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Xelga [282]
3 years ago
5

Please EXPLAIN how to get this answer so I can do it with other problems. A straight answer would be great, but I have to do abo

ut 30 more problems like these, so I'd like to know how to do it with an explanation! It doesn't have to be long, just so I can understand it. (This is for geometry by the way)

Mathematics
1 answer:
OLEGan [10]3 years ago
6 0

Answer:

PQ = 3.58, and RQ = 10.4

Step-by-step explanation:

We are given the hypotenuse of the triangle, and an angle. Use sin and cos to solve.  

Hypotenuse = 11,

Opposite side is PQ

Adjacent side is RQ

x = 19

Sin x  = (opposite side)/(hypotenuse)

Cos x = (adjacent side)/(hypotenuse)

For PQ, this is the side opposite to the angle, so use sin,

Sin 19 = x/11

   11(Sin 19) = x

        3.58 = x        (rounded to the nearest hundredth)            

For RQ, this is the side adjacent to the angle, so use cos,

Cos 19 = x/11

   11(Cos 19) = x

        10.4 = x        (rounded to the nearest hundredth)        

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Calculate the rate of change from the graph below
qaws [65]

Answer:

The rate of change is 0.75 gallons per minute

Step-by-step explanation:

The rate of change of the linear relation is represented by the slope of the line which represents this relation

The rule of the slope of a line is

m = Δy/Δx, where

  • Δy is the vertical change
  • Δx is the horizontal change

From the given graph

∵ The line passes through points (0, 0) and (4, 3)

∴ Δx = 4 - 0 = 4

∴ Δy = 3 - 0 = 3

→ Use the rule of the slope above to find the slope of the line

∴ m = \frac{3}{4} = 0.75

∵ The x-axis represents the time in minutes

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7 0
3 years ago
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Stella [2.4K]
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7 0
3 years ago
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il63 [147K]

Answer:

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6 0
3 years ago
Solve. Show work please. Passing through (-4,-1) and (3,4)
kvasek [131]
What are you asking exactly
3 0
3 years ago
H=−4.9t2+25t
lina2011 [118]

ANSWER

5

EXPLANATION

The equation that expresses the approximate height h, in meters, of a ball t seconds after it is launched vertically upward from the ground is

h(t) =  - 4.9 {t}^{2}  + 25t

To find the time when the ball hit the ground,we equate the function to zero.

- 4.9 {t}^{2}  + 25t = 0

Factor to obtain;

t( - 4.9t + 25) = 0

Apply the zero product property to obtain,

t = 0 \: or \:  \:  - 4.9t + 25 = 0

t = 0 \:  \: or \:  \: t =  \frac{ - 25}{ - 4.9}

t=0 or t=5.1 to the nearest tenth.

Therefore the ball hits the ground after approximately 5 seconds.

4 0
3 years ago
Read 2 more answers
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