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never [62]
3 years ago
9

The track club runs the 50-yard dash. How many inches do they run?

Mathematics
2 answers:
kiruha [24]3 years ago
8 0

Answer:

1800in

Step-by-step explanation:

3ft=1yd

3x50=150ft

12in=1ft

12x150=1800in

nignag [31]3 years ago
8 0

Answer:

Step-by-step explanation:

1 yard = 36 inches

50 yards = 36 x 50 = 1800 inches

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The graph shows the solution to which system of inequalities?
Misha Larkins [42]

Answer:

The orange one is the answer

Step-by-step explanation:

The easiest way to find the solution is to plug in one of the coordinates that is located in the shaded area into each equation. If the equations make sense, as they do in the orange option, you have your answer.

Hope this helps you!

7 0
3 years ago
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3 years ago
Given: △ABC; AB=BC, m∠BDA = 60°, BD=4 cm, BD ⊥ BA . Find: DC, AC.
vekshin1

Answer:

DC = 10.93 cm ,  AC = 9.8 cm

Step-by-step explanation:

From trigonometry;

⇒ Tan 60 = AB/BD

⇒AB = BD Tan 60 ( where BD = 4 cm )

⇒ AB = 6.93 cm

Also, AB=BC , therefore;

⇒ BC = 6.93 cm

⇒ Cos 60 = BD/AD

⇒ AD = BD/ Cos 60 = 4/Cos 60

⇒ AD = 8 cm

From Pythagoras theorem;

⇒ AC^{2} = AB^{2} + BC^{2} = (6.93)^{2} + (6.93)^{2}

⇒ AC = \sqrt{96.05} = 9.80 cm

⇒ DC = BD + BC = 4 + 6.93

⇒ DC = 10.93 cm

8 0
3 years ago
Someone please help me with #2
mario62 [17]

<em>Good luck, you'll need it :)</em>

6 0
3 years ago
Please help me with these, oh sweet jesus
Lelechka [254]

Answer:

77.  \cot^{6} x = \cot^{4} x \csc^{2}x - \cot^{4} xProved

78.  \sec^{4}x \tan^{2} x = \sec^{2}x [\tan^{2}x + \tan^{4}x ] Proved

79. \cos^{3} x\sin^{2} x = [\sin^{2}x - \sin^{4}x] \cos x Proved.

80. \sin^{4}x - \cos^{4}x = 1 - 2\cos^{2}x + 2 \cos^{4} x Proved.

Step-by-step explanation:

77. Left hand side

= \cot^{6} x

= \cot^{4} x \times \cot^{2} x

= \cot^{4}x [\csc^{2}x - 1]  

{Since we know, \csc^{2} x - \cot^{2}x = 1}

= \cot^{4} x \csc^{2}x - \cot^{4} x  

= Right hand side (Proved)

78. Left hand side

= \sec^{4}x \tan^{2} x

= \sec^{2} x [1 + \tan^{2}x] \tan^{2} x  

{Since \sec^{2}x - \tan^{2}x = 1}

= \sec^{2}x [\tan^{2}x + \tan^{4}x ]

= Right hand side (Proved)

79. Left hand side  

= \cos^{3} x\sin^{2} x

= \cos x[1 - \sin^{2} x] \sin^{2} x

{Since \sin^{2}x + \cos^{2} x = 1}

= [\sin^{2}x - \sin^{4}x] \cos x

= Right hand side

80. Left hand side  

= \sin^{4}x - \cos^{4}x

= [\sin^{2}x + \cos^{2}x]^{2} - 2\sin^{2} x \cos^{2}x

{Since \sin^{2}x + \cos^{2} x = 1}

= 1 - 2\cos^{2} x[1 - \cos^{2}x ]

= 1 - 2\cos^{2}x + 2 \cos^{4} x

= Right hand side. (Proved)

7 0
3 years ago
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