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muminat
3 years ago
12

Solve the following systems of equations using substitution: x = 6 y = 2x - 3

Mathematics
2 answers:
Stolb23 [73]3 years ago
6 0

Answer:

y=9

Step-by-step explanation:

2(6)-3=9

Anvisha [2.4K]3 years ago
5 0
The answer is y=9
Y=2x-3
Y=2(6)-3
Y=12-3
Y=9
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The problem 10bf + 25bg – 21p – 14pq cannot be factored because it cannot be simplifed further. This is because there are no common factors between in any of the terms.

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3 years ago
NGC is a hot air balloon in the sky from her spot on the ground. The angle of elevation from Angie to the balloon is 40°. If she
Goshia [24]

Answer:

The distance from hot air balloon to the ground  AC=50.1457 ft.

Step-by-step explanation:

Labelled diagram of given scenario is shown below.

Given that,

An angle of elevation of Hot air balloon by Angie is 40°.

When she  stepped back 200 ft then angle of elevation was 10°.

Height of Angie is  5.5 ft.

To find: How far off the ground is a hot air balloon.

So, from figure

Height of Angie (BC) = 5.5ft

  In triangle ΔABD,

         ⇒                    tan {40\si {\degree}} = \frac{AB}{BD}

         ⇒                      AB = tan (40) \times BD          ....................(1)

Now, In triangle ΔABE

         ⇒                  tan (10)= \frac{AB}{200+BD}

         ⇒                 AB = tan(10)\times (200+BD)

Here, substituting the value AB from Equation (1) we get,

                        tan(10)\times (200+BD)= tan(40)\times BD

                      tan(10)\times 200+tan(10)\times BD= tan(40)\times BD

        ⇒         BD\times (tan(40)-tan(10))=tan(10)\times 20        

        ⇒                   BD\times 0.6628 = 35.2654

        ⇒                       BD = \frac{35.2654}{0.6628} =53.2067 ft        

Now, finding the value of AB from equation (1)

                                 AB= tan(40)\times 53.2067=0.8391\times 53.2067

                                 AB=44.6457 ft              

Therefore Length of AC= AB+BC = (44.6457 + 5.5) ft

                                  AC=50.1457 ft.

Hence,

The distance from hot air balloon to the ground  AC=50.1457 ft.                

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