2/11 because there is 11 letters which would be the denominator and then there is 2 I's which would be the numerator
6/10
300/500
30/50
60/100
12/20
Answer:
x = 15
Step-by-step explanation:
the easiest way to do this is 5 times 3 and with that we get 15 :)
Have an amazing day!!
Please rate and mark brainliest!!
Answer:
-8 + x ≥ 9
Step-by-step explanation:
Let's look at your sentence:
The <u>sum of -8 and a number</u> is <u>at least 9 </u>what is the equation look like
Sum is the result of adding
At least 9 means that the number is 9 or greater.
So what you have here is an inequality expression:
Sum of -8 and a number
-8 + x
x here is the "number" since we do not know the value we will use "x" to represent that number.
At least 9
At least means that it is equal or greater, so we use this expression ≥
and so we use put that all together
-8 + x ≥ 9
Answer:
At first, we divide the parallelogram into two triangles by joining any two opposite vertices. These two triangles are exactly the same (congruent) and thus have equal areas. The area of the parallelogram is the summation of the individual areas of the two triangles. We drop a perpendicular from a vertex to its opposite side to get an expression for the height of the triangles. The area of the individual triangle is 12×base×height12×base×height .The area of the parallelogram being twice the area of the triangle, thus becomes after evaluation base×heightbase×height .
Complete step by step answer:
The parallelogram can be divided into two triangles by constructing a diagonal by joining any two opposite vertices.


In the above figure, ΔABDΔABD and ΔBCDΔBCDare the two such triangles. These two triangles have:
AB=CDAB=CD (as opposite sides of a parallelogram are equal)
AD=BCAD=BC (opposite sides of a parallelogram are equal)
BDBD is common
Thus, the two triangles are congruent to each other by SSS axiom of congruence. Since, the areas of two congruent triangles are equal,
⇒area(ΔABD)=area(ΔBCD)⇒area(ΔABD)=area(ΔBCD)
Now, we need to find the area of ΔABDΔABD . We draw a perpendicular from DD to the side ABAB and name it as DEDE . Thus, ΔABDΔABD is now a triangle with base ABAB and height DEDE .
Then, the area of the ΔABD
