Answer:
We can use linear system of equation with one variable.
Step-by-step explanation:
It is given that,Nina is 10 years younger than Deepak.
Deepak is 3 times as old as Nina.
<u>To solve this using linear equation</u>
Let d be the age of Deepak and n be the age of NIna
we can write,
n = (d - 10) ---(1)
d = 3n
d = 3(d - 10) --------(2)
d = 3d - 30
3d - d = 30
2d = 30
d = 30/2 = 15
n - d - 10 = 15 - 10 = 5
Therefore the age of Deepak = 15 and age of Nina = 5
Answer:
<h2><u>Solution</u><u> </u><u>:</u><u>-</u></h2>
We know that
V = πr²h
400 = 22/7 × r² × 8
400 × 7 = 22 × 7r²
2800 = 154r²
2800/154 = r²
18 ≈ r²
4.2 = r
Find what number it is on the bottom to make the point
1. M is the midpoint of LN and O is the midpoint of NP. This makes the triangle MNO equal to half of LNP. Then you can get this equation
MO= (1/2) LP
If you insert MO = 2x +6 and LP = 8x – 20 the calculation would be:
2x+6= (1/2)( 8x-20)
2x+6= 4x-10
2x-4x= -10 - 6
-2x= -16
x=8
2. Centroid is the point that intersects with three median lines of the triangle. The centroid should divide the median lines into 1:2 ratio. In AC lines, A located in the base so A.F:FC would be 1:2
Then, the answer would be:
A.F= 1/(1+2) * AC
A.F= 1/3 * 12= 4
FC= 2/(1+2) * AC
FC= 2/3 * 12= 8
3. Since
∠BAD=∠DAC
∠ABD=∠ACD
AD=AD
The triangle ABD and ACD are similar. You can get this equation
BD=DC
x+8= 3x+12
x-3x= 12-8
-2x=4
x=-2
DC=3x+12= 3(-2) +12= 6
4. Orthocenter made by intersection of triangle altitude
A
BC lines slope would be (-4)-(-1)/1-4= -3/-3= 1. The altitude line slope would be -1, the function would be:
y=-x +a
0= 1+a
a=-1
y=-x-1
B
AC lines slope would be (-4)-(-1)/1-0= -3. The altitude line slope would be 1/3, the function would be:
y=1/3x+a
-1=1/3(4)+a
a=-7/3
y=1/3x - 7/3
C
BC lines slope would be (-1)-(-1)/4 = 0/4.
The line would be
0=x+a
a=-1
0=x-1
x=1
y=-x-1 = 1/3x-7/3
-x-(1/3x)=-7/3 +1
-4/3x= -4/3
x=1
y=-x-1
y=-1-1= -2
The orthocenter would be (1,-2)
5.
a. Circumcenter: the intersection of perpendicular bisector lines<span>
b. Incenter: the intersection of bisector lines
c. Centroid: </span>the intersection of median lines<span>
d. Orthocenter: </span>the intersection of altitude lines
