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inna [77]
2 years ago
6

tle="\lim_{x \to 0 (\frac{x(x-2)}{2-2e^2x} )" alt="\lim_{x \to 0 (\frac{x(x-2)}{2-2e^2x} )" align="absmiddle" class="latex-formula">
Help evaluting this limit
Mathematics
1 answer:
krek1111 [17]2 years ago
7 0

Answer: 0

Step-by-step explanation:

Substituting in x=0, we get

\frac{0(0-2)}{2-2e^{2}(0)}=0

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Find an equation of the plane that is parallel to the line of intersection of the planes ( x + 3y + z = 3 ) ( 3x - y - z = 0 ) a
NeTakaya

Answer:

Wow I got the first step

are you looking for the

intersection of the line perpendicular

to plane 1 thorough the origin and plane 2

for the first part???

7 0
3 years ago
How is the answer 2.7
Darya [45]

Answer:

2.7

Step-by-step explanation:

Its 2.7 because if you use the calculator app on your phone or a calculator you can see that you use .09 30 times. that's why I reccomend a calculator. Or you can try cross multiplication.

3 0
3 years ago
What digit makes each number divisible by nine?<br><br> _7,302
Alborosie

Answer:

Not completely sure but hopefully these will help.

Step-by-step explanation:

I'm kind of confused what you're asking for. If you could clarify that would be great. If you're trying to make it divisible by 9, you can either do 3, 6 or 9. If you want the first two digits: 2,5,8. Please clarify so I can help

8 0
4 years ago
There are 15 identical pens in your drawer, nine of which have never been used. On Monday, yourandomly choose 3 pens to take wit
DaniilM [7]

Answer: p = 0.9337

Step-by-step explanation: from the question, we have that

total number of pen (n)= 15

number of pen that has never been used=9

number of pen that has been used = 15 - 9 =6

number of pen choosing on monday = 3

total number of pen choosing on tuesday=3

note that the total number of pen is constant (15) since he returned the pen back .

probability of picking a pen that has never been used on tuesday = 9/15 = 3/5

probability of not picking a pen that has never been used on tuesday = 1-3/5=2/5

probability of picking a pen that has been used on tuesday = 6/15 = 2/5

probability of not picking a pen that has not been used on tuesday= 1- 2/5= 3/5

on tuesday, 3 balls were chosen at random and we need to calculate the probability that none of them has never been used .

we know that

probability of ball that none of the 3 pen has never being used on tuesday = 1 - probability that 3 of the pens has been used on tuesday.

to calculate the probability that 3 of the pen has been used on tuesday, we use the binomial probability distribution

p(x=r) = nCr * p^{r} * q^{n-r}

n= total number of pens=15

r = number of pen chosen on tuesday = 3

p = probability of picking a pen that has never been used on tuesday = 9/15 = 3/5

q = probability of not picking a pen that has never been used on tuesday = 1-3/5=2/5

by slotting in the parameters, we have that

p(x=3) = 15C3 * (\frac{2}{5})^{3} * (\frac{3}{5})^{12}

p(x=3) = 455 * 0.4^{3} * 0.6^{12}

p(x=3) = 455 * 0.064 * 0.002176

p(x=3) = 0.0633

thus probability that 3 of the pens has been used on tuesday. = 0.0633

probability of ball that none of the 3 pen has never being used on tuesday  = 1 - 0.0633 = 0.9337

3 0
3 years ago
How would you solve these 2 maths problems?
sweet-ann [11.9K]

Answer:

see explanation

Step-by-step explanation:

(9)

Expand and simplify right side and compare coefficients of like terms on left side.

(x - 2)(x² + ax + b) + c

= x³ + ax² + bx - 2x² - 2ax - 2b + c

= x³ + x²(a - 2) + x(b - 2a) - 2b + c

compare with

x³ + 2x² - 3x + 4

x² terms

a - 2 = 2 ( add 2 to both sides )

a = 4

x terms

b - 2a = - 3 ← substitute a = 4

b - 8 = - 3 ( add 8 to both sides )

b = 5

constant terms

- 2b + c = 4 ← substitute b = 5

- 10 + c = 4 ( add 10 to both sides )

c = 14

Thus a = 4, b = 5 and c = 14

-------------------------------------------------

(10)

Since both functions cross the x- axis at - 2 then (- 2, 0) satisfies both, that is

f(- 2) = (-2)^{4} + a(- 2)³ + b(- 2)² + 36(- 2) + 144 = 0, that is

- 16 - 8a + 4b - 72 + 144 = 0

- 8a + 4b + 56 = 0

- 8a + 4b = - 56 → (1)

and

g(- 2) = (-2)^{4} + (a + 3)(- 2)³ - 23(- 2)² + (b + 10)(- 2) + 40 = 0, that is

- 16 - 8(a + 3) - 92 - 2(b + 10) + 40 = 0

- 16 - 8a - 24 - 92 - 2b - 20 + 40 = 0

- 8a - 2b - 112 = 0

- 8a - 2b = 112 → (2)

Subtract (1) from (2) term by term

- 6b = 168 ( divide both sides by - 6 )

b = - 28

Substitute b = - 28 into (2)

- 8a + 56 = 112 ( subtract 56 from both sides )

- 8a = 56 ( divide both sides by - 8 )

a = - 7

Thus a = - 7 and b = - 28

5 0
3 years ago
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