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3 years ago

An angle bisector is a ray starting at the vertex of an angle and divides the angle into two obtuse angles

Mathematics

1 answer:

AVprozaik [17]3 years ago

8 0

False. An angle bisector is a bisector starting at the vertex and cutting an angle in half. It could be obtuse angles, but not always. It is usually two smaller angles. It is a ray, though.

I hope this helps!
~kaikers

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Pls help I’ve been stuck for days

kolezko [41]

Answer:

Step-by-step explanation:

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3 years ago

Solve please:<br><br> 1/5 (25-5a) = 4-a

Liono4ka [1.6K]

Answer: There is no solution.

Step-by-step explanation:

1.) 1 / 5 (25) + 1 / 5 (-5a) = 4 - a

2.) 5 - a = 4 - a

3.) 5 - a + a = 4 - a + a

4.) 5 = 4

The variable a is completely removed, leaving the final answer as 5 = 4, which isn't true.

8 0

3 years ago

Read 2 more answers

Evaluate the expression below at x = 5. x/35 + 5x^2

kramer

Answer:

876/7

Step-by-step explanation:

Just plug the number 5 into the equation and solve. Remember your Order of Operations (PEMDAS).

4 0

2 years ago

Can you guys help please im so confused

marissa [1.9K]

Answer:

a) see below

b) 40x20 meters

Step-by-step explanation:

Write down what you know:

- The area of the enclosure is length*width, so

$A = x \cdot y$

- The length of the fencing is 80 meters, so

$2y + x = 80$

Now we have to combine these two equations above, and get rid of y in the process.

First rewrite the second as:

$y = 40 - \frac12 x$

Then substitute for y in the first:

$A = x (40 - \frac12 x) = 40x - \frac12x^2$

b) To maximize A, find the zero of the first derivative:

$A'(x) = 40 - x = 0 \implies x=40$

So y = (80-40)/2 = 20 meters.

8 0

3 years ago

Find the derivative of the function at P 0 in the direction of A. f(x,y,z) = 3 e^x cos(yz), P0 (0, 0, 0), A = - i + 2 j + 3k

Alik [6]

The derivative of $f(x,y,z)$ at a point $p_0=(x_0,y_0,z_0)$ in the direction of a vector $\vec a=a_x\,\vec\imath+a_y\,\vec\jmath+a_z\,\vec k$ is

$\nabla f(x_0,y_0,z_0)\cdot\dfrac{\vec a}{\|\vec a\|}$

We have

$f(x,y,z)=3e^x\cos(yz)\implies\nabla f(x,y,z)=3e^x\cos(yz)\,\vec\imath-3ze^x\sin(yz)\,\vec\jmath-3ye^x\sin(yz)\,\vec k$

and

$\vec a=-\vec\imath+2\,\vec\jmath+3\,\vec k\implies\|\vec a\|=\sqrt{(-1)^2+2^2+3^2}=\sqrt{14}$

Then the derivative at $p_0$ in the direction of $\vec a$ is

$3\,\vec\imath\cdot\dfrac{-\vec\imath+2\,\vec\jmath+3\,\vec k}{\sqrt{14}}=-\dfrac3{\sqrt{14}}$

3 0

4 years ago