What are all the names that apply to <br> √24

Answer this Q QUIQLY PLZ

MariettaO [177]

Answer: 3. HAVE AN AMAZING DAY

8 0

3 years ago

N a right triangle ΔABC, the length of leg AC = 5 ft and the hypotenuse AB = 13 ft. Find:

kirza4 [7]

First find the length of leg BC by using the Pythagorean theorem BC=132+52=12 ft if we find the midpoint of BC which is 6 ft we find where the angle bisector would touch the BC. From there we construct another triangle ANC where AC is equal to 5 ft and NC is equal to 6 ft where we can again use the Pythagorean theorem to find the length of the hypotenuse which is the angle bisector AN=Angle Bisector=62+52<span>=7.8102 ft.</span>

5 0

3 years ago

Spam e-mail containing a virus is sent to 1000 e-mail addresses. After 1 second, a recipient machine broadcasts 10 new spam e-ma

pickupchik [31]

Answer:

Answer explained below

Step-by-step explanation:

Spam e-mail containing a virus is sent to 1000 e-mail addresses. After 1 second, a recipient machine broadcasts 10 new spam e-mails containing the same virus, after which the virus disables itself on that machine. (1) Write a recursive definition (i.e. recurrence relation) to show how many spam emails will be sent out after n seconds. (2) Solve the recurrence relation. (3) How many e-mails are sent at the end of 20 seconds

1.START T=0.....

1000 EMAILS SENT AND RECEIVED BY 1000 M/CS.

T=1.....

EACH OF THE M/C SENDS 10 NEW MAILS ....

.................................

LET M[N] BE THE NUMBER OF MAILS SENT OUT AFTER N SECONDS.

SO , EACH OF THESE M/CS WILL SEND 10 MAILS IN NEXT 1 SECOND.

HENCE NUMBER OF MAILS SENT IN N+1 SECONDS=M[N+1]=

M[N+1]=10*M[N].......................1

THIS IS THE RECURRENCE RELATION.....

2.SOLUTION .....

M[N+1]=10M[N]=10*10M[N-1]=10*10*10M[N-2]=........

M(N+1)=[10^1][M(N)]=[10^2][M(N-1)]=[10^3][M(N-2)]=..........=[10^N][M(1)]=[10^(N+1)][M(0)]

M[N+1]=[10^(N+1)][1000]=[10^(N+1)][10^3]=[10^(N+4)]......................................2

THIS IS THE NUMBER OF MAILS SENT AFTER N+1 SECONDS .....OR ....

M[N]=[10^(N+3)].............................................3

..................IS THE SOLUTION FOR NUMBER OF MAILS SENT AFTER N SECONDS.....

3.AFTER N=20 SECONDS , THE ANSWER IS ....

M[20]=10^(20+3)=10^23

8 0

3 years ago

Read 2 more answers

Find the missing number for 1/3 = 3:

RSB [31]

1/3=3:9
since 1×3=3
3×3=9

7 0

3 years ago

Use the work shown to find the solutions of the quadratic equation. x2 – x –3/4=0 x2 – x = 3/4

bija089 [108]

Answer:

x = - $\frac{1}{2}$ , x = $\frac{3}{2}$

Step-by-step explanation:

Given

x² - x - $\frac{3}{4}$ = 0

Multiply through by 4 to clear the fraction

4x² - 4x - 3 = 0

Consider the factors of the product of the coefficient of the x² term and the constant term which sum to give the coefficient of the x- term

product = 4 × - 3 = - 12 and sum = - 4

The factors are + 2 and - 6

Use these factors to split the x- term

4x² + 2x - 6x - 3 = 0 ( factor the first/second and third/fourth terms )

2x(2x + 1) - 3(2x + 1) = 0 ← factor out (2x + 1) from each term

(2x + 1)(2x - 3) = 0

Equate each factor to zero and solve for x

2x + 1 = 0 ⇒ 2x = - 1 ⇒ x = - $\frac{1}{2}$

2x - 3 = 0 ⇒ 2x = 3 ⇒ x = $\frac{3}{2}$

8 0

3 years ago

What ﻿are ﻿all ﻿the ﻿names ﻿that ﻿apply ﻿to ﻿ √24

What are all the names that apply to √24