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3 years ago

Can someone for once ACTUALLY ANSWER THIS!I REALLY NEED HELP

Mathematics

1 answer:

ivolga24 [154]3 years ago

5 0

Answer: r - radius of a circle

d - diagonal of a square inscribed in a circle

a - side of a square inscribed in a circle

shaded area - area of a square inscribed in a circle

The answer will basically be 9

Step-by-step explanation:

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PLS HELP ASAP FEJKRKJGR My bad dont mind the blue bubble

kondor19780726 [428]

Answer:

Step-by-step explanation:

yellow

6+6+3=15

for green

4+4+2=10

8 0

3 years ago

Read 2 more answers

Surface area measures the area of the faces. Area is a blank dimensional measurement, so it is measured in square units.

dlinn [17]

Answer:

2d or 2 dimensional is the answer i think

Step-by-step explanation:

since area is measured for 2d shapes and surface area is just finding the area of each side separately and adding it together

6 0

3 years ago

K + 4(5k +6) = - 123 <br> HELP ASAP

Anni [7]

Answer:

-7

Step-by-step explanation:

-123-(6x4=24)=-147 -147 divided by (4x5k+1k=21k)=-7

4 0

2 years ago

) f) 1 + cot²a = cosec²a

notsponge [240]

Answer:

It is an identity, proved below.

Step-by-step explanation:

I assume you want to prove the identity. There are several ways to prove the identity but here I will prove using one of method.

First, we have to know what cot and cosec are. They both are the reciprocal of sin (cosec) and tan (cot).

$\displaystyle \large{\cot x=\frac{1}{\tan x}}\\\displaystyle \large{\csc x=\frac{1}{\sin x}}$

csc is mostly written which is cosec, first we have to write in 1/tan and 1/sin form.

$\displaystyle \large{1+(\frac{1}{\tan x})^2=(\frac{1}{\sin x})^2}\\\displaystyle \large{1+\frac{1}{\tan^2x}=\frac{1}{\sin^2x}}$

Another identity is:

$\displaystyle \large{\tan x=\frac{\sin x}{\cos x}}$

Therefore:

$\displaystyle \large{1+\frac{1}{(\frac{\sin x}{\cos x})^2}=\frac{1}{\sin^2x}}\\\displaystyle \large{1+\frac{1}{\frac{\sin^2x}{\cos^2x}}=\frac{1}{\sin^2x}}\\\displaystyle \large{1+\frac{\cos^2x}{\sin^2x}=\frac{1}{\sin^2x}}$

Now this is easier to prove because of same denominator, next step is to multiply 1 by sin^2x with denominator and numerator.

$\displaystyle \large{\frac{\sin^2x}{\sin^2x}+\frac{\cos^2x}{\sin^2x}=\frac{1}{\sin^2x}}\\\displaystyle \large{\frac{\sin^2x+\cos^2x}{\sin^2x}=\frac{1}{\sin^2x}$