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3 years ago

Solve the equation x^2-4x+85=0

1 answer:

3 0

Can't factor
use quadratic
for
ax^2+bx+c=0
x= $\frac{-b+/- \sqrt{b^2-4ac} }{2a}$

given
1x^2-4x+85=0
a=1
b=-4
c=85

x= $\frac{-(-4)+/- \sqrt{(-4)^2-4(1)(85)} }{2(1)}$
x= $\frac{4+/- \sqrt{16-340} }{2}$
x= $\frac{4+/- \sqrt{-324} }{2}$
x= $\frac{4+/- (\sqrt{-1})(\sqrt{324}) }{2}$
remember that √-1=i
x= $\frac{4+/- i(\sqrt{324}) }{2}$
x= $\frac{4+/- i(18) }{2}$
x=2+/-9i

x=2+9i or x=2-9i

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Suppose an airline policy states that all baggage must be box shaped with a sum of length, width, and height not exceeding 114

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Answer:

Step-by-step explanation:

Represent the length of one side of the base be s and the height by h. Then the volume of the box is V = s^2*h; this is to be maximized.

The constraints are as follows: 2s + h = 114 in. Solving for h, we get 114 - 2s = h.

Substituting 114 - 2s for h in the volume formula, we obtain:

V = s^2*(114 - 2s), or V = 114s^2 - 2s^3, or V = 2*(s^2)(57 - s)

This is to be maximized. To accomplish this, find the first derivative of this formula for V, set the result equal to 0 and solve for s:

----- = 2[(s^2)(-1) + (57 - s)(2s)] = 0 = 2s^2(-1) + 114s - 2s^2

Simplifying this, we get dV/ds = -4s^2 + 114s = 0. Then either s = 28.5 or s = 0.

Then the area of the base is 28.5^2 in^2 and the height is 114 - 2(28.5) = 57 in

and the volume is V = s^2(h) = 46,298.25 in^3

7 0

3 years ago

Uhh I need help what is equivalent to 1 to 4 ????

Nataly_w [17]

Answer:

2/8

Step-by-step explanation:

8 0

3 years ago

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lozanna [386]

Answer:

The correct answer is A

Step-by-step explanation:

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5 0

3 years ago

Read 2 more answers

N = {X:X<br>is an odd number, 21<x<3<br>XEN]

SIZIF [17.4K]

So it’s saying that the following set in roster form: (1) N = (x :x is an odd number, 21< × ≤ 35,×€ N)?

It seems to be that the Roster form would do not do anything but the thing is that this listing would give us the output of a set from this problem.

Your answer would be, N = Set of Odd Numbers such that x is greater than 21 and less than or equal to 35 = {23, 25, 27, 29, 31, 33, 35}.

5 0

3 years ago

A library subscribes to two different weekly news magazines, each of which is supposed to arrive in Wednesday's mail. In actuali

Dominik [7]

Answer:

y 0 1 2 3

P(Y=y) 0.0676 0.3549 0.3875 0.19

Step-by-step explanation:

P(Wed) = 0.26

P(Thurs) = 0.39

P(Fri) = 0.25

P(Sat) = 0.10

Y = No. of days beyond Wednesday it takes for both magazines to arrive i.e. 0,1,2,3

Y=0 means the magazines will arrive on Wednesday

Y=1 means the magazines will arrive till Thursday

Y=2 means the magazines will arrive till Friday

Y=3 means the magazines will arrive till Saturday

The possible combinations for Y are

Y(W,W) Y(W,T) Y(W,F) Y(W,S)

Y(T,W) Y(T,T) Y(T,F) Y(T,S)

Y(F,W) Y(F,T) Y(F,F) Y(F,S)

Y(S,W) Y(S,T) Y(S,F) Y(S,S)

So, we can classify these possible outcomes as Y=0,1,2,3.

Y(0) = Y(W,W) (both magazines take 0 days to arrive beyond Wednesday)

Y(1) = Y(W,T), Y(T,T), Y(T,W) (both magazines take 1 day to arrive beyond Wednesday)

Y(2) = Y(W,F), Y(T,F), Y(F,F) Y(F,W) Y(F,T) (both magazines arrive till Friday)

Y(3) = Y(W,S), Y(T,S), Y(F,S), Y(S,W), Y(S,T), Y(S,F), Y(S,S) (both magazines arrive till Saturday)

To calculate the PMF, we need to calculate the probability for each of the points in Y(0,1,2,3).

Y(0) = Y(W,W)

= 0.26 x 0.26

Y(0) = 0.0676

Y(1) = Y(W,T) + Y(T,T) + Y(T,W)

= (0.26 x 0.39) + (0.39 x 0.39) + (0.39 x 0.26)

= 0.1014 + 0.1521 + 0.1014

Y(1) = 0.3549

Y(2) = Y(W,F) + Y(T,F) + Y(F,F) + Y(F,W) + Y(F,T)

=(0.26 x 0.25) + (0.39 x 0.25) + (0.25 x 0.25) + (0.25 x 0.26) + (0.25 x 0.39)

= 0.065 + 0.0975 + 0.0625 + 0.065 + 0.0975

Y(2) = 0.3875

Y(3) = Y(W,S) + Y(T,S) + Y(F,S) + Y(S,W) + Y(S,T) + Y(S,F) + Y(S,S)

= (0.26 x 0.10) + (0.39 x 0.10) + (0.25 x 0.10) + (0.10 x 0.26) + (0.10 x 0.39) + (0.10 x 0.25) + (0.10 x 0.10)

= 0.026 + 0.039 + 0.025 + 0.026 + 0.039 + 0.025 + 0.010

Y(3) = 0.19

y 0 1 2 3

P(Y=y) 0.0676 0.3549 0.3875 0.19

The PMF plot is attached as a photo here.

7 0

3 years ago