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jok3333 [9.3K]
3 years ago
12

Please help!! I'll make you Brainlest!! Thanks!!

Mathematics
1 answer:
nirvana33 [79]3 years ago
4 0
16. The open circle is excluded and closed is included.
 x≤-6 
x>2

17. The open circle is excluded so the inequality would be: -3<x<2
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Simplify this (2x)^{3}
Margaret [11]

Answer:

8x^3

Step-by-step explanation:

(2x)3

=(2x)3

=2x*2x*2x

=8x^3

7 0
3 years ago
F(n)=45⋅(4/5)^n−1<br> complete the recursive formula of f(n)
Radda [10]

F(1) = 45

f(n) = f(n-1) * 4/5

Step-by-step explanation:

f(n) = 45 x (4/5)ⁿ⁻¹

f(1) = 45 x (4/5)¹⁻¹ = 45 x (4/5)⁰ = 45 x 1 = 45

f (n-1) = 45 x (4/5)ⁿ⁻¹⁻¹ = 45 x (4/5)ⁿ⁻²

f(n) = f(n-1) * (4/5)¹ = 45 x (4/5) ⁿ⁻²⁺¹ = 45 x (4/5)ⁿ⁻¹

7 0
3 years ago
Find x! Please help!!
a_sh-v [17]

{(8 + x)}^{2}  =  {8}^{2}  +  {15}^{2}  \\ {(8 + x)}^{2}  =  64  +  225 \\  {(8 + x)}^{2}  = 289 \\ {(8 + x)}^{2}  =  {17}^{2}  \\ 8 + x = 17 \\ so \: x = 17 - 8 = 9
5 0
3 years ago
Let V be the event that a computer contains a virus, and let W be the event that a computer contains a worm. Suppose P(V) = 0.17
ArbitrLikvidat [17]

Answer: 0.82

Step-by-step explanation:

The probability of the computer not containing neither a virus nor a worm is expressed as P(V^{C}∩W^{C}) , where P(V^{C}) is the probability that the event V doesn't happen and P(W^{C}) is the probability that the event W doesn't happen.

P(V^{C})= 1-P(V) = 1-0.17 = 0.83

P(W^{C})=1-P(W) = 1-0.05 = 0.95

Since V^{C} and W^{C} aren't mutually exclusive events, then:

P(V^{C}∪W^{C}) = P(V^{C}) + P(W^{C}) - P(V^{C}∩W^{C})

Isolating the probability that interests us:

P(V^{C}∩W^{C})= P(V^{C}) + P(W^{C}) - P(V^{C}∪W^{C})

Where P(V^{C}∪W^{C}) = 1 - 0.04 = 0.96

Finally:

P(V^{C}∩W^{C}) = 0.83 + 0.95 - 0.96 = 0.82

5 0
3 years ago
Calculate the perimeter
Leto [7]

Answer:

48m

Step-by-step explanation:

12m - 4m = 8m / 2 = 4

12m + 4m + 4m + 4m + 4m + 8m + 8m + 4m = 48

3 0
3 years ago
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