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Nat2105 [25]
2 years ago
5

Zubin drew a circle with right triangle PRQ inscribed in it, as shown below: The figure shows a circle with points P, Q, and R o

n it forming an inscribed triangle. Side PQ is a chord through the center, and angle R is a right angle. Arc QR measures 50 degrees. If the measure of arc QR is 50°, what is the measure of angle PQR?
Mathematics
1 answer:
iris [78.8K]2 years ago
7 0

The measure of angle PQR when Zubin drew a circle with right triangle will be 65°.

<h3>How to find the measure of the angle?</h3>

From the information given, when Zubin drew a circle with right triangle, arc QR measures 50 degrees.

Therefore, the measure of angle PQR will be:

= 180° - 90° - (1/2 × 50°)

= 180° - 90° - 25°

= 65°

In conclusion, the angle is 65°.

Learn more about circle on:

brainly.com/question/24375372

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Ping's Ice Cream Palace offers a special sundae that contains over 2 kilograms of ice cream plus any number of toppings. Write a
hjlf

Let "x" represent the weight of the toppings. We know that we can have any number of toppings. This means that one may ask for no toppings at all too.

Now, we have been told that "S" is the weight of the special sundae in kilograms. This definitely included the "mandatory" 2 kilograms of ice cream. Therefore, S will be at-least equal to 2.

Thus, the inequality that describes S, the weight of the special sundae in kilograms at Ping's Ice Cream Palace is given as:

S\geq 2+x kilograms.

It can be seen that as x increases, S increases too and if an order does not want any toppings in it then the weight of the special sundae will be a minimum of 2 kg which is the weight of the ice cream.

7 0
3 years ago
Read 2 more answers
Differentiate with respect to X <br><img src="https://tex.z-dn.net/?f=%20%5Csqrt%7B%20%5Cfrac%7Bcos2x%7D%7B1%20%2Bsin2x%20%7D%20
Mice21 [21]

Power and chain rule (where the power rule kicks in because \sqrt x=x^{1/2}):

\left(\sqrt{\dfrac{\cos(2x)}{1+\sin(2x)}}\right)'=\dfrac1{2\sqrt{\frac{\cos(2x)}{1+\sin(2x)}}}\left(\dfrac{\cos(2x)}{1+\sin(2x)}\right)'

Simplify the leading term as

\dfrac1{2\sqrt{\frac{\cos(2x)}{1+\sin(2x)}}}=\dfrac{\sqrt{1+\sin(2x)}}{2\sqrt{\cos(2x)}}

Quotient rule:

\left(\dfrac{\cos(2x)}{1+\sin(2x)}\right)'=\dfrac{(1+\sin(2x))(\cos(2x))'-\cos(2x)(1+\sin(2x))'}{(1+\sin(2x))^2}

Chain rule:

(\cos(2x))'=-\sin(2x)(2x)'=-2\sin(2x)

(1+\sin(2x))'=\cos(2x)(2x)'=2\cos(2x)

Put everything together and simplify:

\dfrac{\sqrt{1+\sin(2x)}}{2\sqrt{\cos(2x)}}\dfrac{(1+\sin(2x))(-2\sin(2x))-\cos(2x)(2\cos(2x))}{(1+\sin(2x))^2}

=\dfrac{\sqrt{1+\sin(2x)}}{2\sqrt{\cos(2x)}}\dfrac{-2\sin(2x)-2\sin^2(2x)-2\cos^2(2x)}{(1+\sin(2x))^2}

=\dfrac{\sqrt{1+\sin(2x)}}{2\sqrt{\cos(2x)}}\dfrac{-2\sin(2x)-2}{(1+\sin(2x))^2}

=-\dfrac{\sqrt{1+\sin(2x)}}{\sqrt{\cos(2x)}}\dfrac{\sin(2x)+1}{(1+\sin(2x))^2}

=-\dfrac{\sqrt{1+\sin(2x)}}{\sqrt{\cos(2x)}}\dfrac1{1+\sin(2x)}

=-\dfrac1{\sqrt{\cos(2x)}}\dfrac1{\sqrt{1+\sin(2x)}}

=\boxed{-\dfrac1{\sqrt{\cos(2x)(1+\sin(2x))}}}

5 0
3 years ago
9n - 15 for n = 5 help me please
MaRussiya [10]

If n = 5, then 9n = 45

45 - 15 = 30, which is the same thing as 6n


7 0
3 years ago
Solve for R= 2(r-4)=-4(r-2)+4
motikmotik

Answe

(R,r)=(-2/5,10/3)

hope it helps.

4 0
3 years ago
Read 2 more answers
0Aphysical education class is doing jumping jacks for 60 seconds. How many seconds remain if 18 seconds haveSed?24Maiklstand tot
Vaselesa [24]

If the initial time was 60 seconds and 18 seconds have passed,

60-18=42

There would be 42 seconds left.

8 0
2 years ago
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