}{e^6}$

8 0

3 years ago