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3 years ago

Tell whether the equation has one, zero, or infinitely many solutions.2(x-1/2)=8+2x

Mathematics

2 answers:

docker41 [41]3 years ago

4 0

Answer: none

Step-by-step explanation:

icang [17]3 years ago

3 0

Answer:

no solutions

Step-by-step explanation:

i hope this helps

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What do the following terms mean: is equal the same as = or x or - or +

JulsSmile [24]

Blue: = because the equal sign is stating that it’s the same.

7 0

2 years ago

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help me brainlist if you right Mona is inviting 10 friends to a party. Each friend wants 5 cookies and each box has 10 cookies.

seropon [69]

Answer:

its 5 boxes

Step-by-step explanation:

5 0

3 years ago

Use polar coordinates to find the volume of the given solid. Inside both the cylinder x2 y2 = 1 and the ellipsoid 4x2 4y2 z2 = 6

Anton [14]

The Volume of the given solid using polar coordinate is: $\frac{-1}{6} \int\limits^{2\pi}_ {0} [(60) ^{3/2} \; -(64) ^{3/2} ] d\theta$

V= $\frac{-1}{6} \int\limits^{2\pi}_ {0} [(60) ^{3/2} \; -(64) ^{3/2} ] d\theta$

<h3>What is Volume of Solid in polar coordinates?</h3>

To find the volume in polar coordinates bounded above by a surface z=f(r,θ) over a region on the xy-plane, use a double integral in polar coordinates.

Consider the cylinder, $x^{2}+y^{2} =1$ and the ellipsoid, $4x^{2}+ 4y^{2} + z^{2} =64$

In polar coordinates, we know that

$x^{2}+y^{2} =r^{2}$

So, the ellipsoid gives

$4{(x^{2}+ y^{2)} + z^{2} =64$

4( $r^{2}$ ) + $z^{2}$ = 64

$z^{2}$ = 64- 4( $r^{2}$ )

z=± $\sqrt{64-4r^{2} }$

So, the volume of the solid is given by:

V= $\int\limits^{2\pi}_ 0 \int\limits^1_0{} \, [\sqrt{64-4r^{2} }- (-\sqrt{64-4r^{2} })] r dr d\theta$

= $2\int\limits^{2\pi}_ 0 \int\limits^1_0 \, r\sqrt{64-4r^{2} } r dr d\theta$

To solve the integral take, $64-4r^{2}$ = t

dt= -8rdr

rdr = $\frac{-1}{8} dt$

So, the integral $\int\ r\sqrt{64-4r^{2} } rdr$ become

= $\int\ \sqrt{t } \frac{-1}{8} dt$

= $\frac{-1}{12} t^{3/2}$

= $\frac{-1}{12} (64-4r^{2}) ^{3/2}$

so on applying the limit, the volume becomes

V= $2\int\limits^{2\pi}_ {0} \int\limits^1_0{} \, \frac{-1}{12} (64-4r^{2}) ^{3/2} d\theta$

= $\frac{-1}{6} \int\limits^{2\pi}_ {0} [(64-4(1)^{2}) ^{3/2} \; -(64-4(2)^{0}) ^{3/2} ] d\theta$

V = $\frac{-1}{6} \int\limits^{2\pi}_ {0} [(60) ^{3/2} \; -(64) ^{3/2} ] d\theta$

Since, further the integral isn't having any term of $\theta$ .

we will end here.

The Volume of the given solid using polar coordinate is: $\frac{-1}{6} \int\limits^{2\pi}_ {0} [(60) ^{3/2} \; -(64) ^{3/2} ] d\theta$

Learn more about Volume in polar coordinate here:

brainly.com/question/25172004

#SPJ4

3 0

2 years ago

What is the measure of angle 3 if the measure of

ELEN [110]

Answer:

Angle 3 = 86.8

Angle 6 = 96.4

Angle 7 = 93.2

Step-by-step explanation:

Given :

angle 6 = 4x + 10

angle 7 = 2x + 40

=> angle 6 + angle 7 = 180° { linear pair }

=> 4x+10 +2x+40 = 180

=> 6x+50 = 180

=> 6x = 180-50

=> 6x = 130

=> x = 130/6

=> x = 21.6

so ,the measure of angle 6 = 4x + 10 = 4(21.6) +10 = 86.4+10 = 96.4

the measure of angle 7 = 2x + 40 = 2(21.6) +40 = 43.2+50 = 93.2

now angle 7 = angle 4 + angle 5

and angle 3 + angle 4 + angle 5 = 180°

so, angle 7 + angle 3 = 180

=> 93.2 + angle 3 = 180

=> angle 3 = 180 - 93.2

=> angle 3 = 86.8

8 0

2 years ago

If f(x) = -x + 5 and g(x) = x^2, what is (g° (-1)?

ipn [44]

Answer:

$(g\circ f)(-1)=36$

Step-by-step explanation:

We are given the two functions:

$f(x)=-x+5\text{ and } g(x)=x^2$

And we want to find:

$(g\circ f)(-1)$

This is equivalent to:

$=g(f(-1))$

So, evaluating f(-1) first, we acquire:

$f(-1)=-(-1)+5=6$

Then by substitution:

$=g(6)$

So:

$=(6)^2=36$

Therefore:

$(g\circ f)(-1)=36$

Your answer is indeed correct.

7 0

3 years ago

Read 2 more answers

Tell whether the equation has one, zero, or infinitely many solutions.2(x-1/2)=8+2x​

Tell whether the equation has one, zero, or infinitely many solutions.2(x-1/2)=8+2x