<h3>
Answer: Midpoint</h3>
Explanation:
The drawings you made are constructions of the perpendicular bisector. The "perpendicular" refers to the fact the second line meets the first line at a right angle (aka 90 degree angle). The "bisector" portion indicates the segment has been bisected, which is mathematical way of saying "cut in half".
The key here is the "bisector" portion. Because we've cut a segment like AB in half, this means segment AC and CB are the same length. The point C is the midpoint of AB. Point C is formed by intersecting the perpendicular bisector and the original segment. I'm referring to drawing (a), but the same idea applies to drawing (b) as well. I recommend using another letter than C for the second drawing.
Answer:
8
Step-by-step explanation:
we know that
If two figures are similar, then the ratio of its volumes is equal to the scale factor elevated to the cube
Remember that all the spheres are similar
Let
z ----> the scale factor
V_1 ---> the volume of the large sphere
V_2 ---> the volume of the small sphere
<em>Find the scale factor</em>
we know that
The radius of the large sphere is double the radius of the small sphere
so
The scale factor is 2
substitute the value oz z
The volume of the large sphere is 8 times the volume of the small sphere
Hola. La derivada de 2x =2 , 8x = 8 , 10x² = 20x. Pienso que te ayudar por este exercicio.
This is overdetermined. Without looking at the ratio (except for the order of the sides) the actute angles in a right triangle are complementary so the answer must be:
Answer: A=35°, C=90°, B=55°
We can do it the way they like.
7+18+11 =36
so to get 180 degrees we need to scale the ratio elements by a factor of 5.
7:18:11=35:90:55
A is the smallest. We see C is indeed a right angle and B is complementary to A.
It would be written as:
5.65 *10^-4
Remember, when writing scientific notation for a decimal, the power of 10 is always negative. In this case, since we're basically moving the decimal point for places to the right, the power of 10 is -4
Hope this helped! :)
