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Igoryamba
3 years ago
12

What is 5 + (-8) I need to put it on a number line

Mathematics
2 answers:
Naily [24]3 years ago
8 0

Answer: -3

Step-by-step explanation: 5-8 is -3 so just go 3 to the left from 0 and plug in the point or number

wlad13 [49]3 years ago
7 0

Answer:

5 + (-8)= (-3)

Step-by-step explanation:

If you do keep,change,change you will get

  • 5-8=(-3)

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Find the conjugate of the complex number, use division:<br> 2+i<br> ——<br> 1+i
Fittoniya [83]

I assume you mean divide using the conjugate to rationalize the denominator and express the result in standard rectangular form.

\dfrac{2+i}{1+i} = \dfrac{2+i}{1+i} \cdot \dfrac{1-i}{1-i} = \dfrac{2(1)-i^2-2i+i}{1^2+1^2} = \frac 3 2 - \frac 1 2 i


4 0
3 years ago
Can someone find the surface area for this? :)
AysviL [449]

Answer:

84 :)

Step-by-step explanation:

bottom surface area= 6m^{2}

top surface area=6m^{2}

lateral surface area= 72m^{2}

add this together you get TOTAL surface area of 84

3 0
3 years ago
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dedylja [7]
Right there 125677 on the left you go 4 ahhbsn

6 0
2 years ago
(WORTH 95 POINTS) PLEASE HELP I CANNOT FIGURE IT OUT AND ITS SUPPOSED TO BE TURNED IN ALREADY!
yanalaym [24]

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5 0
3 years ago
Find the probability of the following events , when a dice is thrown once:
Fudgin [204]

Answer:

Step-by-step explanation:

s a die is rolled once, therefore there are six possible outcomes, i.e., 1,2,3,4,5,6.

(a) Let A be an event ''getting a prime number''.

Favourable cases for a prime number are 2,3,5,

i.e., n(A)=3

Hence P(A)=n(A)n(S)=36=12

(b) Let A be an event ''getting a number between 3 and 6''.

Favourable cases for events A are 4 or 5.

i.e., n(A)=2

P(A)=n(A)n(S)=26=13

(c) Let A be an event ''a number greater than 4''.

Favourable cases of events A are 5, 6.

i.e., n(A)=2

P(A)=n(A)n(S)=26=13

(d) Let A be the event of getting a number at most 4.

∴ A={1,2,3} ⇒ n(A)=4,n(S)=6

∴ Required probability =n(A)n(S)=42=23

(e) Let A be the event of getting a factor of 6.

∴ A={1,2,36} ⇒ n(A)=4,n(A)=6

∴ Required probability =46=23

(ii) Since, a pair of dice is thrown once, so there are 36 possible outcomes. i.e.,

(a) Let A be an event ''a total 6''. Favourable cases for a total of 6 are (2,4), (4,2), (3,3), (5,1), (1,5).

i.e., n(A)=5

Hence P(A)=n(A)n(S)=536

(b) Let A be an event ''a total of 10n. Favourable cases for total of 10 are (6,4), (4,6), (5,5).

i.e., n(A)=5

P(A)=n(A)n(S)=336=112

(c) Let A be an event ''the same number of the both the dice''. Favourable cases for same number on both dice are (1,1), (2,2), (3,3), (4,4), (5,5), (6,6).

i.e., n(A)=6

P(A)=n(A)n(S)=636=16

(d) Let A be an event ''of getting a total of 9''. Favourable cases for a total of 9 are (3,6), (6,3), (4,5), (5,4).

i.e., n(A)=4

P(A)=n(A)n(S)=436=19

(iii) We have, n(S) = 36

(a) Let A be an event ''a sum less than 7'' i.e., 2,3,4,5,6.

Favourable cases for a sum less than 7 ar

7 0
3 years ago
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