Part 1)
area of triangle=b*h/2
b=2.9 cm
h=9 cm
so
area=2.9*9/2-----> 13.05 cm²
the answer part 1) is 13.05 cm²
Part 2)
area of the figure=area of rectangle+area of triangle
find the area of rectangle
area rectangle=b*h
b=12 ft
h=?
tan 45=1
tan 45=h/(22-10)----------> h/10=1-------> h=10 ft
area of rectangle=12*10-----> 120 ft²
area of triangle=b*h/2
b=10 ft
h=10 ft
area of triangle=10*10/2----> 50 ft²
area of the figure=120+50----> 170 ft²
the answer Part 2) is 170 ft²
Part 3)
<span>the area of kite is half the product of the diagonals.
</span>Area=(d1*d2)/2
d1=5+5---> 10 ft
d2=16+8----> 24 ft
area=(10*24)/2----> 120 ft²
the answer Part 3) is 120 ft²
Part 4)
the area of rhombus is half the product of the diagonals.
Area=(d1*d2)/2
d1=6+6---> 12 m
d2=6+6----> 12 m
area=(12*12)/2----> 72 m²
the answer part 4) is 72 m²
Part 5)
area of the figure=6*area of one triangle
area of triangle=b*h/2
b=4 cm
is an equilateral triangle
applying the Pythagoras theorem
h²=4²-2²-----> h²=12-----> h=2√3 cm
area of triangle=4*2√3/2----> 4√3 cm²
area of hexagon=6*(4√3)----> 24√3 cm²
the answer Part 5) is 24√3 cm²
Answer:
Hello!
First you will want to get both of equations into slope-intercept form. That way, you can easily determine the y-intercept of the first equation and the slope of the second equation.
When 2 lines have the same slope, that means they will be parallel.
y+4x-10=0 ; add 10 to each side
y+4x=10 ; subtract 4x from each side
y = -4x + 10 ; y-intercept = 10
1x + 7y = 6 ; subtract 1x from each side
7y = -1x + 6 ; divide both sides by 7
y = -1/7 x + 6/7 ; slope = -1/7
So, a line with the same y-intercept as the first, and same slope (parallel) as the second would have the equation:
y = -1/7x + 10
Step-by-step explanation:
2 because you have to divide
Given f (x) = 2x, g(x) = x + 4, and h(x) = 5 – x3, find (f + g)(2), (h – g)(2), (f × h)(2),
