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Archy [21]
3 years ago
6

Pease help me ASAP....................

Mathematics
2 answers:
Rasek [7]3 years ago
5 0
A.  
-5 +5(x+4)\\ -5 +5x+20\\ 5x+15

b.
3m+2(5m-7)\\ 3m+10m-14\\ 13m-14

c.
4(6t+9)-10t\\ 24t+36-10t\\ 14t+36

d.
8k-6(3-2k)\\ 8k-18+12k\\ 20k-18
AfilCa [17]3 years ago
3 0
First one, x+4

second one, 13m-7

third one, 14x+36


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-3 + 10y – 8y + 4 – 6 + y simplify
Darya [45]

Answer:

3y - 5

Step-by-step explanation:

We combine like terms. The y terms are 10y, -8y and y which combine to 3y. Think of it like having 10 items, take away 8 items but then add on 1 more. So we'd have 10-8+1 = 2+1 = 3 items overall. Replace "Items" with y and we have 10y-8y+1y = (10-8+1)y = 3y

The other terms are constant terms as they dont have any variables attached to them. These terms are -3, 4 and -6 which combine to -3+4-6 = 1-6 = -5

In summary, the y terms combine to 3y while the constant terms combine to -5. Put together, the fully simplified answer is 3y-5

7 0
3 years ago
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What’s is 4 over 9 and negative 5 over 9
Lelechka [254]
Answer=the fourth bubble
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3 years ago
Ben had twice as many nickels as dimes. Altogether, Ben has $4.20. How many nickels and how many dime did Ben have
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14 quarters and 7 dimes.
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3 years ago
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A complex electronic system is built with a certain number of backup components in its subsystems. One subsystem has four identi
Tcecarenko [31]

Answer:

a. 0.1536

b. 0.9728

Step-by-step explanation:

The probability that a component fails, P(Y) = 0.2

The number of components in the system = 4

The number of components required for the subsystem to operate = 2

a. By binomial theorem, we have;

The probability that exactly 2 last longer than 1,000 hours, P(Y = 2) is given as follows;

P(Y = 2) = \dbinom{4}{2} × 0.2² × 0.8² = 0.1536

The probability that exactly 2 last longer than 1,000 hours, P(Y = 2) = 0.1536

b. The probability that the system last longer than 1,000 hours, P(O) = The probability that no component fails + The probability that only one component fails + The probability that two component fails leaving two working

Therefore, we have;

P(O) = P(Y = 0) + P(Y = 1) + P(Y = 2)

P(Y = 0) = \dbinom{4}{0} × 0.2⁰ × 0.8⁴ = 0.4096

P(Y = 1) = \dbinom{4}{1} × 0.2¹ × 0.8³ = 0.4096

P(Y = 2) = \dbinom{4}{2} × 0.2² × 0.8² = 0.1536

∴ P(O) = 0.4096 + 0.4096 + 0.1536 = 0.9728

The probability that the subsystem operates longer than 1,000 hours = 0.9728

4 0
2 years ago
PLEASSSSSEEE NEED HELP ASAP ALOT OF POINTS BIGGG POINTS PLEASSSS
Elena-2011 [213]
2. x = 8, -4
|x-2|
2 * x-2=12
2x-4=12
2x=16
x=8
-4 - 2 = -6
-6 x 2 = 12
3 0
3 years ago
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