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3 years ago

Please help me!!!! precalc question

Mathematics

1 answer:

dangina [55]3 years ago

7 0

Step-by-step explanation:

1. Ampltiude is 1

The period is

$\frac{2\pi}{3}$

is the period

The phase shift is 0

2. The Ampltiude is 2

The period is 2 pi

The phase shift is 1

3. The amplitude is 3

The period is 2 pi/3

The phase shift is -2

4.The amplitude is 2/3

The period is -10pi

The phase shift is 0

5. The Ampltiude is 1

The period is -2 pi

The phase shift is pi

6. The amplitude is 1/4

The period is -4pi

The phase shift is negative 2pi/3

7. The Ampltiude is 3

The period is 4

The phase shift is 8

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Step 1: $26.95 - $8.95 (which equals $18.00)

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3 years ago

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2 years ago

A(1)=20

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Answer:

-14

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3 0

2 years ago

In rectangle abcd, points p and q lie on side AB and DC respectively. Angle PMQ is a right angle, M is the midpoint of side BC a

nirvana33 [79]

Answer:

PM:MQ = 8:3.

Step-by-step explanation:

$\rm \angle B\hat{M}P + 90^{\circ} + \angle C\hat{M}Q = 180^{\circ}$ ;

$\implies \rm \angle B\hat{M}P + \angle C\hat{M}Q = 90^{\circ}$ ;

$\implies \rm 90^{\circ} - \angle B\hat{M}P = \angle C\hat{M}Q$ .

Also,

$\rm \angle B\hat{P}M = 90^{\circ} - \angle B\hat{M}P$ in right triangle PBM.

Thus $\rm \angle{P\hat{B}M} = \angle C\hat{M}Q$ .

Additionally $\rm \angle \hat{B} = 90^{\circ} = \angle \hat{C}$ .

Therefore $\rm \triangle PBM \sim \triangle MCQ$ .

$\rm \displaystyle BC = 2\;MC$ for M is the midpoint of segment BC.

$\rm \displaystyle PB = \frac{4}{3}BC = \frac{8}{3}MC$ .

$\rm \triangle PBM \sim \triangle MCQ$ implies that

$\displaystyle \rm PM:MQ = PB:MC = 1:\frac{8}{3} = 8:3$ .

8 0

2 years ago