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4 years ago

PLS HELP ME ASAP FOR 1, 2, 3, and 4 (SHOW WORK!!) + LOTS OF POINTS!!

Mathematics

1 answer:

Lera25 [3.4K]4 years ago

5 0

1/2in = 16 mi
2 3/4 in without whole numbers [(denominator ×whole) + numerator] / denominator
4×2 = 8 +3 = 11/4
Find common denominator for 1/2 and 11/4 (common D Is 4)
Multiply 1/2 by 2/2
2/4 in = 16 mi divide this by 2
2/4 ×1/2 = 1/4
16 /2 = 8
1/4 = 8
1/4 goes into 11/4 11 times
Multiply 8 times 11
88 miles (D)

End zone to end zone is = length so his scale is 6 in and in real life it is 360 ft
To find scale divide 360 by 6
1in = 60 ft (B)

160 ft wide
(divide 160 by 60 ft ) fraction
160/60 is a fraction now simplify
16/6
Common factors of both numbers is 2
Simplify by 2
8/3
Make into whole number fraction
3 goes into 8 two times (2×3=6)(2 becomes whole number) while there's is 2 left over( 8 -6=2) (becoming the numerator)
2 2/3 in (B)

120 ft long
80 ft wide

24 in is scale of "long"
Divide 120 by 24
120÷24=5
So 5in in scale equals 1ft in real life
To find width in scale divide 80 by 5
80/5= 16in (A)

If you have any trouble understanding feel free to let me know I will try to explain as best as I can :)

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Answer:

Match each of the trigonometric expressions below with theequivalent non-trigonometric function from the following list.Enter the appropiate letter(A,B, C, D or E)in each blank

A . tan(arcsin(x/8))

B . cos (arsin (x/8))

C. (1/2)sin (2arcsin (x/8))

D . sin ( arctan (x/8))

E. cos (arctan (x/8))

These are the spaces to fill out :

.. ..........x/64 (sqrt(64-x^2))

.............x/sqrt(64+x^2)

.............sqrt(64-x^2)/8

..............x/sqrt(64-x^2)

..............8/sqrt(64+x^2)

A. ........tan(arcsin(x/8)) =......x/sqrt(64-x^2)

B . cos (arsin (x/8)) ....sqrt(64-x^2)/8

Step-by-step explanation:

To solve this we have to find the missing sides to each of the triange discribed in prenthesis thus

A we have the sides of the triangle given by x, 8 and $\sqrt{8^{2} - x^{2} }$ or $\sqrt{64 - x^{2} }$

thus tan(arcsin(x/8)) = $\frac{x}{\sqrt{64 - x^{2} }}$ =

Therefore ........tan(arcsin(x/8)) =......x/sqrt(64-x^2)

Here we have cos = adjacent/hypotenuse where adjacent side is $\sqrt{64 - x^{2} }$ and hypothenuse = 8 we have $\sqrt{64 - x^{2} }$ /8

B . cos (arsin (x/8)) ....sqrt(64-x^2)/8

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