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3 years ago

Neeeeeeed helpppppppppppppp

Mathematics

2 answers:

r-ruslan [8.4K]3 years ago

5 0

Step-by-step explanation:

Simplify the expressions

$2 {}^{3} - 2 {}^{1} = 2 {}^{3} - 2 = 8 - 2 = 6$

$2 + 3 = 5$

${3}^{2} = 9$

So the expressions from least to greatest is

$2 {}^{1} + 3 {}^{1} < 2 {}^{3} - 2 {}^{1} < 3 {}^{2}$

8_murik_8 [283]3 years ago

3 0

Answer:

$2^{1} + 3^{1}$ , $2^{3} - 2^{1}$ , $3^{2}$

Step-by-step explanation:

$3^{2}$ = 9

$2^{3} - 2^{1}$ = 6

$2^{1} + 3^{1}$ = 5

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Igoryamba

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3 years ago

(a) Find parametric equations for the line through (1, 4, 4) that is perpendicular to the plane x − y + 2z = 7. (Use the paramet

Alex777 [14]

Answer:

$x=1+t\\y=4-t\\z=4+2t$

Step-by-step explanation:

A vector perpendicular to the plane ax+by+cz+d=0 is of the form $(a,b,c)$ .

So, a vector perpendicular to the plane x − y + 2z = 7 is $(1,-1,2)$ .

The parametric equations of a line through the point $(x_0,y_0,z_0)$ and parallel to the vector $(a,b,c)$ are as follows:

$x=x_0+at\\y=y_0+bt\\z=z_0+ct$

Put $(x_0,y_0,z_0)=(1,4,4)$ and $(a,b,c)=(1,-1,2)$

Therefore,

$x=1+t\\y=4-t\\z=4+2t$

xy-plane:

Put z = 0 ⇒ t = -2 ⇒x = - 1 , y = 6

So, at point (-1,6,0)

yz-plane:

Put x = 0 ⇒ t = -1 ⇒ y = 5, z =2

So, at point (0,5,2)

xz-plane:

Put y = 0 ⇒ t = 4 ⇒ x = 5, z = 12

So, at point (5,0,12)

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3 years ago