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2 years ago

E^4x-5-e=0 Solve the equation for x

Mathematics

1 answer:

kati45 [8]2 years ago

8 0

Answer:

$x = \frac{1}{4} \times In(5 + e)$

Step-by-step explanation:

${e}^{4x} - 5 - e = 0 \\ {e}^{4x} = 5 + e \\ 4x = In(5 + e) \\$

$\boxed{Answer:{\boxed{\green{x = \frac{1}{4} \times In(5 + e)}}}}$

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what is the mean of this data set? 12, 17, 16, 10, 20, 13, 14, 14, 12, 12, 19, 18. answers: 13.50, 14.00, 14.75, and 15.75

murzikaleks [220]

Answer:

14.75

Step-by-step explanation:

Add them all up, divide by 12

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2 years ago

Please help me with this

sergij07 [2.7K]

Answer:

x=8

Step-by-step explanation:

The ratio of a side of PQRS to the corresponding side in TUVW. is 6:4=3:2.

The ratio of RS:VW=3:2. RS=12, and VW=x. 12:x=3:2. 3 multiplied by 4 is 12, so 2 multiplied by 4 is x. So, the makes x=8.

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3 years ago

Here is a picture of a circle. Each square represents 1 square unit.

TEA [102]

Answer:

because there isnt any full circles. in order to have a square you need 4 corners .they all only have 1 inside that circle.

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3 years ago

A lake polluted by bacteria is treated with an antibacterial chemical. Aftertdays, thenumber N of bacteria per milliliter of wat

jeyben [28]

Answer:

N(1)=50 is a minimum

N(15)=4391.7 is a maximum

Step-by-step explanation:

<u>Extrema values of functions </u>

If the first and second derivative of a function f exists, then f'(a)=0 will produce values for a called critical points. If a is a critical point and f''(a) is negative, then x=a is a local maximum, if f''(a) is positive, then x=a is a local minimum.

We are given a function (corrected)

$N(t) = 20(t^2-lnt^2)+ 30$

$N(t) = 20(t^2-2lnt)+ 30$

(a)

First, we take its derivative

$N'(t) = 20(2t-\frac{2}{t})$

Solve N'(t)=0

$20(2t-\frac{2}{t})=0$

Simplifying

$2t^2-2=0$

Solving for t

$t=1\ ,t=-1$

Only t=1 belongs to the valid interval $1\leqslant t\leqslant 15$

Taking the second derivative

$N''(t) = 20(2+\frac{2}{t^2})$

Which is always positive, so t=1 is a minimum

(b)

$N(1)=20(1^2-2ln1)+ 30$

N(1)=50 is a minimum

(c) Since no local maximum can be found, we test for the endpoints. t=1 was already determined as a minimum, we take t=15

(d)

$N(15)=20(15^2-2ln15)+ 30$

N(15)=4391.7 is a maximum

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2 years ago

Suppose that you have eight cards. Five are green and three are yellow. The five green cards are numbered 1, 2, 3, 4, and 5. The

Marina86 [1]

Answer:

Step-by-step explanation:

Given that you have eight cards. Five are green and three are yellow. The five green cards are numbered 1, 2, 3, 4, and 5. The three yellow cards are numbered 1, 2, and 3. The cards are well shuffled. You randomly draw one card.

G = card drawn is green

Y = card drawn is yellow

E = card drawn is even-numbered

List:

Sample space = {G1, G2, G3, G4, G5, Y1, Y2, Y3}

2) P(G) = 5/8

3) P(G/E) = P(GE)/P(E)

GE = {G2, G4}

Hence P(G/E) = 2/5

4) GE = {G2, G4}

P(GE) = 2/8 = 1/4

5) P(G or E) = P(G)+P(E)-P(GE)

= 5/8 + 3/8-2/8 = 3/5

6) No there is common element as G2 and G4

Cannot be mutually exclusive

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3 years ago