Answer
- 1928 adults
explanation
- if you do 2408 plus 557 that would be 2965 but then 379 come off board so you do 2965 take away 379 that leads to 2586 and if 658 are kids u do 2586 take away 658 which leads to 1928 adults
Answer: the costs would be the same if he orders 29 rose flowers.
Step-by-step explanation:
Let x represent the number of roses that Victor would order such that the cost will be the same with either flower shop.
Dayton Florist charges $3 per rose, plus $12 for the vase. This means that if he orders x rose flowers, the total amount that he would pay is
3x + 12
Tammy's Flowers, in contrast, charges $2 per rose and $41 for the vase. This means that if he orders x rose flowers, the total amount that he would pay is 2x + 41
For the costs to be the same, the number of rose flowers that he orders would be
3x + 12 = 2x + 41
3x - 2x = 41 - 12
x = 29
Answer:
That is the solutions are:
,
Step-by-step explanation:
The period of
or
is
.
Let's look at the first rotation to see when
happens.
This happens at
and also at
. (Notice I just looked at the x-coordinates because that is what cosine is. Sine is the y-coordinate.)
Now to find the rest of the solutions we can just make full rotations either way to get back to those points .
That is the solutions are:


Sets and set operations are ways of organizing, classifying and obtaining information about objects according to the characteristics they possess, as objects generally have several characteristics, the same object can belong to several sets, an example is the subjects of a school , where students (objects) are classified according to the subject they study (set).
The <em>intersection</em> of sets is a new set consisting of those objects that simultaneously possess the characteristics of each intersected set, the intersection of two subjects will be those students who have both subjects enrolled.
The <em>union</em> of sets is a new set consisting of all the objects belonging to the united sets, the union of two subjects will be all students of both courses.
In this case there are three sets B, C and S of which we are given the following information:
Answer
n(BꓵSꓵC)=5
n(BꓵS)=10 – 5 = 5
n(BꓵC)=12 – 5 = 7
n[(BꓵC)ꓴ(BꓵS)ꓴ(CꓵS)]=21 – 5 – 5 – 7 = 4
n(B)=36 – 5 – 5 – 7 = 19
n(S)=30 – 5 – 5 – 4 = 16
n(C)=34 – 5 – 7 – 4 = 18
Y+5= 3(x-4)
y + 5 = 3x - 12
y = 3x - 17
the answer is a