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2 years ago

Using law of sines please show process and answer

Mathematics

1 answer:

Katyanochek1 [597]2 years ago

5 0

Hello,

$\widehat{A}=180^o-24.4^o-103.6^o=52^o\\\\\dfrac{sin(24.4^o)}{37.3} =\dfrac{sin(103.6^o)}{c} \\\\\\c=87.760246\approx{87.8}\\\\\\\dfrac{sin(52^o)}{a} =\dfrac{sin(24.4^o)}{37.3} \\\\\\a=71.1510189...\approx{71.2}\\$

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Simplify the expression

SashulF [63]

Answer:

1/(c - d)

Step-by-step explanation:

Multiply numerator and denominator by cd, then factor the denominator and cancel the common factor. (The result is restricted to c+d ≠ 0.)

$\displaystyle\frac{\frac{1}{c}+\frac{1}{d}}{\frac{c}{d}-\frac{d}{c}}=\frac{d+c}{c^2-d^2}\\\\=\frac{d+c}{(c-d)(d+c)}=\frac{1}{c-d}$

4 0

3 years ago

Read 2 more answers

Help! Please!

fomenos

Answer:

is the board game to BOARD for Neal so you have 1.54

Step-by-step explanation:

its ligit 5.5

6 0

2 years ago

Cubed root x cubed root x2

Yuki888 [10]

Answer:

Final answer is $\sqrt[3]{x^1}\cdot\sqrt[3]{x^2}=x$ .

Step-by-step explanation:

Given problem is $\sqrt[3]{x}\cdot\sqrt[3]{x^2}$ .

Now we need to simplify this problem.

$\sqrt[3]{x}\cdot\sqrt[3]{x^2}$

$\sqrt[3]{x^1}\cdot\sqrt[3]{x^2}$

Apply formula

$\sqrt[n]{x^p}\cdot\sqrt[n]{x^q}=\sqrt[n]{x^{p+q}}$

so we get:

$\sqrt[3]{x^1}\cdot\sqrt[3]{x^2}=\sqrt[3]{x^{1+2}}$

$\sqrt[3]{x^1}\cdot\sqrt[3]{x^2}=\sqrt[3]{x^{3}}$

$\sqrt[3]{x^1}\cdot\sqrt[3]{x^2}=x$

Hence final answer is $\sqrt[3]{x^1}\cdot\sqrt[3]{x^2}=x$ .

7 0

2 years ago

Surface integrals using an explicit description. Evaluate the surface integral \iint_{S}^{}f(x,y,z)dS using an explicit represen

Jobisdone [24]

Parameterize $S$ by the vector function

$\vec r(x,y)=x\,\vec\imath+y\,\vec\jmath+f(x,y)\,\vec k$

so that the normal vector to $S$ is given by

$\dfrac{\partial\vec r}{\partial x}\times\dfrac{\partial\vec r}{\partial y}=\left(\vec\imath+\dfrac{\partial f}{\partial x}\,\vec k\right)\times\left(\vec\jmath+\dfrac{\partial f}{\partial y}\,\vec k\right)=-\dfrac{\partial f}{\partial x}\vec\imath-\dfrac{\partial f}{\partial y}\vec\jmath+\vec k$

with magnitude

$\left\|\dfrac{\partial\vec r}{\partial x}\times\dfrac{\partial\vec r}{\partial y}\right\|=\sqrt{\left(\dfrac{\partial f}{\partial x}\right)^2+\left(\dfrac{\partial f}{\partial y}\right)^2+1}$

In this case, the normal vector is

$\dfrac{\partial\vec r}{\partial x}\times\dfrac{\partial\vec r}{\partial y}=-\dfrac{\partial(8-x-2y)}{\partial x}\,\vec\imath-\dfrac{\partial(8-x-2y)}{\partial y}\,\vec\jmath+\vec k=\vec\imath+2\,\vec\jmath+\vec k$

with magnitude $\sqrt{1^2+2^2+1^2}=\sqrt6$ . The integral of $f(x,y,z)=e^z$ over $S$ is then

$\displaystyle\iint_Se^z\,\mathrm d\Sigma=\sqrt6\iint_Te^{8-x-2y}\,\mathrm dy\,\mathrm dx$

where $T$ is the region in the $x,y$ plane over which $S$ is defined. In this case, it's the triangle in the plane $z=0$ which we can capture with $0\le x\le8$ and $0\le y\le\frac{8-x}2$ , so that we have

$\displaystyle\sqrt6\iint_Te^{8-x-2y}\,\mathrm dx\,\mathrm dy=\sqrt6\int_0^8\int_0^{(8-x)/2}e^{8-x-2y}\,\mathrm dy\,\mathrm dx=\boxed{\sqrt{\frac32}(e^8-9)}$

5 0

2 years ago

The Amazon rainforest is gradually being destroyed by pollution and agricultural and industrial development. For simplicity, sup

KATRIN_1 [288]

The Amazon rainforest is gradually being destroyed by pollution and agricultural and industrial development. For simplicity, suppose that each year, 10% of the remaining forest is destroyed. Assume, also for simplicity, that the present area of the Amazon rainforest is 1,200,000 square miles.
1.a) What will the area of the forest be after 1 year of this destruction process?
1.b) What will the area of the forest be after 2 years of this destruction process?
2) Make a graph showing your results from Question 1 and continuing through 5 years of the destruction process. Include the present situation as a point on your graph.
3) Find a rule for how much rain forest will remain after X years. That is, express the area of the rain forest as a function of X.

7 0

2 years ago