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2 years ago

What is the quotient. 4x^2+25x+12 divided by x+5

Mathematics

2 answers:

VARVARA [1.3K]2 years ago

6 0

The answer is

4x+5-(13/x+5)

AveGali [126]2 years ago

3 0

4x2 - 25x + 27

——————————————

x + 5

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Which of the following is equivalent to 7(2x+1)-2(6x+3)=15

Ivenika [448]

Answer:

2x + 1 = 15

Step-by-step explanation:

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2 years ago

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Identify the surface with the given vector equation. r(s, t) = s sin 4t, s2, s cos 4t

k0ka [10]

Answer:

Circular paraboloid

Step-by-step explanation:

Given ,

$r(s,t)=ssin4t,s^2,scos4t$

Here, these are the respective $x,y,z$ axes components.

Component along x axis $r_i : ssin4t$

Component along y axis $r_j : s^2$

Component along z axis $r_k : scos4t$

We see that , from the parameterised equation , $r_i^2+r_k^2=s^2sin^24t+s^2cos^24t\\r_i^2+r_k^2=s^2\\r_i^2+r_k^2=r_j$

This can also be written as :

$x^2+z^2=y$

This is similar to an equation of a parabola in 1 Dimension.

By fixing the value of z=0,

We get $y=x^2$ which is equation of a parabola curving towards the positive infinity of y-axis and in the x-y plane.

By fixing the value of x=0,

We get $y=z^2$ which is equation of a parabola curving towards positive infinity of y-axis and in the y-z plane.

Thus by fixing the values of x and z alternatively , we get a CIRCULAR PARABOLOID.

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3 years ago

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Bogdan [553]

Step-by-step explanation:

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3 years ago

Typically,the x-value in a function is called the

Lina20 [59]

The answer would be C, Input Value because it is called the independent variable and represents an input value from the domain.

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3 years ago

A box with a square base and an open top is being constructed out of A cm2 of material. If the volume of the box is to be maximi

viktelen [127]

Answer:

Side length = $\sqrt{\frac{A}{3} }$ cm , Height = $\frac{1}{2} \sqrt{\frac{A}{3} }$ cm , Volume = $\frac{A\sqrt{A}}{6\sqrt{3} }$ cm³

Step-by-step explanation:

Assume

Side length of base = x

Height of box = y

total material required to construct box = A ( given in question)

So it can be written as

A = x² + 4xy

4xy = A - x²

$y = \frac{A - x^{2} }{4x}$

Volume of box = Area x height

V = x² ₓ y

V = x² ₓ ( $\frac{A - x^{2} }{4x}$ )

V = $\frac{Ax - x^{3} }{4}$

To find max volume put V' = 0

So taking derivative equation becomes

$\frac{A - 3 x^{2} }{4} = 0$

A = 3 $x^{2}$

$x^{2}$ = $\frac{A}{3}$

x = $\sqrt{\frac{A}{3\\} }$

put value of x in equation 1

y = $\frac{A - \frac{A}{3} }{4\sqrt{\frac{A}{3} } }$

y = $\frac{2 \sqrt{\frac{A}{3} } }{4 \sqrt{\frac{A}{3} } }$

y = $\frac{1}{2} \sqrt{\frac{A}{3} }$

So the volume will be

V = $x^{2}$ × y

Put values of x and y from equation 2 & 3

V = $\frac{A}{3} (\frac{1}{2} \sqrt{\frac{A}{3} } )$

V = $\frac{A\sqrt{A}}{6\sqrt{3} }$

8 0

3 years ago