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tamaranim1 [39]

1 year ago

13

Cars A, B, and C are travelling at constant speeds. Select the car that travels the fastest. Choose 1 answer: Choose 1 answer: (

Choice A) A (Choice B) B Car B travels a distance of ddd kilometers in hhh hours, based on the equation 55h=d55h=d55, h, equals, d. (Choice C) C Car C travels 135135135 kilometers in 333 hours.

1 answer:

Gennadij [26K]1 year ago

6 0

Answer:

correct answer is B

Step-by-step explanation:

comparing the speed at which the three cars travel and also considering the fact that they travel at a constant speed without increase or decrease in speed or acceleration, it gives B the highest speed at 55h per day

thank you.

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3 years ago

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3 years ago

Write each of the following as a function of theta.<br> 1.) sin(pi/4 - theta) 2.) tan(theta+30°)

Paladinen [302]

Step-by-step explanation:

Let x represent theta.

$\sin( \frac{\pi}{4} - x )$

Using the angle addition trig formula,

$\sin(x - y) = \sin(x) \cos(y) - \cos(x) \sin(y)$

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$( \frac{ \sqrt{2} }{2}) \cos(x) - (\frac{ \sqrt{2} }{2} )\sin(x)$

Multiply one side at a time

Replace theta with x , the answer is

$\frac{ \sqrt{2} \cos(x) }{2} - \frac{ \sin(x) \sqrt{2} }{2}$

2. Convert 30 degrees into radian

$\frac{30}{1} \times \frac{\pi}{180} = \frac{\pi}{6}$

Using tangent formula,

$\tan(x + y) = \frac{ \tan(x) + \tan(y) }{1 - \tan(x) \tan(y) }$

$\frac{ \tan(x) + \tan( \frac{\pi}{6} ) }{1 - \tan(x) \tan( \frac{\pi}{6} ) }$

Tan if pi/6 is sqr root of 3/3

$\frac{ \tan(x) + ( \frac{ \sqrt{3} }{3} ) }{1 - \tan(x) (\frac{ \sqrt{3} }{3} ) }$

Since my phone about to die if you later simplify that,

you'll get

$\frac{(3 \tan(x) + \sqrt{3} )(3 + \sqrt{3} \tan(x) }{3(3 - \tan {}^{2} (x) }$

Replace theta with X.

4 0

2 years ago

1. Carter will install fencing all around the flat area of his

MariettaO [177]

Answer:

Amount of fencing required = 75 yards

Step-by-step explanation:

Distance between the two points $(x_1,y_1)$ and $(x_2,y_2)$ is given by the formula,

d = $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Distance between A(3, 6) and B(3, -2) = $\sqrt{(3-3)^2+(6+2)^2}$

= 8 yards

Distance between B(3, -2) and C(-7, 4) = $\sqrt{(3+7)^2+(-2-4)^2}$

= $\sqrt{136}$

= 11.66 ≈ 12 yards

Distance between C(-7, 4) and D(-7, -2) = $\sqrt{(-7+7)^2+(4+2)^2}$

= 6 yards

Distance between D(-7, -2) and E(-3, -2) = $\sqrt{(-7+3)^2+(-2+2)^2}$

= 4 yards

Distance between E(-3, -2) and F(-3, -8) = $\sqrt{(-3+3)^2+(-2+8)^2}$

= 6 yards

Distance between F(-3, -8) and G(3, -8) = $\sqrt{(-3-3)^2+(-8+8)^2}$

= 6 yards

Distance between G(3, -8) and H(10, -12) = $\sqrt{(3-10)^2+(-8+12)^2}$

= $\sqrt{49+16}$

= $\sqrt{65}$

= 8.06 ≈ 8 yards

Distance between H(10, -12) and J(10, 6) = $\sqrt{(10-10)^2+(-12-6)^2}$

= 18 yards

Distance between A(3, 6) and J(10, 6) = $\sqrt{(10-3)^2+(6-6)^2}$

= 7 yards

Since length of fence required = perimeter of the flat area

Perimeter of the given area = 8 + 12 + 6 + 4 + 6 + 6 + 8 + 18 + 7

= 75 yards

Therefore, amount of fencing required = 75 yards

8 0

3 years ago