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dusya [7]
3 years ago
14

What is the area of this figure?

Mathematics
1 answer:
Pachacha [2.7K]3 years ago
7 0
(24*11)+(14*14)+(7*14)+(8*12)=654
I think that this is the answer!
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What is the product of 4/7 and 1/6?
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Answer: 0.09523

Step-by-step explanation:

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What is the quotient?
Zepler [3.9K]

Answer:

= \dfrac{1}{5^{9}}

Step-by-step explanation:

When dividing values with exponents, we simply need to subtract the numerators exponent to the denominators exponent.

So we have:

\dfrac{5^{-6}}{5^{3} }

= 5^{-6 - 3}

= 5^{-9}

Now since we have a negative exponent, we can get the reciprocal of the value to make it positive.

= \dfrac{1}{5^{9}}

7 0
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What is one way to find the total cost of a pair of shoes for $57 with a sales tax of 2%?
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Price: $ 57.00

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3 years ago
Simplify the expression:<br>15 + 3(2 + 8) - 42​
SVETLANKA909090 [29]
The answer is three!!!!!!!!!!!
8 0
3 years ago
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Find the maximum and minimum values attained by f(x, y, z) = 5xyz on the unit ball x2 + y2 + z2 ≤ 1.
Allushta [10]
Check for critical points within the unit ball by solving for when the first-order partial derivatives vanish:
f_x=5yz=0\implies y=0\text{ or }z=0
f_y=5xz=0\implies x=0\text{ or }z=0
f_z=5xy=0\implies x=0\text{ or }y=0


Taken together, we find that (0, 0, 0) appears to be the only critical point on f within the ball. At this point, we have f(0,0,0)=0.

Now let's use the method of Lagrange multipliers to look for critical points on the boundary. We have the Lagrangian

L(x,y,z,\lambda)=5xyz+\lambda(x^2+y^2+z^2-1)

with partial derivatives (set to 0)

L_x=5yz+2\lambda x=0
L_y=5xz+2\lambda y=0
L_z=5xy+2\lambda z=0
L_\lambda=x^2+y^2+z^2-1=0

We then observe that

xL_x+yL_y+zL_z=0\implies15xyz+2\lambda=0\implies\lambda=-\dfrac{15xyz}2

So, ignoring the critical point we've already found at (0, 0, 0),


5yz+2\left(-\dfrac{15xyz}2\right)x=0\implies5yz(1-3x^2)=0\implies x=\pm\dfrac1{\sqrt3}
5xz+2\left(-\dfrac{15xyz}2\right)y=0\implies5xz(1-3y^2)=0\implies y=\pm\dfrac1{\sqrt3}
5xy+2\left(-\dfrac{15xyz}2\right)z=0\implies5xy(1-3z^2)=0\implies z=\pm\dfrac1{\sqrt3}

So ultimately, we have 9 critical points - 1 at the origin (0, 0, 0), and 8 at the various combinations of \left(\pm\dfrac1{\sqrt3},\pm\dfrac1{\sqrt3},\pm\dfrac1{\sqrt3}\right), at which points we get a value of either of \pm\dfrac5{\sqrt3}, with the maximum being the positive value and the minimum being the negative one.
5 0
3 years ago
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