Answer:
Since Darcie wants to crochet a minimum of 3 blankets and she crochets at a rate of 1/5 blanket per day, we can determine how many days she will need to crochet a minimum of 3 blankets following the next steps:
- Finding the number of days needed to crochet one (1) blanket:
\begin{gathered}1=\frac{1}{5}Crochet(Day)\\Crochet(Day)=5*1=5\end{gathered}
1=
5
1
Crochet(Day)
Crochet(Day)=5∗1=5
So, she can crochet 1 blanket every 5 days.
- Finding the number of days needed to crochet three (3) blankets:
If she needs 5 days to crochet 1 blanket, to crochet 3 blankets she will need 15 days because:
\begin{gathered}DaysNeeded=\frac{NumberOfBlankets}{Rate}\\\\DaysNeeded=\frac{3}{\frac{1}{5}}=3*5=15\end{gathered}
DaysNeeded=
Rate
NumberOfBlankets
DaysNeeded=
5
1
3
=3∗5=15
- Writing the inequality
If she has 60 days to crochet a minimum of 3 blankets but she can complete it in 15 days, she can skip crocheting 45 days because:
AvailableDays=60-RequiredDaysAvailableDays=60−RequiredDays
AvailableDays=60-15=45DaysAvailableDays=60−15=45Days
So, the inequality will be:
s\leq 45s≤45
The inequality means that she can skip crocheting a maximum of 45 days since she needs 15 days to crochet a minimum of 3 blankets.
Have a nice day!
Ok so if you did 156-6 (considering it was 6 more you want to subtract it) and then because the girls was doubled the boys (you are now at 150) you must divide that by 3 which will give you 50. But you need to double it (for girls) which is 100 and then you have 50 left over which is for the boys. Boys=50 girls=106
Answer:
26: x=34
28: x=17
30: m=30
Step-by-step explanation:
<u>26:</u> 2x+22=90
subtract 22
2x=68
divide by 2
x=34
<u>28:</u> 18+3x+21=90
add 18 and 21
3x+39=90
subtract 39
3x=51
divide by 3
x=17
<u>30:</u> 43+87+m+20=180
add 43 and 87 and 20
m+150=180
subtract 150
m=30
Answer:
12 : 4.92
Step-by-step explanation:
a= apples
12a=4.92
12 : 4.92
Given:


To find:
The quadrant of the terminal side of
and find the value of
.
Solution:
We know that,
In Quadrant I, all trigonometric ratios are positive.
In Quadrant II: Only sin and cosec are positive.
In Quadrant III: Only tan and cot are positive.
In Quadrant IV: Only cos and sec are positive.
It is given that,


Here cos is positive and sine is negative. So,
must be lies in Quadrant IV.
We know that,



It is only negative because
lies in Quadrant IV. So,

After substituting
, we get





Therefore, the correct option is B.