Answer:
90 sq inches
Step-by-step explanation:
The lateral area is the side area of the pyramid. In this case, there are 3 sides because the base is a triangle. Since the base is an equilateral triangle, all sides are of equal dimensions. So, the sides of the pyramid are also all of the same size, area.
We then just have to figure out the area of one of the side triangles, then multiply it by 3 to get the total lateral area of that pyramid.
We know how to calculate the area of a triangle: A = (b * h)/2
In this case, b = 5 in, h = 12 in. So,
TA = (5 * 12)/2 = 60 / 2 = 30
Each side triangle has an area of 30 sq inches.
LA = 3 * TA (since there are 3 sides)
LA = 3 * 30 = 90 sq inches
Gradient is the slope, which is the letter m in the equation.
The point is written as (x1, y1) so replace x1 and y1 in the equation with the numbers in the given point.
A would be y - 1= 2(x-4)
B: y +5 = 5(x-2)
C: y-1= -3(x+1)
D: y -6 = 1/2(x-1)
E: y +1/4 = -2(x-3:4)
F: y+7 = -1/5(x+ 3)
if x is changed by 2 , the y changes by +3, Option B is the correct answer.
<h3>What is the equation of a Linear Function ?</h3>
The equation of a Linear Function is given by
y = mx +c
here m is the slope , c is the intercept on the y axis
The data is given in the table
To determine the relation , a function needs to be established between the variables.
m = ( y2 -y1 )/(x2-x1)
m = (-7 +10) /(-2+4)
m = 3 / 2
y = (3/2)x + c
at x = 0 , y= -4
-4 = 0 + c
c = -4
The function is
y = (3/2) x - 4
if x is changed by 2 , the y changes by +3
Therefore Option B is the correct answer.
To know more about Linear Function
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Answthe 2 at the bottom go to Not a polynomial
Step-by-step explanation:
Answer:
{-11, -9, -7, -5, -3}
Step-by-step explanation:
Put each domain value into the function to find the corresponding range value. The range is the list of all of those values.
__
range = f(domain)
= 2{-2, -1, 0, 1, 2} -7 = {-4, -2, 0, 2, 4} -7
= {-11, -9, -7, -5, -3} . . . . . the range for the given domain
_____
If you have a lot of function values to find, a spreadsheet or calculator can be helpful.
