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Romashka [77]
3 years ago
8

I will mark brainliest..

Mathematics
1 answer:
earnstyle [38]3 years ago
6 0

Answer:

a)6 {x}^{8}  {y}^{5}  \\ b)4 {x}^{5}  {z}^{8}  \\ c)12 {a}^{9}  {b}^{7}  \\ d)6 {s}^{9}  {t}^{3}

Step-by-step explanation:

Dont forget to apply THE LAW OF INDICES!!

Brainliest would be appreciated :))

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I need help pls❗❗❗Its complicated ​
mel-nik [20]

Answer:

The answer to 36 is -18

Step-by-step explanation:

x=-18 because 3×(-6)=(-6×-6) cancels

3 0
3 years ago
LAST QUESTION NEED HELP PLEASE
Licemer1 [7]

Answer:

g(x)= (x+8)^2 +2

Step-by-step explanation:

f(x) = x^2

We want to translate it 8 units to the left

h(x) = f(x+8)

h(x) = (x+8)^2

We also want to translate up 2 units

g(x) = h(x) +2

g(x)= (x+8)^2 +2

4 0
4 years ago
A warehouse contains ten printing machines, five of which are defective. a company selects four of the machines at random, think
Yuri [45]

This kind of situation is modelled by Bernoulli's formula. It applies everytime there is an experiments with two possible outcomes repeated n times. Each repetition is independent on the others, and we know the probability of the two outcomes p and 1-p. If we want the outcome with probability p to appear k times, the probability is


\binom{n}{k}p^k(1-p){n-k}


In your case, you run the "experiment" 4 times (you choose 4 printers) and want that all of them to be non-defective. A printer is non-defective with probability 1/2, since there are 5 defective and 5 non-defective printers.


So, our model is built with n = k = 4, p = 1-p = 1/2. The probability is


P=\binom{4}{4}\left(\cfrac{1}{2}\right)^4\left(\cfrac{1}{2}\right)^0  = 1 \cdot \left(\cfrac{1}{2}\right)^4 \cdot 1 = \cfrac{1}{16}

6 0
3 years ago
Solve 3/2 + t = 1/2 <br><br><br> t = ___
kiruha [24]

Answer:

-1

Step-by-step explanation

1/2-3/2 = -1

5 0
4 years ago
Read 2 more answers
Please HELP I WILL MARK BRAINLIEST!!!! :)
Serggg [28]

Answer: John is 69 inches tall.

Step-by-step explanation:

There are 12 inches in each foot.

5x12=60

12x3/4= 9

60+9= 69

6 0
3 years ago
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