Answer:
Therefore 'x' is equal to 65.4°
Step-by-step explanation:
In Right Angle Triangle ABC
∠ B = 90°
AC = Hypotenuse = 12
CB = Adjacent Side = 5
To Find:
∠ C = x
Solution:
In Right Angle Triangle ABC Cosine Identity we have
Substituting the values we get
Therefore 'x' is equal to 65.4°
Answer:
B :) (im sure btw)
Step-by-step explanation:
4). x/5 =8
x = 40
5). 2x/3 = 12
2x = 36
x = 36/2 = 18
6). x + 26 =12
x = 12 - 26 = -14
7). 8x + 1 = -23
8x = -23 -1
8x = -24
x = -24/8 = -3
8). 2x + 5 =15
2x = 15 - 5
2x = 10
x = 10/2 = 5
9). 3x - 1 = -29
3x = -29 + 1
3x = -28
x = -28/3 Or 9.33
10). 8x - 12 = 53
8x = 53 + 12
8x = 65
x = 65/8 Or 8.12
11). 4x + 7 = 25
4x = 25 - 7
4x = 18
x =18/4 = 9/2 Or 4.5
<u>Correct </u><u>Inputs </u><u>:-</u>
In ΔABC right angled at A, D and E are points on BC, C such that BD = CD and AD ⊥ BC
Let us know about definition of altitude first. The altitude of a triangle is the perpendicular line segment drawn from the vertex to the opposite side of the triangle.
Median is the line segment from a vertex to the midpoint of the opposite side.
<u>Let us Check all options one by one </u>
- CD is line segment which starts from vertex C but don't falls on opposite side AB thus it is not an altitude.❌
- BA is line segment which starts from vertex B and falls perpendicularly on opposite sides AC and is thus an altitude.✔️
- AD is line segment which starts from vertex A and falls perpendicularly on opposite side BC and is thus an altitude.✔️
- AE is a line segment which starts from vertex A but doesn't falls perpendicularly on opposite side BC and is thus not an altitude.❌
- AD falls on BC with D as mid point because BD = CD and is thus a median. ✔️
Try not to be bored. You can take the attitude that you actually like math, and that you will some day actually be a MathGenius.
t represents an unknown quantity defined by the equation 16 - 2t = 5t + 9.
To find this value, do the following: Add 2t to both sides, and then subtract 9 from both sides. You'll end up with 7t = 7, which yields t = 1.
We call this the "solution" of the equation given.
