Let
x-------------> length of the side of a square
[area of a square]=x*x------> x²
[area of rectangle]=[shorter side]*[<span>longer side]
</span>[shorter side]=0.90 x-------> (<span>is 10% percent less than the length of a side of square)
</span>[longer side]=1.10x-------> (<span>is 10% percent more than the length of a side of square
</span>
[area of rectangle]=(1.10x)*(0.90x)-----> 0.99x²
[the ratio of the area of rectangle to the area of square]=0.99x²/x²=0.99
the answer is
the ratio of the area of rectangle to the area of square is 0.99
Answer:
-21 = 6t
Step-by-step explanation:
3t - 7 = 5t
Subtract 3t from each side
3t-7-3t = 5t-3t
-7 = 2t
Multiply each side by 3 to get 6t
-7*3 = 2t*3
-21 = 6t
It’s d times it three times length width height
Answer:

Step-by-step explanation:
step 1
Find the measure of the arc DC
we know that
The inscribed angle measures half of the arc comprising
![m\angle DBC=\frac{1}{2}[arc\ DC]](https://tex.z-dn.net/?f=m%5Cangle%20DBC%3D%5Cfrac%7B1%7D%7B2%7D%5Barc%5C%20DC%5D)
substitute the values
![60\°=\frac{1}{2}[arc\ DC]](https://tex.z-dn.net/?f=60%5C%C2%B0%3D%5Cfrac%7B1%7D%7B2%7D%5Barc%5C%20DC%5D)


step 2
Find the measure of arc BC
we know that
----> because the diameter BD divide the circle into two equal parts
step 3
Find the measure of angle BDC
we know that
The inscribed angle measures half of the arc comprising
![m\angle BDC=\frac{1}{2}[arc\ BC]](https://tex.z-dn.net/?f=m%5Cangle%20BDC%3D%5Cfrac%7B1%7D%7B2%7D%5Barc%5C%20BC%5D)
substitute the values
![m\angle BDC=\frac{1}{2}[60\°]](https://tex.z-dn.net/?f=m%5Cangle%20BDC%3D%5Cfrac%7B1%7D%7B2%7D%5B60%5C%C2%B0%5D)

therefore
The triangle DBC is a right triangle ---> 60°-30°-90°
step 4
Find the measure of BC
we know that
In the right triangle DBC


substitute the values
