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Over [174]
3 years ago
9

Find the angle measure in degrees

Mathematics
1 answer:
loris [4]3 years ago
7 0

Answer:

82°

Step-by-step explanation:

By inscribed angle theorem:

m\angle QRP =  \frac{1}{2}  \times 164 \degree \\  \\   \huge \red{ \boxed{\therefore \: m\angle QRP =82 \degree}}

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The table shows some values of a function of the form y
kherson [118]

The value of c, the constant of the function y = ax² + bx + c, exists -3.

<h3>What is an equation?</h3>

An equation exists as an expression that indicates the relationship between two or more numbers and variables.

Given that: y = ax² + bx + c

At point (4, 21)

21 = a(4²) + 4b + c .......(1)

At point (5, 32)

32 = a(5²) + 5b + c .........(2)

At points (6, 45)

45 = a(6²) + 6b + c .......(3)

Therefore, the value of a = 1, b = 2 and c = -3.

The value of c, the constant of the function y = ax² + bx + c, exists -3.

To learn more about equations refer to:

brainly.com/question/2972832

#SPJ9

4 0
1 year ago
Pleeaaassseeeees help,
sattari [20]

Hey there!

Since you probably have better things to do, your answer would be A: π(3)²

To back up my answer, the formula to find the area of a circle is π · R² where R = Radius. In that formula then, you just plug everything in. Since the radius is 3, you would put 3 into the "( )". Finally, just plug the rest of the equation in!

<em>I'm open to any question or comment!</em>

<em>God Bless!</em>

<em>-X8lue83rryX</em>

5 0
3 years ago
Using distributive property 7(5x+12)
Anuta_ua [19.1K]
35x+84 is the answer :)
4 0
3 years ago
Read 2 more answers
How do I go about solving (27x^3/8y^9)^5/3? And what is the role of the numerator and denominator?
MrRissso [65]
\left( \frac{27x^3}{8y^9}\right)^ \frac{5}{3}  \\\\\\ =\left( \frac{(3x)^3}{(2y^3)^3}\right)^ \frac{5}{3} \\\\\\ =  \frac{(3x)^{3 \times  \frac{5}{3} }}{(2y^3)^{3 \times  \frac{5}{3} }} \\\\\\ =\frac{(3x)^5}{(2y^3)^{5 }} \\\\\\ =\frac{243x^5}{32y^{15}}

Now, If the exponent was negative like you asked....

\left( \frac{27x^3}{8y^9}\right)^ {-\frac{5}{3}} \\\\\\ =\left( \frac{8y^9}{27x^3}\right)^ {\frac{5}{3}}\\\\\\ =\left( \frac{(2y^3)^3}{(3x)^3}\right)^ \frac{5}{3} \\\\\\ = \frac{(2y^3)^{3 \times \frac{5}{3} }}{(3x)^{3 \times \frac{5}{3} }} \\\\\\ =\frac{(2y^3)^{5 }}{(3x)^5} \\\\\\ =\frac{32y^{15}}{243x^5}

5 0
3 years ago
7. Use the quotient rule to simplify the following expression. Assume that x&gt;0.
NeTakaya
Quotient rule says we can divide 192 and 3 because they are both under the radical. This gives us the square root of 64.

We can also divide x^3 and x to get x^2.

The square root of 64x^2 is 8x
8 0
3 years ago
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