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2 years ago

What is the value of the missing side x?

Mathematics

1 answer:

stich3 [128]2 years ago

4 0

Answer: $3*\sqrt{3}$ = x

Step-by-step explanation:

With our knowledge of a 30 60 triangle we can determine the value a from dividing 6 by 2. Look at my diagram for clarification. From now know the value of a as 3 we can make x = $3*\sqrt{3}$ .

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LaTeX: \sqrt[]{36} 36 2) LaTeX: \sqrt[3]{-8} − 8 3 3) LaTeX: -\sqrt[]{100} − 100 4) LaTeX: \sqrt[3]{27} I'm sorry that's the bes

Mamont248 [21]

Given:

Consider the expression are

1) $\sqrt{36}$

2) $\sqrt[3]{-8}$

3) $-\sqrt{100}$

4) $\sqrt[3]{27}$

To find:

The simplified form of each expression.

Solution:

1. We have,

$\sqrt{36}=\sqrt{6^2}$

$\sqrt{36}=6$

Therefore, the value of this expression is 6.

2. We have,

$\sqrt[3]{-8}=(-8)^{\frac{1}{3}}$

$\sqrt[3]{-8}=((-2)^3)^{\frac{1}{3}}$

$\sqrt[3]{-8}=(-2)^{\frac{3}{3}}$

$\sqrt[3]{-8}=-2$

Therefore, the value of this expression is -2.

3. We have,

$-\sqrt{100}=-\sqrt{10^2}$

$-\sqrt{100}=-10$

Therefore, the value of this expression is -10.

4. We have,

$\sqrt[3]{27}=(27)^{\frac{1}{3}}$

$\sqrt[3]{27}=(3^3)^{\frac{1}{3}}$

$\sqrt[3]{27}=(3)^{\frac{3}{3}}$

$\sqrt[3]{27}=3$

Therefore, the value of this expression is 3.

3 0

3 years ago

What is the area of a cross section that is parallel to face ABCD ?

Hunter-Best [27]

To solve this problem you must apply the proccedure shown below:
1. The problem asks for the area of a cross section that is parallel <span>to face ABCD. As is parallel to that face, you have can calculate its area as following:
A=12 cm x 6 cm
2. Therefore, the result is:
A=72 cm</span>²
The answer is: T<span>he area of a cross section that is parallel to face ABCD is 72 cm</span>².

3 0

3 years ago

Let divf = 6(5 − x) and 0 ≤ a, b, c ≤ 12. (a) find the flux of f out of the rectangular solid 0 ≤ x ≤ a, 0 ≤ y ≤ b, and 0 ≤ z ≤

dusya [7]

Continuing from the setup in the question linked above (and using the same symbols/variables), we have

$\displaystyle\iint_{\mathcal S}\mathbf f\cdot\mathrm d\mathbf S=\iiint_{\mathcal R}(\nabla\cdot f)\,\mathrm dV$
$=\displaystyle6\int_{z=0}^{z=c}\int_{y=0}^{y=b}\int_{x=0}^{x=a}(5-x)\,\mathrm dx\,\mathrm dy\,\mathrm dz$
$=\displaystyle6bc\int_0^a(5-x)\,\mathrm dx$
$=6bc\left(5a-\dfrac{a^2}2\right)=3abc(10-a)$

The next part of the question asks to maximize this result - our target function which we'll call $g(a,b,c)=3abc(10-a)$ - subject to $0\le a,b,c\le12$ .

We can see that $g$ is quadratic in $a$ , so let's complete the square.

$g(a,b,c)=-3bc(a^2-10a+25-25)=3bc(25-(a-5)^2)$

Since $b,c$ are non-negative, it stands to reason that the total product will be maximized if $a-5$ vanishes because $25-(a-5)^2$ is a parabola with its vertex (a maximum) at (5, 25). Setting $a=5$ , it's clear that the maximum of $g$ will then be attained when $b,c$ are largest, so the largest flux will be attained at $(a,b,c)=(5,12,12)$ , which gives a flux of 10,800.