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3 years ago

Someone please help on this problem ?

Mathematics

1 answer:

Wewaii [24]3 years ago

7 0

Answer:

CR = 17 ; PR = 30

Step-by-step explanation:

From the problem hypothesis we know that CD⊥PR or CQ⊥PR(perpendicular) and ∡PCQ ≡∡RCQ (bisector)

so ∡PCQ≡∡RCQ

[CQ]≡[CQ] (common side)

CQ⊥PR

--------------------------------------------

⇒(cathetus - angle case of congruence)⇒ ΔPQC≡ΔRQC so [PQ]≡[QR] (=15) // PR = PQ+QR ⇒ PR=30

from this congruence ⇒ΔPRC = isosceles so PC=CR=17

Hint: in any triangle is bisector of an angle it is perpendicular on the opposite side then triangle is isosceles.

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Vitek1552 [10]

Answer:

5/48

Step-by-step explanation:

1x5=5

8x6=48

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5/48

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4 years ago

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It's a question from real and complex numbers which I can't solve. so someone PLZ HeLp

Semmy [17]

Answer:

$\frac{1}{5}$

Step-by-step explanation:

Using the rules of exponents

$a^{m}$ × $a^{n}$ = $a^{(m+n)}$ , $\frac{a^{m} }{a^{n} }$ = $a^{(m-n)}$ , $(a^m)^{n}$ = $a^{mn}$

Simplifying the product of the first 2 terms

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= $a^{p^2-r^2}$

Simplifying the third term

5( $(a^p+r)^{p-r}$

= 5 $a^{(p+r)(p-r)}$ = 5 $a^{(p^2-r^2)}$

Performing the division, that is

$\frac{a^{(p^2-r^2)} }{5a^{(p^2-r^2)} }$ ← cancel $a^{(p^2-r^2)}$ on numerator/ denominator leaves

= $\frac{1}{5}$

4 0

3 years ago

Point A is located at (5, 10) and point B is located at (20, 25).

Alla [95]

Let's say the point is C, so C partitions AB into two pieces, where AC is at a ratio of 3 and CB is at a ratio of 7, thus 3:7,

$\bf \left. \qquad \right.\textit{internal division of a line segment}
\\\\\\
A(5,10)\qquad B(20,25)\qquad
\qquad 3:7
\\\\\\
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-------------------------------\\\\
{ C=\left(\cfrac{\textit{sum of "x" values}}{\textit{sum of ratios}}\quad ,\quad \cfrac{\textit{sum of "y" values}}{\textit{sum of ratios}}\right)}\\\
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\\\\\\
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