On a separate piece of paper sketch a unit circle with angle 0 in standard position. For what values of θ is the sine increasing? Decreasing? For what values of θ is the cosine increasing? Decreasing? For which angle between 0° and 360° is sine equal to 0? Where is cosine equal to 0? Please help(:

Answers

Answer 1

Answer: 1) Sine is the y-component in the unit circle. It grows from 0 t o1 in the sector 0° to 90 °, then starts to decrease from 1 to -1 until 270° where it starts to increase again from 270° to 360° (0°=.

2) Cosine is the x - component in the unit circle. It decreases from 1 to - 1 in the sector 0° to 180°, and increases from -1 to +1 in the sector 180° to 360°.

3) Sine is equal to 0 at angles 0° and 180°.

4) Cosine is equal to 0 at angles 90° and 270°

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Find the surface area and volume of a cube with sides of 5.5 cm each.

Answers

Answer:

$V= 166.375cm^3$ and $As = 181.5cm^2.$

Step-by-step explanation:

First, let's calculate the volume. The volume of a cube is $V= s^3$ where s is the side. Then,

$V = (5.5cm)^3 = 166.375cm^3$

Now, the surface area will be the sum of the areas of each face. Each face of the cube is a square and the area of a square is $A=s^2$ , also we know that the cube has 6 faces, then the surface area of the cube will be

$As = 6*s^2$ , so

$As = 6*(5.5cm)^2$

$As = 6*30.25cm^2$

$As = 181.5cm^2.$

The equation to find the surface area of a cube is: 2*base+lateral area. The base of a cube is just the area of the square. So you would multiply 5.5*5.5. That would equal 30.25
The equation to find the lateral area is: perimeter*height. So you would add 5.5 four times since its a square. It would equal 22. Then you would multiply it by the height of which is also 5.5. 5.5*22=121.

Now that you have all the information just plug it back in the equation.(side note, because cubes are prisms you have two bases which is why you multiply them by two.) 2*30.25+121=181.5

your awnser is 185.5cm squared

6. Minimum value determined by the formula function f (x) = 2x ²-8x + p was 20. Value f (2) is.7. Shape factor of the quadratic equation 4x ²-13x = -3 is ...
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10. Equation x ²-4x +3 = 0 and x ² +4 x-21 = 0, has a root persekutuan.Akar the alliance is

Answers

$6)\ \ \ f(x)=2x^2-8x+p\nthe\ minimum\ value =20\ \ \ \Leftrightarrow\ \ \ y_(\ of\ vertex)=20\ \ \ \Leftrightarrow\ \ \ - (\Delta)/(2a) =20\n\n\Delta=(-8)^2-4\cdot2\cdot p=64-8p\ \ \Leftrightarrow\ \ - (64-8p)/(2\cdot2) =20\ \ \Leftrightarrow\ \ -16+2p=20\n\n2p=36\ \ \ \Leftrightarrow\ \ \ p=18\ \ \ \Rightarrow\ \ \ \ f(x)=2x^2-8x+18\n\nf(2)=2\cdot2^2-8\cdot2+18=2\cdot4-16+18=8+2=10$

$7)\ the\ shape\ factor\ of\ the\ quadratic\ equation\ 4x^2-13x = -3\n is\ a=4\ \ \ (\ a>0\ \ \ \rightarrow\ \ \ the\ shape\ is\ \cup\ )\n\n8)\ \ \ the\ turning\ point=(-15;3)\ \ \ \Rightarrow\ \ \ f(x)=a(x+15)^2+3\n\n the\ graph\ passes\ through\ the\ point\ (-12.0) \ \Rightarrow\ \ 0=a(-12+15)^2+3\n\n\Rightarrow\ \ \ a\cdot3^2=-3\ \ \ \Rightarrow\ \ \ a=- (3)/(9) =- (1)/(3) \ \ \ \Rightarrow\ \ \ f(x)=- (1)/(3)(x+15)^2+3$

$\Rightarrow\ \ \ f(x)=- (1)/(3)(x^2+30x+225)+3=- (1)/(3)x^2-10x-72\n\n9)\ \ \ 4x^2+px+25=0\n\n\Delta=p^2-4\cdot4\cdot25=p^2-400\n\ntwo\ solutions\ \ \Leftrightarrow\ \ \Delta>0\ \ \Leftrightarrow\ \ p^2-40>0\ \ \Leftrightarrow\ \ (p-20)(p+20)>0\n.\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \Leftrightarrow\ \ \ p\in(-\infty;\ -20)\ \cap\ (20;\ +\infty)\n-------------------------------$

$the\ Vieta's\ formulas\ to\ the\ quadratic\ equation\ ax^2+bx+c=0\n\nx_1+x_2=- (b)/(a) \ \ \ and\ \ \ x_1\cdot x_2= (c)/(a) \n------------------------------\n\nx_1+x_2=- (p)/(4) \ \ \ and\ \ \ x_1\cdot x_2= (25)/(4) \n\nx_1^2+x_2^2=x_1^2+2\cdot x_1\cdot x_2 +x_2^2-2\cdot x_1\cdot x_2 =(x_1+x_2)^2-2\cdot x_1\cdot x_2 \n\nx_1^2+x_2^2=(x_1+x_2)^2-2\cdot x_1\cdot x_2 \ \ \ \Leftrightarrow\ \ \ 12.5=(- (p)/(4) )^2-2\cdot (25)/(4) \n\n$

$12.5= (p^2)/(16) +12.5 \ \ \ \Leftrightarrow\ \ \ (p^2)/(16)=0 \ \ \ \Leftrightarrow\ \ \ p^2=0 \ \ \ \Leftrightarrow\ \ \ p=0\n\n\n10)\ \ \ x^2-4x+3=0\ \ \ and\ \ \ x^2+4x-21=0\n\n x^2-4x+3=x^2+4x-21\ \ \Leftrightarrow\ \ -4x-4x=-21-3\n\n\ \ \Leftrightarrow\ \ -8x=-24\ \ \Leftrightarrow\ \ x=3$

Ohama is landing a plane on the runway. He's trying to decide where he should deploy the plane's landing gear so that the plane comes to a stop exactly at the end of the runway. The runway is 600 yards long, and the plane will travel half the distance with the landing gear than without the landing gear. Where should he deploy the landing gear ?

Answers

Answer:

Let's define:

A = distance traveled before deploying the landing gear

B = distance traveled after deploying the landing gear.

We must have that the sum of those two distances must be equal to 600 yards.

A + B = 600 yd.

And we know that:

" the plane will travel half the distance with the landing gear than without the landing gear."

Then we have that:

B = A/2.

Now we can replace this last equation in the first one:

A + B = 600yd

A + A/2 = 600yd.

(3/2)*A = 600yd.

A = (2/3)*600yd = 400yd.

Ohama should deploy the landing gear 400 yd into the runway.

Please explain how to solve this On Monday,it took Helen 3 hours to do a page of science homework exercises.The next day she did the same number of exercises in 2 hours.If her average rate on Monday was p exercises per hour,what was her average rate the next day, in terms of p?