Answer:
Area of parallelogram is equal to 
Step-by-step explanation:
It is given length pf parallelogram which is equal to base of parallelogram is b=15 cm
Height of parallelogram h = 7.5 cm
We have to find the area of the parallelogram.
Area of parallelogram is equal to multiplication of base and height.
Therefore area of parallelogram is,
, here b is base and h is height.
So 
Therefore area of parallelogram is equal to 
Let the coordinates of the treasure be (x,y).
Let

= distance from the rock to the treasure.
Let

= distance from the tree to the treasure.


We want

= 5/9 in order to locate the treasure.
Note that 5/9 = 0.556 (approx)
Test (11.4, 14.2)
d1 = 14.8, d2 = 8.2, d1/d2 = 1.8 Incorrect
Test (7.6, 8.8)
d1 = 8.2, d2 = 14.8, d1/d2 = 0.554 Correct
Test (5.7, 7.5)
d1 = 6.13, d2 = 16.98, d1/d2 = 0.36 Incorrect
Test (10.2, 12.6)
d1 = 12.8, d2 = 10.2, d1/d2 = 1.255 Incorrect
Answer:
The correct answer is (7.6, 8.8)
The altitude's slope must be perpendicular to that of the line.
The slope of segment BC is 2/5. The slope of the altitude must be perpendicular to 2/5. Since the negative reciprocal of 2/5 is -5/2, then the slope of the altitude is -5/2.
Answer:
w= 9
Step-by-step explanation:

Square both sides:
-4w +61= (w -4)²

Expand:
-4w +61= w² -2(w)(4) +4²
-4w +61= w² -8w +16
Simplify:
w² -8w +16 +4w -61= 0
w² -4w -45= 0
Factorize:
(w -9)(w +5)= 0
w -9= 0 or w +5= 0
w= 9 or w= -5 (reject)
Note:
-5 is rejected since we are only taking the positive value of the square root here. If the negative square root is taken into consideration, then w= -5 would give us -9 on both sides of the equation.
<u>Why </u><u>do </u><u>we </u><u>use </u><u>negative </u><u>square </u><u>root?</u>
When solving an equation such as x²= 4,
we have to consider than squaring any number removes the negative sign i.e., the result of a squared number is always positive.
In the case of x²= 4, x can be 2 or -2. Thus, whenever we introduce a square root, a '±' must be used.
However, back to our question, we did not introduce the square root so we should not consider the negative square root value.