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ExtremeBDS [4]
3 years ago
15

Which of the following is true of the values of x and y in the diagram below? A unit circle is shown. A radius with length 1 for

ms angle theta = StartFraction pi Over 4 EndFraction. The radius intersects point (x, y) on the unit circle. y < x y > x y + x = 1 StartFraction y Over x EndFraction = 1
Mathematics
2 answers:
jok3333 [9.3K]3 years ago
6 0

Answer:

y/x=1

Step-by-step explanation:

took the test

Valentin [98]3 years ago
4 0

Answer:

D

Step-by-step explanation:

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(9b+3)(b-7)=0<br><br> Solve this using factoring, what are the two x’s?
telo118 [61]

Answer:

answer is down below

Step-by-step explanation:

u have ur two brackets.

(9b+3)(b-7)=0

take 1 bracket at a time

9b+3=0

solve it so:

9b+3=0

9b=-3

b= -3 / 9

<u>Ans: b=-1/3</u>

take the other bracket

b-7=0

solve it:

<u>Ans: b=+7 </u>

5 0
3 years ago
25% equivalent to 1/4%
Effectus [21]
Those are equivalent take 100 times 1/4 it’s the same as dividing 4 :)
3 0
3 years ago
Four more then the product of 3 and x is less than 40
dedylja [7]

3x + 4 < 40

Hope this helps! ;)

5 0
3 years ago
What is the distance between the points (59.5, 34.2) and (15.3, 14.9)? Enter your answer rounded to the nearest tenth (0.1).
nirvana33 [79]

Answer:

48.2

Step-by-step explanation:

√(x2-x1)²+(y2-y1)²

√(15.3-59.5)² +(14.9-34.2)²

√(-44.2)² + (-19.3)²

√1953.64+372.49

√2326.13

48.2

5 0
2 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
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