Answer:

Step-by-step explanation:
Given
--- Point
Required
Component of V
From the question, we understand that V is in standard position.
This implies that V starts at the origin (0,0) and ends at (a,b)
So, vector V is:

This gives:


Answer:
1 day ; 10^0
Step-by-step explanation:
Given that:
Order received = 10^6 pages
Number of machines = 100
Daily printing capability by each machine = 10^4
Hence, total output per day = (10^4) * 100 = 10^4 * 10^2 = 10^4+2 = 10^6
Number of days required = number of order / daily output
Number of days = 10^6 / 10^6 = 1
In exponent form:
10^6 / 10^6 = 10^6-6 = 10^0
Answer:
<u>a) x = 3</u>
<u>b) z = 10</u>
<u>c) p = 2</u>
<u>d) x = 7</u>
<u>e) u = 1</u>
Step-by-step explanation:
a) 2x = 6
Despejamos x dividiendo por 2 a amabos lados de la eacuacion.
(2/2)x = 6/2
<u>x = 3</u>
Si remplazamos x en la ecuación original:
2(3)=6
6 = 6
Queda demostrado.
b) 10 + z = 20
Despejamos z restando 10 en amabos lados de la eacuacion.
10-10+z = 20-10
<u>z = 10</u>
Si remplazamos z en la ecuación original:
10 + 10=20
20 = 20
Queda demostrado.
c) p + 9 = 11
Despejamos p restando 9 en amabos lados de la eacuacion.
p + 9 - 9 = 11-9
<u>p = 2</u>
Si remplazamos p en la ecuación original:
2 + 9 = 11
11 = 11
Queda demostrado.
d) 3x + 8 = 29
Despejamos x restando 8 en amabos lados de la eacuacion y luego divideindo por 3 en ambos lados de la ecuación.
3x+8-8 = 29-8
3x = 21
(3/3)x = 21/3
<u>x = 7</u>
Si remplazamos x en la ecuación original:
3(7) + 8 = 29
21 + 8 = 29
29 = 29
Queda demostrado
e) 2u + 8 = 10
Despejamos u restando 8 en amabos lados de la eacuacion y luego divideindo por 2 en ambos lados de la ecuación.
2u+8-8 = 10-8
2x = 2
(2/2)x = 2/2
<u>x = 1</u>
Si remplazamos x en la ecuación original:
2(1) + 8 = 10
2 + 8 = 10
10 = 10
Queda demostrado
Espero te haya sido de ayuda!
Answer:
There are twice as many people who purchased a red car than a black car.
Please give me brainlest
click this link.
https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_24
or copy it then paste it in search.