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3 years ago

What is straight is part of a line and has 2 endpoints

Mathematics

2 answers:

kipiarov [429]3 years ago

8 0

A line segment has two endpoints and is a straight line while a ray has one end point.

Karolina [17]3 years ago

6 0

A line has no endpoints. A line segment is a part of a line that has two end points. A ray is a part of a line that has one end point. hope this helps

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In the graph, the area below f(x) is shaded and labeled A, the area below g(x) is shaded and labeled B, and the area where f(x)

Tcecarenko [31]

Answer:

Step-by-step explanation:

6 0

3 years ago

What’s the answer for this question

Sergeu [11.5K]

Answer:

84 Square Inches.

Area = Side length x Side length

4 0

3 years ago

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A square pyramid has a base with an area of 20 square meters, and its lateral faces have a slant height of x meters. Sydney is c

svetoff [14.1K]

1. Consider the pyramid ABCDE in the figure 1 attached:

Each side of the square base equals $\sqrt{20} = \sqrt{4*5} = \sqrt{4}* \sqrt{5}=2 \sqrt{5}$ (units)

Area of 1 lateral side of pyramid ABCDE is $\frac{1}{2}* 2 \sqrt{5}*x=\sqrt{5}x$

so the total lateral area is $4\sqrt{5}x$ (units squared)

and the total surface area is $20+4\sqrt{5}x$ (units squared)

2. Now consider the second figure KLMNP, Sydney's pyramid:

The lateral area is $4* \frac{1}{2}* 4 \sqrt{5} *2x=16 \sqrt{5}x$

so the total area of the second pyramid is $20+16 \sqrt{5}x$

So 2 things can be said about the areas:

i) the lateral area of Sydney's pyramid is $\frac{16 \sqrt{5}x}{4\sqrt{5}x} =4$ times larger than the lateral area of the original pyramid.

ii) The total area of Sydney's pyramid is $(20+16 \sqrt{5}x)-(20+4\sqrt{5}x)=12\sqrt{5}x$ units squared larger than the total area of the original pyramid.

5 0

2 years ago

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Someone solve these equations with working really appreciate it

nignag [31]

Answer:

a) $t = \sqrt{\frac{d}{4.9} }$

b) t = 0.3 seconds

c) d = 4.9t²

d) d = 11 cm.

Step-by-step explanation:

One test involves catching a 1 m ruler and measuring the distance, d in meters, that the ruler will travel before being caught, the reaction time t in seconds is then calculated using the formula

$t = \sqrt{\frac{2d}{9.8} }$ .......... (1)

a) the simplified expression is

$t = \sqrt{\frac{d}{4.9} }$ ......... (2)

b) If d = 44 cm = 0.44 meters, from equation (2) we get,

t = 0.299 seconds ≈ 0.3 seconds (Approximate)

c) If we make d as the subject of the equation, then

$t^{2} = \frac{d}{4.9}$

⇒ d = 4.9t² .......... (3)

d) Now, if $t = \frac{0.3}{2} = 0.15$ seconds, then from equation (3) we get,

d = 4.9 (0.15)² = 0.11 meters = 11 cm (Answer)

4 0

2 years ago

What is the period of the sinusoid given by y=-4sin( <img src="https://tex.z-dn.net/?f=%20%5Cfrac%7B2%CF%80%7D%7B3%7D%20" id="Te

Sonja [21]

$\bf \qquad \qquad \qquad \qquad \textit{function transformations}
\\ \quad \\
% function transformations for trigonometric functions
\begin{array}{rllll}
% left side templates
f(x)=&{{ A}}sin({{ B}}x+{{ C}})+{{ D}}
\\\\
f(x)=&{{ A}}cos({{ B}}x+{{ C}})+{{ D}}\\\\
f(x)=&{{ A}}tan({{ B}}x+{{ C}})+{{ D}}
\end{array}
\\\\
-------------------\\\\$

$\bf \bullet \textit{ stretches or shrinks}\\
\left. \qquad \right. \textit{horizontally by amplitude } |{{ A}}|\\\\
\bullet \textit{ flips it upside-down if }{{ A}}\textit{ is negative}\\
\left. \qquad \right. \textit{reflection over the x-axis}
\\\\
\bullet \textit{ flips it sideways if }{{ B}}\textit{ is negative}\\
\left. \qquad \right. \textit{reflection over the y-axis}$

$\bf \bullet \textit{ horizontal shift by }\frac{{{ C}}}{{{ B}}}\\
\left. \qquad \right. if\ \frac{{{ C}}}{{{ B}}}\textit{ is negative, to the right}\\\\
\left. \qquad \right. if\ \frac{{{ C}}}{{{ B}}}\textit{ is positive, to the left}\\\\
$

$\bf \bullet \textit{vertical shift by }{{ D}}\\
\left. \qquad \right. if\ {{ D}}\textit{ is negative, downwards}\\\\
\left. \qquad \right. if\ {{ D}}\textit{ is positive, upwards}\\\\
\bullet \textit{function period or frequency}\\
\left. \qquad \right. \frac{2\pi }{{{ B}}}\ for\ cos(\theta),\ sin(\theta),\ sec(\theta),\ csc(\theta)\\\\
\left. \qquad \right. \frac{\pi }{{{ B}}}\ for\ tan(\theta),\ cot(\theta)$

with that template in mind, let's see

$\bf \begin{array}{llll}
y=&-4sin(&\frac{2\pi }{3}x)\\
&A&B
\end{array}\qquad period\implies \cfrac{2\pi }{B}\implies \cfrac{2\pi }{\frac{2\pi }{3}}\implies \cfrac{\frac{2\pi }{1}}{\frac{2\pi }{3}}
\\\\\\
\cfrac{2\pi }{1}\cdot \cfrac{3}{2\pi }\implies 3$

8 0

3 years ago

Read 2 more answers