Answer:
<em><u>hope </u></em><em><u>this</u></em><em><u> answer</u></em><em><u> helps</u></em><em><u> you</u></em><em><u> dear</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u>take </u></em><em><u>care</u></em><em><u> </u></em><em><u>and </u></em><em><u>may</u></em><em><u> u</u></em><em><u> have</u></em><em><u> a</u></em><em><u> great</u></em><em><u> day</u></em><em><u> ahead</u></em><em><u>!</u></em>
Answer: Universal set
Step-by-step explanation:
A universal set is usually represented by U and it is set that has all the elements of a related sets.
For example, if there are two sets which are X and Y. Let's say X = {1,2,3} and Y = {1,a,b,c}, therefore, the universal set will be U = {1,2,3,a,b,c}.
In the above scenario, H(...-2,-10,1,2,...3) is a Universal set.
For the fraction 28/10,
times 10 at 28 and 10,(to get the same denominator"100")
28×10=280
10×10=100
Thus,
28/10 become 280/100
We can plus the number above the denominator"100",which is 39 and 280.It is because they have achieved a same denominator.
39/100+280/100
=(39+280)/100
=319/100
Answer:
<h2>He will use flowers per vase</h2><h2>but will have one left</h2>
Step-by-step explanation:
In this problem we are required to place flowers evenly across 3 flower vases.
Given that we have 25 flower to collect from and distribute evenly.
we are going to divide 25 by 3
We have

<u><em>From the result in the improper fraction we can see the Darren can place 8 flowers evenly of the 25 flowers and there will be just one left over</em></u>
Answer:
CB = 5, 
Step-by-step explanation:
Assuming that the marks next to angle D mean that
, then there is enough information to conclude that the two right triangles are congruent (HA theorem--hypotenuse/angle).
Then all the corresponding parts of the two triangles are congruent, including segments BA and BC, making them the same length.
4x + 1 = 9x - 4
Solving for x,
1 = 5x - 4
5 = 5x
1 = x
Pluggint x = 1 into the two expressions reveals that the length of segment CB is 5.
The two small angles at D are congruent, so
.