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3 years ago

I need help please????

Mathematics

1 answer:

ivanzaharov [21]3 years ago

3 0

-84
21x4=84 plus the negative

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Find the max and min values of f(x,y,z)=x+y-z on the sphere x^2+y^2+z^2=81

Anton [14]

Using Lagrange multipliers, we have the Lagrangian

$L(x,y,z,\lambda)=x+y-z+\lambda(x^2+y^2+z^2-81)$

with partial derivatives (set equal to 0)

$L_x=1+2\lambda x=0\implies x=-\dfrac1{2\lambda}$
$L_y=1+2\lambda y=0\implies y=-\dfrac1{2\lambda}$
$L_z=-1+2\lambda z=0\implies z=\dfrac1{2\lambda}$
$L_\lambda=x^2+y^2+z^2-81=0\implies x^2+y^2+z^2=81$

Substituting the first three equations into the fourth allows us to solve for $\lambda$ :

$x^2+y^2+z^2=\dfrac1{4\lambda^2}+\dfrac1{4\lambda^2}+\dfrac1{4\lambda^2}=81\implies\lambda=\pm\dfrac1{6\sqrt3}$

For each possible value of $\lambda$ , we get two corresponding critical points at $(\mp3\sqrt3,\mp3\sqrt3,\pm3\sqrt3)$ .

At these points, respectively, we get a maximum value of $f(3\sqrt3,3\sqrt3,-3\sqrt3)=9\sqrt3$ and a minimum value of $f(-3\sqrt3,-3\sqrt3,3\sqrt3)=-9\sqrt3$ .

5 0

3 years ago

FInding the length of EB

larisa86 [58]

Answer:

EB=3.3

Step-by-step explanation:

From the diagram, triangle ABC is similar to triangle DBE.

This means that, the corresponding sides are proportional.

It was given that AC=6, DE=4, CB=5.

We want to find EB=x

Since the ratio of corresponding sides are proportional, we have:

$\frac{EB}{CB} =\frac{ED}{AC}$

We substitute to obtain:

$\frac{x}{5} =\frac{4}{6}$

Multiply both sides by 5 to get:

$x=\frac{4}{6}\times 5=\frac{20}{6}=\frac{10}{3}=3.3$

4 0

3 years ago

Find the radius and circumference of a cd with a diameter of 4.75 inches

gayaneshka [121]

<span>To calculate for the radius, we divide the given diameter by 2. That would be 4.75 inches/2. This operation would give us the answer of 2.375 inches. The circumference is calculated by the formula, C = 2πr: C = 2(π)(2.375 inches) = 14.92 inches</span>

6 0

3 years ago

HELP NEEDED. 37 POINTS<br>I just need the answers

Juli2301 [7.4K]

Answer:

Part 1) $P=[2\sqrt{29}+\sqrt{18}]\ units$ or $P=15.01\ units$

Part 2) $P=2[\sqrt{20}+\sqrt{45}]\ units$ or $P=22.36\ units$

Part 3) $P=4[\sqrt{13}]\ units$ or $P=14.42\ units$

Part 4) $P=[19+\sqrt{17}]\ units$ or $P=23.12\ units$

Part 5) $P=2[\sqrt{17}+\sqrt{68}]\ units$ or $P=24.74\ units$

Part 6) $A=36\ units^{2}$

Part 7) $A=20\ units^{2}$

Part 8) $A=16\ units^{2}$

Part 9) $A=10.5\ units^{2}$

Part 10) $A=6.05\ units^{2}$

Step-by-step explanation:

we know that

The formula to calculate the distance between two points is equal to

$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$

Part 1) we have the triangle ABC

$A(0,3),B(5,1),C(2,-2)$

step 1

Find the distance AB

$A(0,3),B(5,1)$

substitute in the formula

$AB=\sqrt{(1-3)^{2}+(5-0)^{2}}$

$AB=\sqrt{(-2)^{2}+(5)^{2}}$

$AB=\sqrt{29}\ units$

step 2

Find the distance BC

$B(5,1),C(2,-2)$

substitute in the formula

$BC=\sqrt{(-2-1)^{2}+(2-5)^{2}}$

$BC=\sqrt{(-3)^{2}+(-3)^{2}}$

$BC=\sqrt{18}\ units$

step 3

Find the distance AC

$A(0,3),C(2,-2)$

substitute in the formula

$AC=\sqrt{(-2-3)^{2}+(2-0)^{2}}$

$AC=\sqrt{(-5)^{2}+(2)^{2}}$

$AC=\sqrt{29}\ units$

step 4

Find the perimeter

The perimeter is equal to

$P=AB+BC+AC$

substitute

$P=[\sqrt{29}+\sqrt{18}+\sqrt{29}]\ units$

$P=[2\sqrt{29}+\sqrt{18}]\ units$

$P=15.01\ units$

Part 2) we have the rectangle ABCD

$A(-4,-4),B(-2,0),C(4,-3),D(2,-7)$

Remember that in a rectangle opposite sides are congruent

step 1

Find the distance AB

$A(-4,-4),B(-2,0)$

substitute in the formula

$AB=\sqrt{(0+4)^{2}+(-2+4)^{2}}$

$AB=\sqrt{(4)^{2}+(2)^{2}}$

$AB=\sqrt{20}\ units$

step 2

Find the distance BC

$B(-2,0),C(4,-3)$

substitute in the formula

$BC=\sqrt{(-3-0)^{2}+(4+2)^{2}}$

$BC=\sqrt{(-3)^{2}+(6)^{2}}$

$BC=\sqrt{45}\ units$

step 3

Find the perimeter

The perimeter is equal to

$P=2[AB+BC]$

substitute

$P=2[\sqrt{20}+\sqrt{45}]\ units$

$P=22.36\ units$

Part 3) we have the rhombus ABCD

$A(-3,3),B(0,5),C(3,3),D(0,1)$

Remember that in a rhombus all sides are congruent

step 1

Find the distance AB

$A(-3,3),B(0,5)$

substitute in the formula

$AB=\sqrt{(5-3)^{2}+(0+3)^{2}}$

$AB=\sqrt{(2)^{2}+(3)^{2}}$

$AB=\sqrt{13}\ units$

step 2

Find the perimeter

The perimeter is equal to

$P=4[AB]$

substitute

$P=4[\sqrt{13}]\ units$

$P=14.42\ units$

Part 4) we have the quadrilateral ABCD

$A(-2,-3),B(1,1),C(7,1),D(6,-3)$

step 1

Find the distance AB

$A(-2,-3),B(1,1)$

substitute in the formula

$AB=\sqrt{(1+3)^{2}+(1+2)^{2}}$

$AB=\sqrt{(4)^{2}+(3)^{2}}$

$AB=5\ units$

step 2

Find the distance BC

$B(1,1),C(7,1)$

substitute in the formula

$BC=\sqrt{(1-1)^{2}+(7-1)^{2}}$

$BC=\sqrt{(0)^{2}+(6)^{2}}$

$BC=6\ units$

step 3

Find the distance CD

$C(7,1),D(6,-3)$

substitute in the formula

$CD=\sqrt{(-3-1)^{2}+(6-7)^{2}}$

$CD=\sqrt{(-4)^{2}+(-1)^{2}}$

$CD=\sqrt{17}\ units$

step 4

Find the distance AD

$A(-2,-3),D(6,-3)$

substitute in the formula

$AD=\sqrt{(-3+3)^{2}+(6+2)^{2}}$

$AD=\sqrt{(0)^{2}+(8)^{2}}$

$AD=8\ units$

step 5

Find the perimeter

The perimeter is equal to

$P=AB+BC+CD+AD$

substitute

$P=[5+6+\sqrt{17}+8]\ units$

$P=[19+\sqrt{17}]\ units$

$P=23.12\ units$

Part 5) we have the quadrilateral ABCD

$A(-1,5),B(3,6),C(5,-2),D(1,-3)$

step 1

Find the distance AB

$A(-1,5),B(3,6)$

substitute in the formula

$AB=\sqrt{(6-5)^{2}+(3+1)^{2}}$

$AB=\sqrt{(1)^{2}+(4)^{2}}$

$AB=\sqrt{17}\ units$

step 2

Find the distance BC

$B(3,6),C(5,-2)$

substitute in the formula

$BC=\sqrt{(-2-6)^{2}+(5-3)^{2}}$

$BC=\sqrt{(-8)^{2}+(2)^{2}}$

$BC=\sqrt{68}\ units$

step 3

Find the distance CD

$C(5,-2),D(1,-3)$

substitute in the formula

$CD=\sqrt{(-3+2)^{2}+(1-5)^{2}}$

$CD=\sqrt{(-1)^{2}+(-4)^{2}}$

$CD=\sqrt{17}\ units$

step 4

Find the distance AD

$A(-1,5),D(1,-3)$

substitute in the formula

$AD=\sqrt{(-3-5)^{2}+(1+1)^{2}}$

$AD=\sqrt{(-8)^{2}+(2)^{2}}$

$AD=\sqrt{68}\ units$

step 5

Find the perimeter

The perimeter is equal to

$P=\sqrt{17}+\sqrt{68}+\sqrt{17}+\sqrt{68}$

substitute

$P=2[\sqrt{17}+\sqrt{68}]\ units$

$P=24.74\ units$

<h3>The complete answer in the attached file</h3>

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8 0

3 years ago

What is the volume of the region bounded by y=sqrt(cosx) from [-pi/2, pi/2] and whose cross sections are isosceles right triangl

swat32

I assume the cross sections are taken perpendicular to the x-axis? This seems more likely than relative to the y-axis as far as easiness of calculation goes.

The base of each triangle is then determined by the distance between $\sqrt{\cos x}$ and the x-axis, or simply $\sqrt{\cos x}$ . Because it's a right triangle, you know the legs' lengths occur in a 1:1 ratio. Since each triangular cross section has one of its legs as its base, the heights must be the same as their bases.

So, the area of any one cross-section is

$A(x)=\dfrac12(\sqrt{\cos x})^2=\dfrac{\cos x}2$

Then the volume of this region would be

$\displaystyle\int_{-\pi/2}^{\pi/2}\frac{\cos x}2\,\mathrm dx=1$

4 0

3 years ago