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3 years ago

[2.5+ 0.187]x10 please help me

Mathematics

1 answer:

MAVERICK [17]3 years ago

6 0

The answer will turn out to be 26.87

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Which right prism would have the same volume as a square prism with a base area of 36 m2 and a height of 3 m?

Neko [114]

Well to find the volume of the square prism, first you need to multiply the base area (36) by its height (3) and you would end up with 108m3 as the volume. You may not be familiar with this because u are most likely taught length times width times height as volume, but what i just did is the same because length times width is the sum of the base area, and height was already given. Anyways, now that we have gotten the square prism volume (108m3) we need to find which prism has the same volume. I’m gonna save u from searching and tell you that it is the rectangular prism because length times width times height (18x2x3) equaled 108 just like the square prism.

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3 years ago

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What is the solution set?<br> O (0-2)<br> O (2.0)<br> O (7,0)<br> O (5,3)

muminat

Answer:

D. (5,3)

Step-by-step explanation:

The solution is (x,y). so in this case, you find the point where both arrows meet.

on the x-axis, which is positive right 5, and positive 3 up, on the y-axis

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3 years ago

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Derek found a function that approximately models the population of iguanas in a reptile garden, where x represents the number of

serious [3.7K]

Answer:

$i(x)=12 \times (1+\frac{0.9}{12})^{12x}$ and growth rate factor is 0.075

Step-by-step explanation:

The function that models the population of iguanas in a reptile garden is given by $i(x)=12 \times (1.9)^{x}$ , where x is the number of years.

Since, $i(x)=12 \times (1.9)^{x}$

i.e. $i(x)=12 \times (1+0.9)^{x}$ .

Therefore, the monthly growth rate function becomes,

i.e. $i(x)=12 \times (1+\frac{0.9}{12})^{x \times 12}$ .

i.e. $i(x)=12 \times (1+\frac{0.9}{12})^{12x}$ .

Hence, the monthly growth rate is i.e. $i(x)=12 \times (1+\frac{0.9}{12})^{12x}$ .

Also, the growth factor is given by $\frac{0.9}{12}$ = 0.075.

Thus, the growth factor to nearest thousandth place is 0.075.

4 0

3 years ago

Use Lagrange multipliers to find the maximum and minimum values of the function subject to the given constraint. (If an answer d

aliya0001 [1]

The Lagrangian

$L(x,y,z,\lambda)=x^2+y^2+z^2+\lambda(x^4+y^4+z^4-13)$

has critical points where the first derivatives vanish:

$L_x=2x+4\lambda x^3=2x(1+2\lambda x^2)=0\implies x=0\text{ or }x^2=-\dfrac1{2\lambda}$

$L_y=2y+4\lambda y^3=2y(1+2\lambda y^2)=0\implies y=0\text{ or }y^2=-\dfrac1{2\lambda}$

$L_z=2z+4\lambda z^3=2z(1+2\lambda z^2)=0\implies z=0\text{ or }z^2=-\dfrac1{2\lambda}$

$L_\lambda=x^4+y^4+z^4-13=0$

We can't have $x=y=z=0$ , since that contradicts the last condition.

(0 critical points)

If two of them are zero, then the remaining variable has two possible values of $\pm\sqrt[4]{13}$ . For example, if $y=z=0$ , then $x^4=13\implies x=\pm\sqrt[4]{13}$ .

(6 critical points; 2 for each non-zero variable)

If only one of them is zero, then the squares of the remaining variables are equal and we would find $\lambda=-\frac1{\sqrt{26}}$ (taking the negative root because $x^2,y^2,z^2$ must be non-negative), and we can immediately find the critical points from there. For example, if $z=0$ , then $x^4+y^4=13$ . If both $x,y$ are non-zero, then $x^2=y^2=-\frac1{2\lambda}$ , and

$xL_x+yL_y=2(x^2+y^2)+52\lambda=-\dfrac2\lambda+52\lambda=0\implies\lambda=\pm\dfrac1{\sqrt{26}}$

$\implies x^2=\sqrt{\dfrac{13}2}\implies x=\pm\sqrt[4]{\dfrac{13}2}$

and for either choice of $x$ , we can independently choose from $y=\pm\sqrt[4]{\frac{13}2}$ .

(12 critical points; 3 ways of picking one variable to be zero, and 4 choices of sign for the remaining two variables)

If none of the variables are zero, then $x^2=y^2=z^2=-\frac1{2\lambda}$ . We have

$xL_x+yL_y+zL_z=2(x^2+y^2+z^2)+52\lambda=-\dfrac3\lambda+52\lambda=0\implies\lambda=\pm\dfrac{\sqrt{39}}{26}$

$\implies x^2=\sqrt{\dfrac{13}3}\implies x=\pm\sqrt[4]{\dfrac{13}3}$

and similary $y,z$ have the same solutions whose signs can be picked independently of one another.

(8 critical points)

Now evaluate $f$ at each critical point; you should end up with a maximum value of $\sqrt{39}$ and a minimum value of $\sqrt{13}$ (both occurring at various critical points).