Answer:
Option (C)
Step-by-step explanation:
From the graph attached,
In ΔABC and ΔA'B'C',
∠A ≅ ∠A'
∠B ≅ ∠B'
Therefore, ΔABC ~ ΔA'B'C'
Corresponding sides of these triangles will be proportional.
Therefore, ratio of the sides, AC : A'C' = 1 : 3 shows that image triangle A'B'C' is a dilated form of pre-image ABC with a scale factor of 3.
Option (C) will be the correct option.
Answer:
Option (B)
Step-by-step explanation:
From the picture attached,
NL and KM are the two lines intersecting at a point P.
Therefore, ∠KPN ≅ LPM [Vertical angles]
In ΔPLM,
m∠LPM + m∠PML + m∠PLM = 180° [Property of a triangle]
m∠LPM + 70° + 60° = 180°
m∠LPM + 130° = 180°
m∠LPM = 180° - 130°
m∠LPM = 50°
Therefore, m∠KPN = 50° [vertical angles]
Option (B) will be the correct option.
wheres the picture
Step-by-step explanation:
<h3>given:</h3>
<h3>to find:</h3>
the radius of the given ball (sphere).
<h3>solution:</h3>
![r = \sqrt[3]{ \frac{3v}{4\pi} }](https://tex.z-dn.net/?f=r%20%3D%20%20%5Csqrt%5B3%5D%7B%20%5Cfrac%7B3v%7D%7B4%5Cpi%7D%20%7D%20)
![r = \sqrt[3]{ \frac{3 \times 905}{4\pi} }](https://tex.z-dn.net/?f=r%20%3D%20%20%5Csqrt%5B3%5D%7B%20%5Cfrac%7B3%20%5Ctimes%20905%7D%7B4%5Cpi%7D%20%7D%20)
<u>therefore</u><u>,</u><u> </u><u>the</u><u> </u><u>radius</u><u> </u><u>of</u><u> </u><u>the</u><u> </u><u>ball</u><u> </u><u>is</u><u> </u><u>6</u><u> </u><u>cm</u><u>.</u>
note: refer to the picture I added on how you can change r as the subject of the formula.
Answer:
Whatever option has a slope of 3/4 and isn't y = 3/4x + 2
Step-by-step explanation:
This means that it has to have a slope of 3/4 (written as 3/4x + b). This is because parallel lines have the same slope; rearranging the first equation gets us y = 3/4x + 2. So, our slope for a parallel line has to be 3/4.
