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2 years ago

Please Hurry Name three angles that sum to 180°.

Mathematics

1 answer:

Sergeu [11.5K]2 years ago

6 0

Answer:

90,90

85,95

5,175

Step-by-step explanation:

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The wind is blowing N 35.0 degrees W at 1.60 * 10^2 mph. A plane has an engine speed of 3.20 * 10^2 mph. Where should the pilot

andreyandreev [35.5K]

Answer: 24.2° SouthWest

Step-by-step explanation:

First step: DRAW A PICTURE of the vectors from head to tail (see image)

I created a perpendicular from the resultant vector to the vertex of the given vectors so I could use Pythagorean Theorem to find the length of the perpendicular. Then I used that value to find the angle of the plane.

Perpendicular (x):

cos 35° = adjacent/hypotenuse

cos 35° = x/160

→ x = 160 cos 35°

Angle (θ):

sin θ = opposite/hypotenuse

sin θ = x/320

sin θ = 160 cos 35°/320

θ = arcsin (160 cos 35°/320)

θ = 24.2°

Direction is down (south) and left (west)

7 0

3 years ago

High demand cars that are also in low supply tend to retain their value better than other cars .The data in the table are for a

valkas [14]

Answer:

3/6

Step-by-step explanation:

8 0

3 years ago

The rule T 5, -0.5° Ro, 1800(x, ) is applied to FGH to

Yuki888 [10]

Answer:

Step-by-step explanation:

Given rule for the multiple translations is,

$T_{5,-0.5}.R_{0.180^{\circ}}(x,y)$

Apply the rule $R_{0,180^{\circ}}$ first.

(x, y) → (-x, -y)

This rule illustrates a rotation of the triangle FGH by 180° about the origin,

Vertices of ΔFGH are,

F → (1, 1)

G → (4, 5)

H → (5, 1)

After rotation vertices of the image triangle are,

F' → (-1, -1)

G' → (-4, -5)

H' → (-5, -1)

Further apply the rule,

$T_{5,-0.5}$

(x, y) → (x + 5, y - 0.5)

By this rule of translation,

F'(-1, -1) → F"{(-1 + 5), (-1 - 0.5)}

→ F"(4, -1.5)

G'(-4, -5) → G"[(-4 + 5), (-5 - 0.5)]

→ G"(1, -5.5)

H'(-5, -1) → H"[(-5 + 5), (-1 -0.5)]

→ H"(0, -1.5)

8 0

2 years ago

*EXTRA PTS* answer #2 - #5

TiliK225 [7]

Answer:

2. 2N

3. -4N

4. 4N

5. 7N

Step-by-step explanation:

4 0

3 years ago

Solve the equation on the interval [0, 27r). 4(sin x)2 - 2 = 0

Ket [755]

$4[sin(x)]^2 - 2 = 0\implies 4[sin(x)]^2=2\implies [sin(x)]^2=\cfrac{2}{4}\implies [sin(x)]^2=\cfrac{1}{2} \\\\\\ sin(x)=\pm\sqrt{\cfrac{1}{2}}\implies sin^{-1}[sin(x)]=sin^{-1}\left( \pm\sqrt{\cfrac{1}{2}} \right)\implies x=sin^{-1}\left( \pm\sqrt{\cfrac{1}{2}} \right)$

$x=sin^{-1}\left( \pm\cfrac{\sqrt{1}}{\sqrt{2}} \right)\implies x=sin^{-1}\left( \pm\cfrac{1}{\sqrt{2}} \right)\implies x=sin^{-1}\left( \pm\cfrac{\sqrt{2}}{2} \right) \\\\[-0.35em] ~\dotfill\\\\ ~\hfill x=\cfrac{\pi }{4}~~,~~\cfrac{3\pi }{4}~~,~~\cfrac{5\pi }{4}~~,~~\cfrac{7\pi }{4}~\hfill$

Check the picture below.

4 0

2 years ago