Answer:
9.6 square inches.
Step-by-step explanation:
We are given that ΔBAC is similar to ΔEDF, and that the area of ΔBAC is 15 inches. And we want to determine the area of ΔDEF.
First, find the scale factor <em>k</em> from ΔBAC to ΔDEF:
Solve for the scale factor <em>k: </em>
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Recall that to scale areas, we square the scale factor.
In other words, since the scale factor for sides from ΔBAC to ΔDEF is 4/5, the scale factor for its area will be (4/5)² or 16/25.
Hence, the area of ΔEDF is:
In conclusion, the area of ΔEDF is 9.6 square inches.
67 because 24 isnot 50 yet so you would keep 67 how it is but if it was 50 or more then you would round it up to 68
You can start with the form
∆y(x -x1) -∆x(y -y1) = 0
Here, we have
∆y = 11-(-3) = 14
∆x = -3-1 = -4
and we can choose (x1, y1) = (1, -3). This gives
14(x -1) -(-4)(y -(-3)) = 0
14x +4y -2 = 0
All these terms have a common factor of 2 that we can remove. Adding 1 to the result puts it in standard form:
7x +2y = 1
The image is missing so i have attached it.
Answer:
Volume = 1.5 litres
Step-by-step explanation:
Using pythagoras theorem, we can get the height (h) of the cylinder
14² + h² = 17²
h² = 289 - 196
h = √93
Now, volume of a cylinder is;
V = πr²h
In the image, r = diameter/2 = 14/2 = 7cm
Thus,
V = π × 7² × √93
V = 1485 cm³
Now, 1 litre = 1000 cm³
Thus, volume = 1485/1000 = 1.485 litres ≈ 1.5 litres
