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2 years ago

How do i find angle measurements in an isosceles triangle??

1 answer:

3 0

In an isosceles triangles the sides that are congruent form two congruent angles...you should be given one of those angle measures or the other one...if you are given one of the angle measure formed by the congruent side you know the one opposite it is the same and you know all the measures of angles in a triangle = 180 so you should be able to find them all

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Differentiation. questions 1 and 2

drek231 [11]

For 11, we start by plugging 1 in, getting 2(1)^2+1-5=-2=y

Plugging 1+m instead, we get 2(1+m)^2+1+m-5=2(1+2m+m^2)+1+m-5
= 2+4m+2m^2+1+m-5= 2m^2+5m-2. We have to find the difference between that and -2, so 2m^2+5m-2 -(-2)=2m^2+5m

Differentiating 12, we use the Power Rule to get -6x^2+14, and plugging x=1 into it, we get -6(1)^2+14=-6+14=18

4 0

3 years ago

If m RDC = 120°, then m DAC=?

monitta

The correct answer to this question is 240

From the image i've seen, the RDC has an angle of 120. So the major arc DC (that passes A and B) has an angle of 240.

So this makes that the measure of the angle of DAC is 240 because of the angle formed by RDC which is 120.

7 0

3 years ago

Read 2 more answers

Which triangles are similar?

Anastaziya [24]

Answer:

i b and d

Step-by-step explanation:

i say this because i see that b the bottom of it is longer that the

top witch causes the number diffrence same for d but but the top is longer

4 0

3 years ago

A circle's diameter is 4 feet.

Soloha48 [4]

Answer:

Step-by-step explanation:

The formula for cirmference is

C= π2 * r

we're using 3.14 as pi and r = radius.

the radius is always half of the diameter.

So in this case since the diamter is 4, the radius is 2. let's substitute these numbers.

C= 3.14 *2*2 and there your answer is 12.56 which rounded to the nearest tenth is 13.

Hope this helps :)

6 0

2 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)

S3: (9/8, 0)

$f(\frac{9}{8},0) = 0$ (Local maximum)

S4: (0, - 9/8)

$f(0,-\frac{9}{8} ) = 0$ (Local maximum)

Saddle point: $(0,0)$

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

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

4 0

3 years ago