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asambeis [7]
3 years ago
11

An object travels back and forth along a straight line. Its velocity, in centimeters per second, is given by the function v(t) =

Equals 13 sine (StartFraction pi Over 45 EndFraction t), where t is time in seconds.
What is the maximum velocity of the object?

0 cm/s
13 cm/s
26 cm/s
90 cm/s

Mathematics
2 answers:
snow_lady [41]3 years ago
7 0

Answer: 13 cm/s , B on Edgenuity

Step-by-step explanation:

natta225 [31]3 years ago
6 0

What is the maximum velocity of the object?

The maximum velocity of the object is 13 cm/s

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50 pts if you answer
andrezito [222]

Answer:

3/20

Step-by-step explanation:

Total trials: 20

Tails and 3 is denoted by 3T

No. of 3T's: 3

Experimental probability: 3/20

7 0
3 years ago
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The exterior angles of triangle UVW are ∠X, ∠Y, and ∠Z, and they are adjacent to ∠U, ∠V, and ∠W, respectively.
Alika [10]

Answer:

m∠W = 67°

Explanation:

According to the triangle sum theorem, three angles added up together in a triangle must be 180°. To find the missing angle:

86°+27°+x°=180°

= 67°

8 0
2 years ago
Translate algebraic expression. <br><br>1) One-fourth of the sum of 6 and s?
zhenek [66]
1/4 (6 + s)

One-fourth (1/4) of the sum (+) of: 6 and s (6 + s)
3 0
3 years ago
A concrete pillar has the shape of a cylinder. It has a diameter of 8 meters and a height of 9 meters. If concrete costs $87 per
Komok [63]

Answer:

Cost, C = $28410.72

Step-by-step explanation:

We have, a concrete pillar has the shape of a cylinder.

Diameter of pillar is 8 m and ts height is 9 m

If concrete costs $87 per cubic meter, it is required to find the concrete cost for the pillar. For this lets find the area of cylindrical shape. So,

A=2\pi rh+2\pi r^2\\\\A=2\times 3.14\times 4\times 9+2\times 3.14\times 4^2\\\\A=326.56\ m^2

The cost of pillar at the rate of $87 per cubic meter will be :

C=326.56\times 87\\\\C=\$28410.72

8 0
3 years ago
Find the value of csc theta, if cos =-3/5; 90 degrees &lt; theta &lt; 180 degrees
Sloan [31]

Answer:\frac{5}{4}

Step-by-step explanation:

Given

\cos \theta =\frac{-3}{5}

and \theta  lies between

90^{\circ}

and for this \theta,\sin and \text{cosec} is Positive as they lie in 2 nd Quadrant

\sin ^2\theta +\cos ^2\theta =1

\sin ^2\theta =1-(\frac{-3}{5})^2

\sin ^2\theta =1-\frac{9}{25}

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\sin \theta =\frac{4}{5}

\therefore \text{cosec}\ \theta =\frac{5}{4}

6 0
2 years ago
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