Answer:
72 sq. mi
Step-by-step explanation:
Breaking this down, we have 2 right triangles with sides of 3, 4, and 5 miles, and 3 rectangles with dimensions 3 x 5, 4 x 5, and 5 x 5 miles. Remember that the area of a triangle is 1/2 x b x h , where b and h are the triangle's base and height. The base and height of the triangles at the bases of the figure are 3 and 4, so each triangle has an area of 1/2 x 3 x 4 = 1/2 x 12 = 6 sq. mi, or 6 + 6 = 12 sq. mi together.
Onto the rectangles, we can find their area by multiplying their length by their width. Since the width of these rectangles is the same for all three - 5 mi - we can make our lives a little easier and just "glue" the lengths together, giving us a longer rectangle with a length of 3 + 4 + 5 = 12 mi. Multiplying the two, we find the area of the rectangles to be 5 x 12 = 60 sq. mi.
Adding this area to the triangle's area gives us a total area of 12 + 60 = 72 sq. mi.
Make 2 eqn, let a be apples and o be oranges:
1. 5a + 4o = 10
2. 5a + 5o = 11
set up an elimination:
1. minus 2.
5a + 4o = 10
- 5a + 5o = 11
____________
o = 1
sub “o” into one of the eqn
5a + 5(1) = 11
5a = 11-5
5a = 6
a = 6/5
a = 1.2
therefore, an orange costs $1.00 and an apple costs $1.20
Answer:
BD = 3 units
Step-by-step explanation:
Since, AD is an angle bisector of ∠BAC,
m∠BAD = m∠CAD = 20°
CD = 3 units
In ΔACD and ΔABD,
m∠BAD = m∠CAD = 20° (Given)
AD ≅ AD [Reflexive property]
Therefore, by H-A property of congruence both the triangles will be congruent.
And by CPCTC,
CD ≅ BD = 3 units
Answer:
30
Step-by-step explanation:
we multiply each number. There are 3 pairs of pants 5 shirts and 2 pairs of shoes, so we multiply 3x5x2 to get 30
She should pick out this outfit...
Answer:
a= 200
b = 210
Step-by-step explanation:
My assumption is, we have to find the length of sides of rectangle
Given
perimeter = 2a + 2b = 820 ft (i) (here a is smaller side and b is larger side)
area = a*b = 42,000 ft^2 (ii)
from eq (1)
2a + 2b = 820
=> 2(a+b) = 820
=> a+b = 820/2
=> a + b = 410
=> a = 410-b (iii)
putting the value of a in eq(ii), we get
(410-b) *b = 42,000
410b - b^2 = 42,000
0 = b^2 - 410b + 42000
b^2 - 410b + 42000 = 0
b^2- 200b- 210b + 42000 = 0
b(b-200)-210(b-200) = 0
(b-200)(b-210) = 0
or
b= 210 and b = 200
if b is larger side than b =210
By putting value of b in eq(iii),
a = 410 -210 = 200
