All categories

4 years ago

I need to know what lines segments and rays are

2 answers:

5 0

A line segment is part of a line that has two endpoints and contains every point between its endpoints. A ray is a line that has one endpoint and continues infinitely in the other direction.

guajiro [1.7K]4 years ago

4 0

A line segment has two end points and a ray has one end point and he other side keeps going

You might be interested in

The corners of a square are cut off two centimeters from each corner to form an octagon. If the octagon is 10 centimeters wide,

zvonat [6]

Answer:

92 cm²

Explanation:

The area of the octagon may be calculated as the difference of the area of the original square and the area of the four corners cut off.

1) Area of the square.

The original square's side length is the same wide of the formed octagon: 10 cm.

So, the area of such square is: (10 cm)² = 100 cm².

2) Area of the four corners cut off.

Since, the corners were cut off two centimeters from each corner, the form of each piece is an isosceles right triangle with legs of 2 cm.

The area of each right triangle is half the product of the legs (because one leg is the base and the other leg is the height of the triangle).

Then, area of one right triangle: (1/2) × 2cm × 2cm = 2 cm².

Since, they are four pieces, the total cut off area is: 4 × 2 cm² = 8 cm².

3) Area of the octagon:

Area of the square - area of the cut off triangles = 100 cm² - 8cm² = 92 cm².

And that is the answer: 92 cm².

4 0

3 years ago

What is the sum of the angle measures in a decagon? 1,800° 1,440° 1,260° 210°

JulijaS [17]

Use the formula (n - 2) * 180 to find the sum of the interior measures of a polygon.

n stands for the number of sides that the polygon has, so substitute 10 for n since a decagon has 10 sides.

(10 - 2) * 180, start by solving inside the parentheses and subtracting 10 and 2.

(8) * 180, multiply.

B. 1,440 is your answer.

3 0

3 years ago

Read 2 more answers

Differentiating a Logarithmic Function in Exercise, find the derivative of the function. See Examples 1, 2, 3, and 4.

mote1985 [20]

Answer:

$\frac{d}{dx}\left(\ln \left(\frac{x}{x^2+1}\right)\right)=\left(\ln{\left(\frac{x}{x^{2} + 1} \right)}\right)^{\prime }=\frac{-x^2+1}{x\left(x^2+1\right)}$

Step-by-step explanation:

To find the derivative of the function $y(x)=\ln \left(\frac{x}{x^2+1}\right)$ you must:

Step 1. Rewrite the logarithm:

$\left(\ln{\left(\frac{x}{x^{2} + 1} \right)}\right)^{\prime }=\left(\ln{\left(x \right)} - \ln{\left(x^{2} + 1 \right)}\right)^{\prime }$

Step 2. The derivative of a sum is the sum of derivatives:

$\left(\ln{\left(x \right)} - \ln{\left(x^{2} + 1 \right)}\right)^{\prime }}={\left(\left(\ln{\left(x \right)}\right)^{\prime } - \left(\ln{\left(x^{2} + 1 \right)}\right)^{\prime }\right)$

Step 3. The derivative of natural logarithm is $\left(\ln{\left(x \right)}\right)^{\prime }=\frac{1}{x}$

${\left(\ln{\left(x \right)}\right)^{\prime }} - \left(\ln{\left(x^{2} + 1 \right)}\right)^{\prime }={\frac{1}{x}} - \left(\ln{\left(x^{2} + 1 \right)}\right)^{\prime }$

Step 4. The function $\ln{\left(x^{2} + 1 \right)}$ is the composition $f\left(g\left(x\right)\right)$ of two functions $f\left(u\right)=\ln{\left(u \right)}$ and $u=g\left(x\right)=x^{2} + 1$

Step 5. Apply the chain rule $\left(f\left(g\left(x\right)\right)\right)^{\prime }=\frac{d}{du}\left(f\left(u\right)\right) \cdot \left(g\left(x\right)\right)^{\prime }$

$-{\left(\ln{\left(x^{2} + 1 \right)}\right)^{\prime }} + \frac{1}{x}=- {\frac{d}{du}\left(\ln{\left(u \right)}\right) \frac{d}{dx}\left(x^{2} + 1\right)} + \frac{1}{x}\\\\- {\frac{d}{du}\left(\ln{\left(u \right)}\right)} \frac{d}{dx}\left(x^{2} + 1\right) + \frac{1}{x}=- {\frac{1}{u}} \frac{d}{dx}\left(x^{2} + 1\right) + \frac{1}{x}$

Return to the old variable:

$- \frac{1}{{u}} \frac{d}{dx}\left(x^{2} + 1\right) + \frac{1}{x}=- \frac{\frac{d}{dx}\left(x^{2} + 1\right)}{{\left(x^{2} + 1\right)}} + \frac{1}{x}$

The derivative of a sum is the sum of derivatives:

$- \frac{{\frac{d}{dx}\left(x^{2} + 1\right)}}{x^{2} + 1} + \frac{1}{x}=- \frac{{\left(\frac{d}{dx}\left(1\right) + \frac{d}{dx}\left(x^{2}\right)\right)}}{x^{2} + 1} + \frac{1}{x}=\frac{1}{x^{3} + x} \left(x^{2} - x \left(\frac{d}{dx}\left(1\right) + \frac{d}{dx}\left(x^{2}\right)\right) + 1\right)$

Step 6. Apply the power rule $\frac{d}{dx}\left(x^{n}\right)=n\cdot x^{-1+n}$

$\frac{1}{x^{3} + x} \left(x^{2} - x \left({\frac{d}{dx}\left(x^{2}\right)} + \frac{d}{dx}\left(1\right)\right) + 1\right)=\\\\\frac{1}{x^{3} + x} \left(x^{2} - x \left({\left(2 x^{-1 + 2}\right)} + \frac{d}{dx}\left(1\right)\right) + 1\right)=\\\\\frac{1}{x^{3} + x} \left(- x^{2} - x \frac{d}{dx}\left(1\right) + 1\right)\\$

$\frac{1}{x^{3} + x} \left(- x^{2} - x {\frac{d}{dx}\left(1\right)} + 1\right)=\\\\\frac{1}{x^{3} + x} \left(- x^{2} - x {\left(0\right)} + 1\right)=\\\\\frac{1 - x^{2}}{x \left(x^{2} + 1\right)}$

Thus, $\frac{d}{dx}\left(\ln \left(\frac{x}{x^2+1}\right)\right)=\left(\ln{\left(\frac{x}{x^{2} + 1} \right)}\right)^{\prime }=\frac{-x^2+1}{x\left(x^2+1\right)}$

3 0

3 years ago

If it starts out at -10 degrees, what will the temperature be if it rises 30 degrees?

ladessa [460]

-10 + 30

so the answer is 20 degrees

6 0

3 years ago

-6/7 × (-3/8) answer?/

Mumz [18]

Answer:

Step-by-step explanation:

0.32142857142

3 0

3 years ago

Read 2 more answers