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Greeley [361]
3 years ago
8

Draw linearequation of x=4 and y=5.find the area formed by 2 graph and the axes

Mathematics
1 answer:
RSB [31]3 years ago
7 0
Its a triangle. 
<span>Area = 1/2bh
b=11
h=4
a = 1/2(11*4) = 1/2(44)= 22

Are looking for that?

</span>
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Find the solution to the system of equations. <br> y= -5x+6<br><br> y=3x-2<br><br> x=<br> y=
Neporo4naja [7]

Answer:

x=1, y=1. (1, 1).

Step-by-step explanation:

-5x+6=3x-2

-5x-3x+6=-2

-8x+6=-2

-8x=-2-6

-8x=-8

8x=8

x=8/8

x=1

y=3(1)-2=3-2=1

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Please help me please I don’t understand and my teacher said I got it wrong for numbers 1 2 and 3
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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]

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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]

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