Answer:
A.
B.199 hours
Step-by-step explanation:
Let carpenters work for x hours and painters work for y hours.
Nelson must spend money <$7500
Carpenters charge per hour=$28
Painters charge per hour=$32
A.We have to write inequalities which represents this situation
According to question
B.If hires carpenters work for total time=40 hours
We have to find the maximum number of hours for which painters can work without exceeding the budget.
Substitute the value in above inequality equation then, we get
Subtracting 1120 on both sides
Divide by 32 on both sides then we get
Hence,painters can work for maximum 199 hours.
Answer: See attached
Step-by-step explanation:
[] Since the inequality uses greater than (>) and <em>not</em> greater than or equal to (≥), the line will be dashed.
[] The inequality is greater than (>) so we will shade above the line as values "above" are greater.
Step-by-step explanation:
The rate by which Rita is paid first 8 hours = $16
The next per hour rate after 8 hours = times the usual rate.
Now, 1.5 times $16 = 1.5 x 1 6 = $24
So, for each hour worked after 8th hour, Rita is paid $24 per hour.
Now, the intervals she has worked are:
(a) from 8:15 to 12: 45 pm
So, the number of hours worked = 45 minutes + 45 minutes + 3 hours
= 4 hours 30 minutes .... (1) (as 60 minutes = 1 hour)
(b) break of 45 minutes after 12: 45 = 12: 45 + 45 minutes
= 12 hour 90 minutes = 1 : 30 P M
So, she started working at 1: 30 and worked till 6: 15
So, the number of hours worked = 30 minutes + 15 minutes +4 hours
= 4 hours 45 minutes .... (2)
So, by adding (1) and (2) the total hours worked by Rita
= 4 hours 30 minutes + 4 hours 45 minutes
= 8 hours 75 minutes = 9 hours 15 minutes
9 hours 15 minutes = ( 8 + 1 .25 ) ( 15 minutes = 0.25 of an hour)
So, the payment is ( 8 x (16) + 1.25 (24) )
= 128 + 30 = 158
Hence, the total money earned by Rita = $158
La división sintetica sirve para el cociente entre polinomios, la cual es particularmente util para cuando tenemos un numerador de grado alto y un denominador de grado 1.
Veremos que si, en la división sintética solo son tomados en cuenta los coeficientes del divisor.
Para explicarla, usaremos un ejemplo.
x^3 - 2x^2 + 5x + 3 será el <u>numerador</u>
x - 2 será el <u>denominador.</u>
Para poder aplicar la división sintetica, el maximo exponente en el <u>denominador </u>debe ser uno.
Así, nuestro cociente es:
Procedemos a igualar el <u>denominador</u> a cero:
x - 2 = 0
x = 2
Ahora veremos los coeficientes del <u>numerador</u> (el <u>numerador </u>es el <u>divisor</u>).
Lo que debemos hacer es tomar cada coeficiente (comenzando por los que multiplican a los terminos con mayor exponente) y multiplicarlos por este número.
En este caso, el primer coeficiente es 1.
2*1 = 2
Ahora le sumamos esto al proximo coeficiente, que es -2:
-2 + 2 = 0
Multiplicamos esto por el valor de x, que es 2.
2*0 = 0
Ahora sumamos esto al proximo coeficiente, que es 5
5 + 0 = 5
Multiplicamos esto por el valor de x, que es 2:
5*2 = 10
Sumamos esto al ultimo coeficiente, que es 3.
10 + 3 = 13
Este ultimo valor es el residuo, y los primeros 3 números que obtuvimos son los coeficientes del cociente, entonces el cociente será:
1*x^2 + 0*x + 10 = x^2 + 10
Como vimos, solo usamos los coeficientes del numerador/divisor para realizar esta operación, por lo que si, solo los coeficientes del divisor son tomados en cuenta.
Si quieres aprender más, puedes leer: