Options:
a. Nikki's proposed placement will light a greater area than Dylan's placement.
b. Dylan's proposed placement will light a greater area than Nikki's placement.
c. Both proposed placements will light the same sized area.
d. Nikki's proposed placement will light more than half the yard.
e. Dylan's proposed placement will light more than half the yard.
f. Dylan's proposed placement will light exactly half of the yard.
g. Nikki's proposed placement will light less than half of the yard.
Answer:
C) Both proposed placements will light the same sized area.
F) Dylan's proposed placement will light exactly half of the yard.
Step-by-step explanation:
The area of a triangle is (base x height) / 2, and both lights illuminate the same base and height = (60 x 38) / 2 = 1,140 sq ft
Both Dylan's and Nikki's proposed placement will lit exactly half of the yard. The yard's total area = 60 x 38 = 2,280 sq ft, which is twice the area lit by the lights.
·_21.21
25l546. 25 goes into 54 twice with a remainder of 4
l50↓
-----
l 46 25 goes into 46 once with 21 left
l 25
------
21
60% if there are only 10 white and black balls
<span>First we need to find the GCF (greatest common factor)</span>
<span>Factors of 42 are:
1, 2, 3, 6, 7, 14, 21, 42</span>
<span>Factors of 48 are:
1, 2, 3, 4, 6, 8, 12, 16, 24, 48</span>
<span>GCF is 6</span>
<span>To simplify you divide numerator and denominator into the GCF</span>
<span>42 ÷ 6 = 7</span>
<span>48 ÷ 6 = 8</span>
<span>42/48 = 7/8</span>
<span>Hope this helps. :)
</span>
The cost of children’s ticket is $ 5
<h3><u>Solution:</u></h3>
Let "c" be the cost of one children ticket
Let "a" be the cost of one adult ticket
Given that adult ticket to a museum costs 3$ more than a children’s ticket
<em>Cost of one adult ticket = 3 + cost of one children ticket</em>
a = 3 + c ------ eqn 1
<em><u>Given that 200 adult tickets and 100 children tickets are sold, the total revenue is $2100</u></em>
200 adult tickets x cost of one adult ticket + 100 children tickets x cost of one children ticket = 2100
200a + 100c = 2100 ------ eqn 2
<em><u>Let us solve eqn 1 and eqn 2 to find values of "a" and "c"</u></em>
Substitute eqn 1 in eqn 2
200(3 + c) + 100c = 2100
600 + 200c + 100c = 2100
600 + 300c = 2100
300c = 1500
<h3>c = 5</h3>
Thus the cost of children’s ticket is $ 5
