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algol [13]
3 years ago
5

Part 1: Using complete sentences, compare the key features and graphs of sine and cosine. What are their similarities and differ

ences?
Part 2: Using these similarities and differences, how would you transform f(x) = 3 sin(4x - π) + 4 into a cosine function in the form f(x) = a cos(bx - c) + d?
Mathematics
1 answer:
Alenkasestr [34]3 years ago
3 0
Part 1: Similarities of sine and cosine graphs are that both has an amplitude of 1 and a period of 2pi.
A differences between sin and cos graphs includes that sin graph passes through point (0, 0).

The required cosine graph is <span>f(x) = 3 cos(4x - 3π/2) + 4</span>
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The graph of the function f(x) = (x-3)(x + 1) is shown.
Drupady [299]

Answer:

x < -1

Step-by-step explanation:

Draw a parabola (the graph) passing trough two x-intercepts -1 and 3 and open up.

8 0
2 years ago
A cubical swimming pool is of side 12m is filled with water up to a depth of 2m. What is the quantity of water in the pool in li
andrezito [222]

Volume is Length x Width x height.

The pool is cubical, so the length and width are the same dimension.

First find the volume of the pool that is filled by water:

Volume = 12 x 12 x 2 = 288 cubic meters.

Now convert cubic meters to liters:

1 cubic meter = 1,000 liters

288 x 1000 = 288,000 liters

7 0
3 years ago
Solve the simultaneous equations x2 + y2 = 10 and x – 3y + 10 = 0.
ch4aika [34]

Answer:

x = - 1, y = 3  or     (-1 , 3)

Step-by-step explanation:

x – 3y + 10 = 0              x = 3y - 10    ... plug in

x² + y² = 10

(3y-10)² + y² = 10

10y² - 60y +90 = 0

y² - 6y + 9 = 0

(y - 3)² = 0

y = 3

x = - 1

3 0
3 years ago
Answer for a lot of points!
earnstyle [38]

Given :

  • ZC = 90°

  • CD is the altitude to AB.

  • \angleA = 65°.

To find :

  • the angles in △CBD and △CAD if m∠A = 65°

Solution :

In Right angle △ABC,

we have,

=> ACB = 90°

=> \angleCAB = 65°.

So,

=> \angleACB + \angleCAB+\angleZCBA = 180° (By angle sum Property.)

=> 90° + 65° + \angleCBA = 180°

=> 155° +\angleCBA = 180°

=> \angleCBA = 180° - 155°

=> \angleCBA = 25°.

In △CDB,

=> CD is the altitude to AB.

So,

=> \angle CDB = 90°

=> \angleCBD = \angleCBA = 25°.

So,

=> \angleCBD + \angleDCB = 180° (Angle sum Property.)

=> 90° +25° + \angleDCB = 180°

=> 115° + \angleDCB = 180°

=> \angleDCB = 180° - 115°

=> \angleDCB = 65°.

Now, in △ADC,

=> CD is the altitude to AB.

So,

=> \angleADC = 90°

=>\angle CAD =\angle CAB = 65°.

So,

=> \angleADC + \angleCAD +\angleDCA = 180° (Angle sum Property.)

=> 90° + 65° + \angleDCA = 180°

=> 155° +\angleDCA = 180°

=> \angleDCA = 180° - 155°

=> \angleDCA = 25°

Hence, we get,

  • \angleDCA = 25°
  • \angleDCB = 65°
  • \angleCDB = 90°
  • \angleACD = 25°
  • \angleADC = 90°.
7 0
3 years ago
Mr smith and mr stein were driving to a buisness meeting 250 miles from their office. Mr Smith drove the first x miles, then Mr
Sergeeva-Olga [200]
250 divided by x = D (the number of miles mr Stein drove) i think that’s the answer
8 0
3 years ago
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