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2 years ago

What is the answer I have to type in the yellow boxes?

Mathematics

2 answers:

kap26 [50]2 years ago

7 0

$\\ \rm\bull\rightarrowtail y=(1x^2)+3(x)$

$\\ \rm\bull\rightarrowtail y=1(-6)^2+3(-6)$

$\\ \rm\bull\rightarrowtail y=1(36)+(-18)$

Note the rule

(+)(-)=(-)

$\\ \rm\bull\rightarrowtail y=36-18$

$\\ \rm\bull\rightarrowtail y=18$

hichkok12 [17]2 years ago

5 0

Answer:

36 -18

Step-by-step explanation:

This is slightly confusing, since the question says "find the y coordinate if x = -4" yet underneath if appears they want x = -6 !

If x = -6

Substitute -6 into the equation in place of x

⇒ y = 1 × (-6)² + 3(-6)

Apply the rule (-a)² = a² and a(-b) = -ab:

⇒ y = 1 × (6)² + -3×6

⇒ y = 1 × 36 + -18

Using order of operations, carry out the multiplication first:

⇒ y = 36 + -18

Then the addition (subtraction)

⇒ y = 36 - 18

⇒ y = 18

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3 years ago

Consider F and C below. F(x, y, z) = yz i + xz j + (xy + 6z) k C is the line segment from (2, 0, −2) to (5, 4, 2) (a) Find a fun

WITCHER [35]

Answer:

Required solution (a) $f(x,y,z)=xyz+3z^2+C$ (b) 40.

Step-by-step explanation:

Given,

$F(x,y,z)=yz \uvec i +xz\uvec j+(xy+6z)\uvex k$

(a) Let,

$F(x,y,z)=yz \uvec i +xz\uvec j+(xy+6z)\uvex k=f_x \uvec{i} +f_y \uvex{j}+f_z\uvec{k}$

Then,

$f_x=yz,f_y=xz,f_z=xy+6z$

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Differentiate this with respect to y we get,

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$g'(y,z)=0\implies g(y,z)=0+h(z)\implies g(y,z)=h(z)$

Then,

$f(x,y,z)=xyz+h(z)$

Again differentiate with respect to z we get,

$f_z=xy+h'(z)=xy+6z$

on compairing we get,

$h'(z)=6z\implies h(z)=3z^2+C$ (By integrating h'(z)) where C is integration constant. Hence,

$f(x,y,z)=xyz+3z^2+C$

(b) Next, to find the itegration,

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3 years ago

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Jobisdone [24]

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7 0

3 years ago

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