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Pepsi [2]
3 years ago
8

Pls tell the answer fast please​

Mathematics
2 answers:
Marianna [84]3 years ago
8 0

Let, ∠ZYQ = ∠QYP = a°

Since, XYP is a straight line,

⇒ ∠XYZ + ∠ZYP = 180° (Linear Pair)

⇒∠ZYP = 180°- ∠XYZ

⇒∠ZYP = 180° - 64°

⇒∠ZYP = 116°

As it's given, YQ is the bisector of ∠ZYP,

⇒∠ZYQ = ∠QYP = ½ ZYP

= ½ × 116°

= 58°

⇒∠XYQ = ∠XYZ + ∠ZYQ

= 64° + 58°

= 112°

[Escape (∠XYQ value) if ya want because in Question it's not mentioned to find the value of ∠XYQ but we need need to find the value of ∠ZYQ, in order to find the reflex of it]

Now, We've to also find the reflex of

∠ZYQ,

⇒ Reflex of ∠ZYQ = 360° - ∠ZYQ

= 360° - 58°

= 302°

jekas [21]3 years ago
7 0

Answer:

If I had to take an educated guess I'd say 126° but I'm not 100% sure.

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UkoKoshka [18]

Solution

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(f/g)(x) = \frac{f(x)}{g(y)}

(f/g)(x) = \frac{-20x^{2}+14x +12 }{5x - 6}

<u>Step 1: </u>Now we have to factorize the numerator.

f(x) = -20x^2 + 14x + 12

Factor out -2, we get

= -2 (10x^2 - 7x - 6)

Now we can factorize 10x^2 - 7x - 6                  

f(x) = -2(2x + 1) (5x - 6)

<u>Step 2: </u>Plug in the factors

(f/g)(x) = \frac{-2 (2x +1)(5x - 6)}{(5x - 6)}

<u>Step 3:</u> Cancel out the common factor (5x - 6) from the numerator and the denominator, we get

(f/g)(x) = -2(2x +1) = -4x -2

Since -4x -2 is linear expression, the domain is all the real numbers.

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skelet666 [1.2K]

Answer:

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Step-by-step explanation:

The specific contents of any of these registers at any point in time <em>depends on the architecture of the computer</em>. If we make the assumption that the only interface registers connected to memory are the memory address register (MAR) and the memory data register (MDR), then <em>all memory transfers of any kind</em> will use both of these registers.

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5. The PC is incremented to 02.

6. The contents of the PC are copied to the MAR.

7. A Memory Read operation is performed, and the contents of memory at address 02 are copied to the MDR. (Contents are the SUB 05 instruction.)

8. The MDR contents are decoded and the value 05 is placed in the MAR.

9. A Memory Read operation is performed and the contents of memory at address 05 are copied to the MDR. (Contents are the value 3.)

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12. The contents of the PC are copied to the MAR.

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16. The PC is incremented to 04.

17. Instruction fetch and decoding continues. This program will go "off into the weeds", since there is no Halt instruction. Results are unpredictable.

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