Answer:
i agree with the other person who answers this question
D=rt for t
÷r both sides
d/r=t
A
5-y/2 = 9
First, we do the inverse operation and subtract 5 from both sides.
5 - y/2 = 9
-5 -5
-y/2 = 4
Now, we do the inverse operation ( multiply by the reciprocal)
-y/2 = 4
-y/2 * -2 = 4* -2
y= -8
y=-8
Now we have our final answer: y=-8
Check our work
5 - - 8/2= 9
5+4=9
9=9
OUR ANSWER IS CORRECT! :)
The area, in square inches, outside the smaller region, but inside the larger region is 99π
<h3>How to determine the area, in square inches, outside the smaller region, but inside the larger region?</h3>
The given parameters are:
Radius, r1 = 1 inch
Radius, r2 = 10 inches
The area, in square inches, outside the smaller region, but inside the larger region is calculated as
Area = π(R^2 - r^2)
This gives
Area = π(10^2 - 1^2)
Evaluate the difference
Area = 99π
Hence, the area, in square inches, outside the smaller region, but inside the larger region is 99π
Read more about area at:
brainly.com/question/17335144
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The description below proves that the perpendicular drawn from the vertex angle to the base bisect the vertex angle and base.
<h3>How to prove an Isosceles Triangle?</h3>
Let ABC be an isosceles triangle such that AB = AC.
Let AD be the bisector of ∠A.
We want to prove that BD=DC
In △ABD & △ACD
AB = AC(Thus, △ABC is an isosceles triangle)
∠BAD =∠CAD(Because AD is the bisector of ∠A)
AD = AD(Common sides)
By SAS Congruency, we have;
△ABD ≅ △ACD
By corresponding parts of congruent triangles, we can say that; BD=DC
Thus, this proves that the perpendicular drawn from the vertex angle to the base bisect the vertex angle and base.
Read more about Isosceles Triangle at; brainly.com/question/1475130
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