Answer:
One Triangle = 2.09 in²
Two Triangles = 4.18 in²
Rectangle = 17.48 in²
Total area of whole trapezoid = 21.66 in²
Step-by-step explanation:
Since it was not clarified which region is shaded we will just find the area of each individual part of the shape.
Let's start with the triangles.
1. To find the area of a triangle, the formula is 
. It is given that the base of one triangle is equal to 1.1 in and the height is equal to 3.8 in., so in the equation, it would look like:
in²
2. So now that we know one triangle is equal to 2.09 in², we now know that the other triangle is equal to the same area. To find the total of the two triangles you need to multiply the area by 2:
in²
Moving on to the rectangle...
1. To find the area of the rectangle we need to use the formula base times height or b x h. It is given that the height is 3.8 in while the length is 4.6 in. So in the equation it would look like:
in²
Now to find the total area of all shapes combined...
1. To do this, we just need to add up all the areas we found, so...
17.48 + 4.18 = 21.66 in²
Answer:
Ryan could have answered multiple choice answers and true or false answers rather than knowledge answer like written prompts
Step-by-step explanation:
Hope this helped:)
Answer:
Step-by-step explanation:
(2x + 1)(ax + b) = 6x² - 5x + c
Expand the left side and put that half into standard form:
2ax² + 2bx + ax + b = 6x² - 5x + c
2ax² + (2b + a)x + b = 6x² - 5x + c
So now look at the x terms. We have "2a" x² on left and 6 x² on right, so:
2a = 6
a = 3
Then look at the x terms, those coefficients will then be equal:
2b + a = -5
2b + 3 = -5
2b = -8
b = -4
and finally, then non-x terms are equal:
b = c
-4 = c
so:
a = 3, b = -4, c = -4
Answer:
11 + (5 + 2)² = 11 + 7²
= 11 + 49
= 60
Step-by-step explanation:
Star with the brackets.
5+2=7
7² = 49
11 + 49 = 60
Answer:
ICm= international federation midwives
Step-by-step explanation:
there examples are:
i) Representing midwifery as sociation for works supports
ii) strengthen professional associations of midwives on a global basis.
