What’s the distance between (3,4) , (7,2)

Mathematics

1 answer:

Reptile [31]3 years ago

3 0

Answer:

$D= 4.47$

Step-by-step explanation:

Here we are given with two coordinates and asked to determine the distance between them.

Here we are going to use the distance formula, which is given as under

$D=\sqrt{(y_2-y_1)^2+(x_2-x_1)^2}$

Where

$(x_1,y_1) = (3,4)$

$(x_2,y_2) = (7,2)$

Replacing these values in the distance formula

$D =\sqrt{(7-3))^2+(2-4)^2}$

$D =\sqrt{(4)^2+(-2)^2}$

$D =\sqrt{16+4}$

$D =\sqrt{20}$

$D= 4.47$

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You travel from point A to point B in a car moving at a constant speed of 70 km/h. Then you travel the same distance from point

tankabanditka [31]

Answer:

Yes, the average speed for the entire trip from A to C is equal to $80\frac{km}{h}$

Step-by-step explanation:

The average speed of an object is defined as the distance traveled divided by the time elapsed. Velocity is a vector quantity, and average velocity can be defined as the displacement divided by the time. For the special case of straight line motion in the x direction, the average velocity takes the form:

$V_a_v_e_r_a_g_e=\frac{x_2-x_1}{t_2-t_1}=\frac{Δx}{Δt}$

If the beginning and ending velocities for the motion are known, and the acceleration is constant, the average velocity can also be expressed as:

$V_a_v_e_r_a_g_e=\frac{V_1+V_2}{2}$

We Know that:

$V_1=70\frac{km}{h} \\\\V_2=90\frac{km}{h}$

Replacing the values:

$V_a_v_e_r_a_g_e=\frac{70+90}{2} =\frac{160}{2}=80\frac{km}{h}$

5 0

3 years ago

Calculate the area of triangle ABC with altitude CD, given A (6,0) B(1,5) C (2,0) and D (4,2)

Elina [12.6K]

So hmm check the picture below

the height or altitude is CD
and the base is AB

how long are those? well $\bf \textit{distance between 2 points}\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
% (a,b)
A&({{ 6}}\quad ,&{{ 0}})\quad 
% (c,d)
B&({{ 1}}\quad ,&{{ 5}})\\\\
C&({{ 2}}\quad ,&{{ 0}})\quad 
% (c,d)
D&({{ 4}}\quad ,&{{ 2}})
\end{array}\qquad 
% distance value
d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2}
\\\\\\
AB=\sqrt{(1-6)^2+(5-0)^2}\qquad \qquad CD=\sqrt{(4-2)^2+(2-0)^2}$