Answer:
AH = 1 or 4
CH = 4 or 1
Step-by-step explanation:
An altitude divides a right triangle into similar triangles. That means the sides are in proportion, so ...
AH/BH = BH/CH
AH·CH = BH²
The problem statement tells us AH + CH = AC = 5, so we can write
AH·(5 -AH) = BH²
AH·(5 -AH) = 2² = 4
This gives us the quadratic ...
AH² -5AH +4 = 0 . . . . in standard form
(AH -4)(AH -1) = 0 . . . . factored
This equation has solutions AH = 1 or 4, the values of AH that make the factors be zero. Then CH = 5-AH = 4 or 1.
1-(28-21)/28=.075
Answer is D
Answer:
equation is: width = area / length
width = 62.5 yd
Explanation:
For the first rectangle:
We have:
length = 75 yd and width = 60 yd
Therefore:
area = length * width
area = 75 * 60
area = 4500 yd²
For the second rectangle:
We have:
area = area of first rectangle = 4500 yd²
length = 72 yd
area = length * width
width = area / length ............> The required equation
width = 4500 / 72
width = 62.5 yd
Hope this helps :)
Answer:
x=6 . m<PQS=82 m<SQR=61 :)
Step-by-step explanation:
(13x+4) + (10x-1) = 141
combine like terms
23x+3=141
subtract 3 from both sides
23x=138
divide both sides by 23
x=6
substitute x into both original equations
m<PQS=13(6)+4
m<PQS=78+4
m<PQS= 82
m<SQR=10(6)+1
m<SQR=60+1
M<SQR=61
