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3 years ago

In what way is geometry related to basic math.

2 answers:

6 0

<span> Geometry is the study of points, lines, angles, surfaces, and solids.</span>Geometry is all about shapes<span> and their properties.
</span>And to do so you must use a ruler to identify them right?

Hope this helps!

Sergeeva-Olga [200]3 years ago

3 0

I think geometry is related to basic math, not only because it uses numbers, but because you need to know basic math in order to understand geometry. And many things involving basic math is used to solve may geometric problems, such as a ruler like ChanyaTalley said.

Hope this helps! Click the thanks!

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Find out the answers to this

meriva

Answer:

$\tan \theta = - \frac{1}{5} = - 0.2$

$\cos \theta = 0.98$

$\sin \theta = - 0.196$

Step-by-step explanation:

It is given that $\cot \theta = - 5$ and $\theta$ is in the fourth quadrant.

So, only $\cos \theta$ will have positive value and $\sin \theta$ , $\tan \theta$ will have negative value.

Now, $\cot \theta = - 5$

⇒ $\tan \theta = \frac{1}{\cot \theta} = -\frac{1}{5}$ (Answer)

We know, that $\sec^{2} \theta - \tan^{2} \theta = 1$

⇒ $\sec \theta = \sqrt{1 + \tan^{2} \theta } = \sqrt{1 + (- \frac{1}{5} )^{2} } = 1.019$

{Since, $\cos \theta$ is positive then $\sec \theta$ will also be positive}

⇒ $\cos \theta = \frac{1}{\sec \theta} = \frac{1}{1.0198} = 0.98$ (Answer)

We know, that $\csc^{2} \theta - \cot^{2} \theta = 1$

⇒ $\csc \theta = - \sqrt{1 + \cot^{2} \theta } = - \sqrt{1 + (- 5 )^{2} } = - 5.099$

{Since, $\sin \theta$ is negative then $\csc \theta$ will also be negative}

⇒ $\sin \theta = \frac{1}{\csc \theta} = \frac{1}{- 5.099} = - 0.196$ (Answer)

4 0

3 years ago

After school, Hugo helps out at a bunny shelter called Hoppy's Home. He cleans cages and feeds all 8 bunnies at the shelter. In

amm1812

Answer:

25.3

1_ draw a triangle

the shortest side is a which will be 8(bunnies)

diagonal will be c ( what you need to find )

the last side will be b.

so the equation is a2+b2=c2 ( 2=squared)

that means..... 8x8+24x24=c

8x8=64

24x24=576

then add

576+64=640

the find square root

√640=25.2982212813

simplify to 25.3

you know your answer is correct because c is always the biggest side

5 0

3 years ago

Find the critical points of the function f(x, y) = 8y2x − 8yx2 + 9xy. Determine whether they are local minima, local maxima, or

NARA [144]

Answer:

Saddle point: $(0,0)$

Local minimum: $(\frac{3}{8}, -\frac{3}{8})$

Local maxima: $(0,-\frac{9}{8})$ , $(\frac{9}{8},0)$

Step-by-step explanation:

The function is:

$f(x,y) = 8\cdot y^{2}\cdot x -8\cdot y\cdot x^{2} + 9\cdot x \cdot y$

The partial derivatives of the function are included below:

$\frac{\partial f}{\partial x} = 8\cdot y^{2}-16\cdot y\cdot x+9\cdot y$

$\frac{\partial f}{\partial x} = y \cdot (8\cdot y -16\cdot x + 9)$

$\frac{\partial f}{\partial y} = 16\cdot y \cdot x - 8 \cdot x^{2} + 9\cdot x$

$\frac{\partial f}{\partial y} = x \cdot (16\cdot y - 8\cdot x + 9)$

Local minima, local maxima and saddle points are determined by equalizing both partial derivatives to zero.

$y \cdot (8\cdot y -16\cdot x + 9) = 0$

$x \cdot (16\cdot y - 8\cdot x + 9) = 0$

It is quite evident that one point is (0,0). Another point is found by solving the following system of linear equations:

$\left \{ {{-16\cdot x + 8\cdot y=-9} \atop {-8\cdot x + 16\cdot y=-9}} \right.$

The solution of the system is (3/8, -3/8).

Let assume that y = 0, the nonlinear system is reduced to a sole expression:

$x\cdot (-8\cdot x + 9) = 0$

Another solution is (9/8,0).

Now, let consider that x = 0, the nonlinear system is now reduced to this:

$y\cdot (8\cdot y+9) = 0$

Another solution is (0, -9/8).

The next step is to determine whether point is a local maximum, a local minimum or a saddle point. The second derivative test:

$H = \frac{\partial^{2} f}{\partial x^{2}} \cdot \frac{\partial^{2} f}{\partial y^{2}} - \frac{\partial^{2} f}{\partial x \partial y}$

The second derivatives of the function are:

$\frac{\partial^{2} f}{\partial x^{2}} = 0$

$\frac{\partial^{2} f}{\partial y^{2}} = 0$

$\frac{\partial^{2} f}{\partial x \partial y} = 16\cdot y -16\cdot x + 9$

Then, the expression is simplified to this and each point is tested:

$H = -16\cdot y +16\cdot x -9$

S1: (0,0)

$H = -9$ (Saddle Point)

S2: (3/8,-3/8)

$H = 3$ (Local maximum or minimum)

S3: (9/8, 0)

$H = 9$ (Local maximum or minimum)

S4: (0, - 9/8)

$H = 9$ (Local maximum or minimum)

Unfortunately, the second derivative test associated with the function does offer an effective method to distinguish between local maximum and local minimums. A more direct approach is used to make a fair classification:

S2: (3/8,-3/8)

$f(\frac{3}{8} ,-\frac{3}{8} ) = - \frac{27}{64}$ (Local minimum)