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Vikentia [17]
3 years ago
11

I need help with this problem

Mathematics
2 answers:
topjm [15]3 years ago
8 0
The answer to this problem is C
aleksandr82 [10.1K]3 years ago
3 0
The answer to this question is x/2
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Can I get help on number 7!!!!!
Ratling [72]

Answer:

C =(43/55)R; C - R = 8t; 48 mi  

Step-by-step explanation:

Let C = the distance a coyote can run in t hours.

Let R = the distance a rabbit can run in t hours.

(1) C = 43t

(2) R = 35t

A. <em>Two equivalent expressions </em>

(a)   C/R = 43/35

           C = (43/55)R

(b) C – R = 43t – 35t

    C – R  = 8t

B. <em>Distance advantage of coyote </em>

Let the difference D = C - R

D = 8t

D = 8 × 6

D = 48 mi

In 6 h, the coyote can run 48 mi further than the rabbit.

3 0
3 years ago
Whats the product of 4.39x0.5
rusak2 [61]

Answer:

2.195

Step-by-step explanation:

Use a calculator

3 0
3 years ago
Read 2 more answers
A 50 ft flagpole is mounted to the top of a building. If the angle of elevation from a spot on the street to the top of the pole
monitta

Answer:

the height of the building = 91.67161722 feet.

Step-by-step explanation:

Suppose the height of the building (BC) = X feet.

A 50 ft flagpole (AB) is mounted to the top of a building.

So, height of the top of flag above the ground (AC) = (X+50) feet.

If the angle of elevation from a spot (P) on the street to the top of the pole is 58 degrees and the angle of elevation from the same spot (P) to the bottom of the pole is 46 degrees.

It means ∠APC = 58° and ∠BPC = 46°.

Considering Right triangle ΔACP, cot(∠APC) = PC / AC.

PC = AC*cot(∠APC) = (X+50)*cot(58°)

Considering Right triangle ΔBCP, cot(∠BPC) = PC / BC.

PC = BC*cot(∠BPC) = X*cot(46°)

We have PC = PC.

X*cot(46°) = (X+50)*cot(58°)

0.965688774 * X = 0.624869351 * (X+50)

(0.965688774 - 0.624869351) * X = 0.624869351 * 50

0.340819422 * X = 31.2434676

X = 31.2434676 / 0.340819422 = 91.67161722 feet.

Hence, the height of the building = 91.67161722 feet.

6 0
3 years ago
Find two other pairs of polar coordinates of the given polar coordinate, one with r &gt; 0 and one with r &lt; 0.
navik [9.2K]

Answer:

A) (5, 7π/4):   (5, 15π/4) and (-5, 3π/4)

B) (−6, π/2):   (−6, 5π/2) and (6, -π/2)

C) (5, −2):       (5, 2π-2) and (-5, -π-2)

Step-by-step explanation:

To find pair of coordinates for (r>0), add 2π to corresponding θ. For (r<0) subtract π from given angle to find second pair of coordinate.

A)  (5, 7π/4)

For (r>0)

θ =  7π/4 + 2π

θ =  15π/4

Point is (5, 15π/4)

For (r<0)

θ =  7π/4 - π

θ =  3π/4

Point is (-5, 3π/4)

As it can be seen in Fig 1

(5, 7π/4)=(5, 15π/4)=(-5, 3π/4)

B)  (−6, π/2)

For (r>0)

θ =  π/2 + 2π

θ =  5π/2

Point is (6, 5π/2)

For (r<0)

θ =  π/2 - π

θ =  -π/2

Point is (-6, -π/2)

As shown in fig 2

(-6, π/2) = (6, 5π/2) = (-6, -π/2)

C) (5, −2)

For (r>0)

θ =  -2 + 2π

Point is (5, 2π-2)

For (r<0)

θ =  -2 - π

Point is (-5, -π)

As shown in Fig. 3

(5, −2) = (5, 2π-2) = (-5, -π)

5 0
4 years ago
Ran 2 miles Jesse ran four times as far there are 5280 feet in a mile how many feet did Jesse run
ella [17]
2 times 4= 8

8 times 5280= 42240
5 0
3 years ago
Read 2 more answers
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