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Sidana [21]
3 years ago
5

What is the solution of d²–12 = 0?​

Mathematics
1 answer:
Mkey [24]3 years ago
6 0

d²-12=0

d²=12

d=√12

d=3.46..

Hope it helps! :)

Brainliest?

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Read 2 more answers
Tan(2 sin^-1 0.4)<br> Find the Exact value
stiv31 [10]

Answer:

tan(2u)=[4sqrt(21)]/[17]

Step-by-step explanation:

Let u=arcsin(0.4)

tan(2u)=sin(2u)/cos(2u)

tan(2u)=[2sin(u)cos(u)]/[cos^2(u)-sin^2(u)]

If u=arcsin(0.4), then sin(u)=0.4

By the Pythagorean Identity, cos^2(u)+sin^2(u)=1, we have cos^2(u)=1-sin^2(u)=1-(0.4)^2=1-0.16=0.84.

This also implies cos(u)=sqrt(0.84) since cosine is positive.

Plug in values:

tan(2u)=[2(0.4)(sqrt(0.84)]/[0.84-0.16]

tan(2u)=[2(0.4)(sqrt(0.84)]/[0.68]

tan(2u)=[(0.4)(sqrt(0.84)]/[0.34]

tan(2u)=[(40)(sqrt(0.84)]/[34]

tan(2u)=[(20)(sqrt(0.84)]/[17]

Note:

0.84=0.04(21)

So the principal square root of 0.04 is 0.2

Sqrt(0.84)=0.2sqrt(21).

tan(2u)=[(20)(0.2)(sqrt(21)]/[17]

tan(2u)=[(20)(2)sqrt(21)]/[170]

tan(2u)=[(2)(2)sqrt(21)]/[17]

tan(2u)=[4sqrt(21)]/[17]

7 0
3 years ago
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