All Triangles are equilateral ( each side = 4)
Area of a Triangle is (B x H)/2. B being the base ( 4) & H the altitude.
WE know that altitude in an equilateral triangle, H = ( side x√3)/2
So H= (4√3)/2 = 2√3
The area of one triangle = (4 x 2√3)/2 = 4√3
And for the 3 triangles the lateral area = (4√3) x 3 = 12√3
Answer:
Step-by-step explanation:
Start from left one first
12x & 70
Area as Sum= 12x+70
Area as Product = 12(x+5)
Right side
X & 4
Area as Sum= 5X+20
Area as Product = 5(X+4)
Answer:
The slope intercept form of both given equations is : y = - 3 x - 4.
Step-by-step explanation:
Here, the given equations are:
y +7 = -3 ( x - 1 )
and 3 x + y = - 4
Now,the SLOPE INTERCEPT FORM of any given equation is given as:
y = m x + C : here, C = Y - intercept, m = Slope
Consider equation (1):
y +7 = -3 ( x - 1 ) ⇒ y + 8 = - 3 x + 3
or, y = -3x + 3 - 7 = -3x - 4
⇒ y = -3x -4
Hence, the slope-intercept form of the given equation is y = -3x -4.
Consider equation (2):
3 x + y = - 4 ⇒ y = -4 - 3 x
⇒ y = -3 x - 4
Hence, the slope-intercept form of the given equation is y = -3x -4.
Answer:
a) Average velocity at 0.1 s is 696 ft/s.
b) Average velocity at 0.01 s is 7536 ft/s.
c) Average velocity at 0.001 s is 75936 ft/s.
Step-by-step explanation:
Given : If a ball is thrown straight up into the air with an initial velocity of 70 ft/s, its height in feet after t seconds is given by 
.
To find : The average velocity for the time period beginning when t = 2 and lasting. a. 0.1 s.
, b. 0.01 s.
, c. 0.001 s.
Solution :
a) The average velocity for the time period beginning when t = 2 and lasting 0.1 s.
b) The average velocity for the time period beginning when t = 2 and lasting 0.01 s.
c) The average velocity for the time period beginning when t = 2 and lasting 0.001 s.
Answer:
fac.k nou ok your mather xd
Step-by-step explanation:
