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lana [24]
3 years ago
12

Someone please help me

Mathematics
1 answer:
timama [110]3 years ago
6 0
A CANNOT DETERMINE, since you have no other data at all
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HALP ME PLEASE VERY IMPORTANT ASSIGNMENT!!!
german

Answer:

wut?

Step-by-step explanation:

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2 years ago
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Please help me this is urgent
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Answer:

99.9

Step-by-step explanation:

8 0
3 years ago
The Saxena family plans to install a light to illuminate part of their rectangular yard. Nikki and Dylan each proposed a differe
monitta

Options:

a. Nikki's proposed placement will light a greater area than Dylan's placement.

b.  Dylan's proposed placement will light a greater area than Nikki's placement.

c.  Both proposed placements will light the same sized area.

d. Nikki's proposed placement will light more than half the yard.

e. Dylan's proposed placement will light more than half the yard.

f.  Dylan's proposed placement will light exactly half of the yard.

g. Nikki's proposed placement will light less than half of the yard.

Answer:

C) Both proposed placements will light the same sized area.

F) Dylan's proposed placement will light exactly half of the yard.

Step-by-step explanation:

The area of a triangle is (base x height) / 2, and both lights illuminate the same base and height = (60 x 38) / 2 = 1,140 sq ft

Both Dylan's and Nikki's proposed placement will lit exactly half of the yard. The yard's total area = 60 x 38 = 2,280 sq ft, which is twice the area lit by the lights.

3 0
3 years ago
If tanA=a <br>then find sin4A-2sin2A/ sin4A+2sin2A​
anygoal [31]

Answer:

The value of the given expression is

\frac{sin4A-2sin2A}{sin4A+2sin2A}=-a^2

Step by step Explanation:

Given that tanA=a

To find the value of \frac{sin4A-2sin2A}{sin4A+2sin2A}

Let us find the value of the expression :

\frac{sin4A-2sin2A}{sin4A+2sin2A}=\frac{2cos2Asin2A-2sin2A}{2cos2Asin2A+2sin2A} ( by using the formula sin2A=2cosAsinA here A=2A)  

=\frac{2sin2A(cos2A-1)}{2sin2A(cos2A+1)}

=\frac{(cos2A-1)}{(cos2A+1)}

=\frac{(-(1-cos2A))}{(1+cos2A)}(using  sin^2A+cos^2A=1  here A=2A)

=\frac{-(sin^2A+cos^2A-(cos^2A-sin^2A))}{sin^2A+cos^2A+(cos^2A-sin^2A)}(using cos2A=cos^2A-sin^2A here A=2A)

=\frac{-(sin^2A+cos^2A-cos^2A+sin^2A)}{sin^2A+cos^2A+(cos^2A-sin^2A)}

=\frac{-(sin^2A+sin^2A)}{cos^2A+cos^2A}

=\frac{-2sin^2A}{2cos^2A}

=-\frac{sin^2A}{cos^2A}

=-tan^2A  ( using tanA=\frac{sinA}{cosA} here A=2A )

 =-a^2 (since tanA=a given )

Therefore \frac{sin4A-2sin2A}{sin4A+2sin2A}=-a^2

6 0
3 years ago
Tape Diagrams and Writing Equations
notka56 [123]

Answer:

3x + 16 = 22.75

3 = 22.75 - 16

<u>3</u><u>×</u><u> </u>= <u>6</u><u>.</u>75

3 3

x= 2.25

8 0
2 years ago
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