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romanna [79]
3 years ago
6

3. Given: BAM is a right angle. If mBAM=4x+2, then solve for the value of x.

Mathematics
1 answer:
Dafna11 [192]3 years ago
4 0

Answer:

3. x = 17

4. a. m<NMP = 48°

b. m<NMP = 60°

Step-by-step explanation:

3. Given that <BAM = right angle, and

m<BAM = 4x + 22, set 90° equal to 4x + 22 to find x.

4x + 22 = 90

Subtract 22 from both sides

4x + 22 - 22 = 90 - 22

4x = 68

Divide both sides by 4

4x/4 = 68/4

x = 17

4. a. m<NMQ = right angle (given)

m<PMQ = 42° (given)

m<PMQ + m<NMP = m<NMQ (angle addition postulate)

42 + m<NMP = 90 (substitution)

m<NMP = 90 - 42 (subtracting 42 from each side)

m<NMP = 48°

b. m<NMQ = right angle (given)

m<NMP = 2*m<PMQ

Let m<PMQ = x

m<NMP = 2*x = 2x

2x + x = 90° (Angle addition postulate)

3x = 90

x = 30 (dividing both sides by 3)

m<PMQ = x = 30°

m<NMP = 2*m<PMQ = 2*30

m<NMP = 60°

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Answer:

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Step-by-step explanation:

Sea 'p (x)' el número que no es divisible por 4, 6, 9, 11 y 12 de manera que cuando se divide por estos números da un resto igual, tenemos;

Por teorema del resto, tenemos;

p (x) = (x - a) · Q (x) + R

Dónde;

p (x) = El número, que se divide

Q (x) = El cociente

(x - a) = El divisor

R = El resto

Dado que el número más pequeño es 4, el resto, 0 <R <4

Para el número entero más pequeño arriba (x - a) · Q (x), tenemos;

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Observamos que el mínimo común múltiplo de 4, 6, 9, 11 y 12 = 396, por lo tanto, podemos tener;

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p (x) = 396 + 1 = 397

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El número que no es divisible por 4, 6, 9, 11 y 12, de manera que cuando se divide por estos números da residuos iguales, p (x) = 397

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