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stepladder [879]
3 years ago
6

I need help please. daad

Mathematics
1 answer:
lapo4ka [179]3 years ago
8 0

Answer:

DASSDASDSADA

Step-by-step explanation:DADSA

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Factor the trinomial:<br> 5x^2 + 12x + 7
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<h2>Hey mate,</h2><h3>I have attached the answer below</h3><h3>Hope it will help you...</h3>

7 0
3 years ago
Jeremy is taking a photography class, but he doesn't own a camera. The class organizers will rent him a camera for $6 per day. J
Mazyrski [523]
Solving the given inequality for d, we get...

6d + 15 < 50
6d + 15-15 < 50-15 <<-- subtract 15 from both sides
6d < 35
6d/6 < 35/6 <<--- divide both sides by 6
d < 5.83

Which means that d can be any of the values in this set: {0, 1, 2, 3, 4, 5}

The smallest d can be is 0. In this scenario, Jeremy pays the $15 registration but doesn't rent the camera at all
The largest d can be is 5. In this scenario, Jeremy rents the camera for 5 days
Any larger value of d is not allowed as it would make the total cost go over $50
Notice how I'm rounding down regardless how close 5.83 is to 6

8 0
3 years ago
Read 2 more answers
Write the Rational Expression in simplest form. <img src="https://tex.z-dn.net/?f=%5Cfrac%7Bx%5E2-x-6%7D%7B24-5x-x%5E2%7D" id="T
Liula [17]

Answer:

-(x+2)/(x+8)

Step-by-step explanation:

( x^2 -x-6)

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24 - 5x -x^2

Factor out a minus sign from the denominator

( x^2 -x-6)

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-( x^2 +5x -24)

Factor the numerators and the denominators

( x-3) (x+2)

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-( x+8)(x-3)

Cancel like terms

  x+2

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-(x+8)

4 0
3 years ago
The number of customers waiting for gift-wrap service at a department store is an rv X with possible values 0, 1, 2, 3, 4 and co
julia-pushkina [17]

Answer:

P[X=3,Y=3] = 0.0416

Step-by-step explanation:

Solution:

- X is the RV denoting the no. of customers in line.

- Y is the sum of Customers C.

- Where no. of Customers C's to be summed is equal to the X value.

- Since both events are independent we have:

                         P[X=3,Y=3] = P[X=3]*P[Y=3/X=3]

              P[X=3].P[Y=3/X=3] = P[X=3]*P[C1+C2+C3=3/X=3]

        P[X=3]*P[C1+C2+C3=3/X=3] = P[X=3]*P[C1=1,C2=1,C3=1]        

              P[X=3]*P[C1=1,C2=1,C3=1]  = P[X=3]*(P[C=1]^3)

- Thus, we have:

                        P[X=3,Y=3] = P[X=3]*(P[C=1]^3) = 0.25*(0.55)^3

                        P[X=3,Y=3] = 0.0416

6 0
2 years ago
Part 1
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Because December is really cold
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