Answer:
Condition A.
A rectangle with four right angles
There can be many quadrilaterals satisfying this condition.
Condition B.
A square with one side measuring 5 inches
There can be only one quadrilateral satisfying this condition.
Condition C.
A rhombus with one angle measuring 43°
There can be many quadrilaterals satisfying this condition.
Condition D.
A parallelogram with one angle measuring 32°
There can be many quadrilaterals satisfying this condition.
Condition E.
A parallelogram with one angle measuring 48° and adjacent sides measuring 6 inches and 8 inches.
There can be only one quadrilateral satisfying this condition.
Condition F.
A rectangle with adjacent sides measuring 4 inches and 3 inches.
There can be only one quadrilateral satisfying this condition
Step-by-step explanation:
<u>Answer:</u>
It's X=2
<u>Working out:</u>
<u>(left one):</u>
2(x+3)=10
2x+6=10
2x=10-6
2x=4
x=2
Working out:
(right one):
2(x+3) must be all divided by 2 so it will be x+3 and same for 10, it will be 5 so...
x+3=5
x=5-3
x=2
Hope it helps ^---^
F(x)=0
f(x)=y
so where y=0, the x value is the x in f(x)
basically, where the line crosses or touches the x axis, those ar the x values that make the function true
As is the case for any polynomial, the domain of this one is (-infinity, +infinity).
To find the range, we need to determine the minimum value that f(x) can have. The coefficients here are a=2, b=6 and c = 2,
The x-coordinate of the vertex is x = -b/(2a), which here is x = -6/4 = -3/2.
Evaluate the function at x = 3/2 to find the y-coordinate of the vertex, which is also the smallest value the function can take on. That happens to be y = -5/2, so the range is [-5/2, infinity).
Answer: 12
Step-by-step explanation:
Given:
A = 452
r = √A/3
Step 1: Substitute 452sqf for A
r = √452/3
Step 2: Solve
452 ÷ 3 = 150.66∞
√150.66∞ = 12.27∞
12.27∞ rounded to the nearest integer equal 12
