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adell [148]
3 years ago
7

Given: line DR tangent to Circle O. If m angle C = 63°, then m angle BDR =

Mathematics
2 answers:
myrzilka [38]3 years ago
6 0
Based on the given figure above, we can say that the measure of angle BDR is also 63°. R is meant to be tangent to the circle so, this makes the angle BDR equal to C. Angle C is half of the arc as well as BDR. Hope this is the answer that you are looking for. 
emmasim [6.3K]3 years ago
5 0

Answer:

m∠BDR=63°  

Step-by-step explanation:

It is given that  line DR tangent to Circle O and m∠BCD=63°, then the measure of the angle BDR can be determined as:

Since, angle BDR is made between the chord and the tangent, and we know that the angle between a chord and a tangent through one of the end points of the chord is equal to the angle in the alternate segment, therefore

m∠BCD=m∠BDR=63°

Hence, the measure of angle BDR=63°.

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the zero of the quadratic polynomial x^2+99x+127 are a both positive b both negative c one +ve one-ve d both equal. please help​
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Step-by-step explanation:

By Descrates' Rule of Signs,

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What is the side length, in inches, of the pets
tatiyna

Candy draws a square design with a side length of x inches for the window at the pet shop. She takes the design to the printer and asks for a sign that has an area of 16x2 – 40x + 25 square inches. What is the side length, in inches, of the pet shop sign?

Answer:

the length of the sign is 4x-5 inches

Step-by-step explanation:

Given

Area of the square of design = 16x^{2} -40x+25

First we find the roots of equation 16x^{2} -40x+25=0

The roots of the quadratic equation ax^{2} +bx^{2} +c=0 are given by

x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}

where a=16, b=-40, c=25

x=\frac{40\pm\sqrt{(-40)^2-4\times 16\times 25}}{2\times 16}

x=\frac{40\pm\sqrt{1600-1600}}{32}

x=\frac{40\pm\sqrt{0}}{32}

x=\frac{40}{32}

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That is, the factors of the polynomial 16x^{2} -40x+25 are 4x-5 and 4x-5.

So, Area of the square design = 16x^{2} -40x+25 = (4x-5)^{2}

Area of a square = Length^2

Thus, the length of the sign is 4x-5 inches

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