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ICE Princess25 [194]
3 years ago
12

Find the value of x.

Mathematics
1 answer:
Dafna1 [17]3 years ago
3 0

Answer:

16

Step-by-step explanation:

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Need help ASAP Math, Equations
soldier1979 [14.2K]
1/4 (a) = (2/3)

a = (2/3) * (4)

a=8/3 

a= 2 2/3
3 0
3 years ago
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Which statements are true about these lines? Select three options.
shepuryov [24]
RS is perpendicular to MN and PQ.


We can use the slopes of these lines to determine the answer.
Slope is given by the formula
m=.

Using the coordinates for M and N, we have:
m=.

Since PQ is parallel to MN, its slope will be as well, since parallel lines have the same slope.

Using the coordinates for points T and V in the slope formula, we have
m=.

This is not parallel to MN or PQ, since the slopes are not the same.
We can also say that it is not perpendicular to these lines; perpendicular lines have slopes that are negative reciprocals (they are opposite signs and are flipped). This is not true of TV either.

Using the coordinates for R and S in the slope formula, we have
m=. Comparing this to the slope of RS, it is flipped and the sign is opposite; they are negative reciprocals, so they are perpendicular.
8 0
4 years ago
Read 2 more answers
Which expressions are equivalent to when x0? Check all that apply.
Genrish500 [490]
We have that

\frac{(x+4)}{3} / \frac{6}{x} = \frac{x*(x+4)}{3*6} \\ \\ = \frac{( x^{2} +4x)}{18}

therefore

case a) 
\frac{(x+4)}{3} * \frac{x}{6}
Is equivalent

case b) 
\frac{6}{x} * \frac{(x+4)}{3}
Is not equivalent

case c) 
\frac{x}{6} * \frac{(x+4)}{3}
Is  equivalent

case d) 
\frac{(2 x^{2} +4x)}{6}
Is not equivalent

case e) 
\frac{(2 x^{2} +4x)}{18}
Is equivalent

Hence

the answer is

\frac{(x+4)}{3} * \frac{x}{6}

\frac{x}{6} * \frac{(x+4)}{3}

\frac{(2 x^{2} +4x)}{18}
3 0
3 years ago
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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

Integrating f_x we get,

f(x,y,z)=xyz+g(y,z)

Differentiate this with respect to y we get,

f_y=xz+g'(y,z)

compairinfg with f_y=xz of the given function we get,

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,

\int_C \vec{F}.dr=\int_C \nabla f. d\vec{r}=f(5,4,2)-f(2,0,-2)=(52+C)-(12+C)=40

3 0
3 years ago
Hi people I need help on this question can y’all help me pls!!!!!! Thanks
skad [1K]
I’m pretty sure it’s 43.96 I may be wrong though!
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3 years ago
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