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4 years ago

Which best describes a scalene triangle?

Mathematics

2 answers:

FinnZ [79.3K]4 years ago

6 0

Answer:

All three sides of the triangle are different lengths.

Step-by-step explanation:

hope this helps :)

professor190 [17]4 years ago

6 0

Answer:

All three sides of the triangle are different lengths.

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When 3 is substituted for x, in the equation y=x-7, what is y?

garri49 [273]

Answer:

-4

Step-by-step explanation:

y=x-7

plug in 3

y=3-7

y=-4

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4 years ago

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3 years ago

8.3 + 3.5y − 0.5(16y − 5)

olganol [36]

8.3 + 3.5y − 0.5(16y − 5) = -4.5y + 10.8

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3 years ago

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What is the strategy when figuring out the IQR and the 5-number summary for the data

vivado [14]

The strategy when figuring out the IQR and the 5-number summary for the data include arrangement from lowest to highest.

<h3>What is 5-number summary?</h3>

This is used to summarize distribution of data obtained or given in a sample and they include:

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2 years ago

The displacement (in centimeters) of a particle moving back and forth along a straight line is given by the equation of motion s

kotykmax [81]

Answer:

(i) [1,2]

$V_a_v_g=\frac{s(2)-s(1)}{2-1} =\frac{2-(-2)}{1} =4cm/s$

(ii) [1,1,1]

$V_a_v_g=\frac{s(1.1)-s(1)}{1.1-1} =\frac{-3.447-(-2)}{0.1} =-14.47cm/s$

(iii)

$V_a_v_g=\frac{s(1.01)-s(1)}{1.01-1} =\frac{-2.156-(-2)}{0.01} =-15.6cm/s$

(iv)

$V_a_v_g=\frac{s(1.001)-s(1)}{1.001-1} =\frac{-2.016-(-2)}{0.001} =-15.698cm/s$

(b)

$\frac{ds}{dt} \left \{ {t=1}} \right =-15.7076327\approx-15.7cm/s$

Step-by-step explanation:

$s(t)=5sin(\pi t)+2cos(\pi t)$

First. Let's find the displacement for every given time:

t=1s

$s(1)=5sin(\pi)+2cos(\pi)=-2cm=-0.02m$

t=2s

$s(2)=5sin(\pi *2)+2cos(\pi *2)=2cm=0.02m$

t=1.1s

$s(1.1)=5sin(\pi *1.1)+2cos(\pi *1.1)\approx -3.447cm\approx-0.03447m$

t=1.01s

$s(1.101)=5sin(\pi *1.101)+2cos(\pi *1.101)\approx -2.156cm\approx-0.02156m$

t=1.001s

$s(1.001)=5sin(\pi *1.001)+2cos(\pi *1.001)\approx -2.016cm\approx-0.02016m$

Now, the average velocity can be found as:

$V_a_v_g=\frac{S_f-S_i}{t_f-t_i}$

Where:

$S_f=Final\hspace{3}displacement\\S_i=Initial\hspace{3}displacement\\t_f=Final\hspace{3}time\\t_i=Initial\hspace{3}time\\$

Hence:

(i) [1,2]

$V_a_v_g=\frac{s(2)-s(1)}{2-1} =\frac{2-(-2)}{1} =4cm/s$

(ii) [1,1,1]

$V_a_v_g=\frac{s(1.1)-s(1)}{1.1-1} =\frac{-3.447-(-2)}{0.1} =-14.47cm/s$

(iii)

$V_a_v_g=\frac{s(1.01)-s(1)}{1.01-1} =\frac{-2.156-(-2)}{0.01} =-15.6cm/s$

(iv)

$V_a_v_g=\frac{s(1.001)-s(1)}{1.001-1} =\frac{-2.016-(-2)}{0.001} =-15.698cm/s$

(b) In order to estimate the instantaneous velocity of the particle for t=1, we need to find its derivative:

$\frac{ds}{dt} \left \{ {t=1}} \right. =5\pi cos(\pi t) -2\pi sin(\pi t)=5\pi cos(\pi (1)) -2\pi sin(\pi (1)) \\\\\frac{ds}{dt} \left \{ {t=1}} \right =-15.7076327\approx-15.7cm/s$

4 0

3 years ago