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Salsk061 [2.6K]

3 years ago

12

In the high jump event at the city track meet ronald successfully jumped 4 feet and 6 inches. Luis successfully jumped 3 feet an

d 9 inches. who had the higher jump and by how much

1 answer:

Genrish500 [490]3 years ago

8 0

Answer:

Ronald by 9 inches

Step-by-step explanation:

Ronald jumped a total of 54inches

Luis jumped a total of 45 inches

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Solve tan 10 - tan 50 +tan 70 with trigonometry.

Jobisdone [24]

Below is the solution, I hope it helps.

<span>i) tan(70) - tan(50) = tan(60 + 10) - tan(60 - 10)

= {tan(60) + tan(10)}/{1 - tan(60)*tan(10)} - {tan(60) - tan(10)}/{1 + tan(10)*tan(60)}

ii) Taking LCM & simplifying with applying tan(60) = √3, the above simplifies to:

= 8*tan(10)/{1 - 3*tan²(10)}

iii) So tan(70) - tan(50) + tan(10) = 8*tan(10)/{1 - 3*tan²(10)} + tan(10)

= [8*tan(10) + tan(10) - 3*tan³(10)]/{1 - 3*tan²(10)}

= [9*tan(10) - 3*tan³(10)]/{1 - 3*tan²(10)}

= 3 [3*tan(10) - tan³(10)]/{1 - 3*tan²(10)}

= 3*tan(30) = 3*(1/√3) = √3 [Proved]

[Since tan(3A) = {3*tan(A) - tan³(A)}/{1 - 3*tan²(A)},
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3 years ago

0.03is 10 times as much as

PSYCHO15rus [73]

Answer:

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Step-by-step explanation:

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Find the surface area of the following figure

Julli [10]

Answer:

<h2>510.4 ft²</h2>

Step-by-step explanation:

We have:

two trapezoids with bases 15ft and 7ft and height 5ft.

four rectangles 5ft × 11ft, 15ft × 11ft, 9.4ft × 11ft and 7ft × 11ft.

The formula of an area of a trapezoid:

$A_t=\dfrac{b_1+b_2}{2}\cdot h$

b₁, b₂ - bases

h - height

Substitute:

$A_t=\dfrac{15+7}{2}\cdot5=\dfrac{22}{2}\cdot5=(11)(5)=55\ ft^2$

The formula of an area of a rectangle:

$A_r=lw$

l - length

w - width

The dimensions of rectangle l × w

Subtitute:

$A_1=(5)(11)=55\ ft^2\\\\A_2=(15)(11)=165\ ft^2\\\\A_3=(9.4)(11)=103.4\ ft^2\\\\A_4=(7)(11)=77\ ft^2$

The surface area of the figure:

[tex]S.A.=2A_t+A_1+A_2+A_3+A_4\\\\S.A.=2(55)+55+165+103.4+77=510.4\ ft^2[/text]

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As we know,

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