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mamaluj [8]
3 years ago
7

1) What is 5/7 x 3/5? A)2/3 B)2/9 C)3/7 D)5/4

Mathematics
2 answers:
Yuliya22 [10]3 years ago
8 0

Answer: 3/7

Step-by-step explanation:

Tpy6a [65]3 years ago
5 0

Answer:

C)3/7

Step-by- step explanation:

5/7 x 3/5= 0.42857142857

3/7 = 0.42857142857

plz mark brainleist

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CaHeK987 [17]
Since q(x) is inside p(x), find the x-value that results in q(x) = 1/4

\frac{1}{4} = 5 - x^2\ \Rightarrow\ x^2 = 5 - \frac{1}{4}\ \Rightarrow\ x^2 = \frac{19}{4}\ \Rightarrow \\
x = \frac{\sqrt{19} }{2}

so we conclude that
q(\frac{\sqrt{19} }{2} ) = 1/4

therefore

p(1/4) = p\left( q\left(\frac{ \sqrt{19} }{2} \right)  \right)

plug x=\sqrt{19}/2 into p( q(x) ) to get answer

p(1/4) = p\left( q\left( \frac{ \sqrt{19} }{2} \right) \right)\ \Rightarrow\ \dfrac{4 - \left(  \frac{\sqrt{19} }{2}\right)^2 }{ \left(  \frac{\sqrt{19} }{2}\right)^3 } \Rightarrow \\ \\ \dfrac{4 - \frac{19}{4} }{ \frac{19\sqrt{19} }{8}} \Rightarrow \dfrac{8\left(4 - \frac{19}{4}\right) }{ 8 \cdot \frac{19\sqrt{19} }{8}} \Rightarrow \dfrac{32 - 38}{19\sqrt{19}} \Rightarrow \dfrac{-6}{19\sqrt{19}} \cdot \frac{\sqrt{19}}{\sqrt{19}}\Rightarrow

\dfrac{-6\sqrt{19} }{19 \cdot 19} \\ \\ \Rightarrow  -\dfrac{6\sqrt{19} }{361}

p(1/4) = -\dfrac{6\sqrt{19} }{361}
3 0
3 years ago
P and Q are two points on the line x - y + 1 = 0 and are at distance 5 from the origin. Find the area of the triangle OPQ. ​
Alecsey [184]

Answer:

P and Q are two points on the line x-y+1=0 and are at a distant of 5 units from the origin. Find the area of triangle POQ.

Step-by-step explanation:

P and Q are the intersection points of

x-y+1 = 0 and the circle x^2 + y^2 = 25

sub y = x+1 into the circle

x^2 + (x+1)^2 = 25

x^2 + x^2 + 2x + 1 - 25 = 0

x^2 + x - 12 = 0

(x+4)(x-3) = 0

x = 3 or x = -4

y = 4 or y = -3

so P(3,4) and Q(-4,3) are our two points

Height of triangle.

h = |0 - 0 + 1|/√2 = 1/√2

PQ = √( (-7)^2 + 1^2) = √50 = 5√2

area POQ = (1/2)(1/√2)(5√2) = 5/2 square units

hope this helped

7 0
2 years ago
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