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Lunna [17]
3 years ago
11

Joshua started cycling at 5:15 pm By 8:09 pm, he has covered a distance of 7250 mWhat was Joshua's average speed during that tim

e? mh
Mathematics
1 answer:
vivado [14]3 years ago
5 0

The average speed of Joshua during that time is 2500 m/h.

Explanation:

It is given that Joshua started cycling at 5:15 pm. By 8:09 pm he has covered a distance of 7250 m.

The total time taken by Joshua from 5:15 pm to 8:09 pm is

2 { h } 54 \text { min }=2 \mathrm{h}+\frac{54}{60} {h}

Dividing we get,

2 { h } 54 \text { min }=2 \mathrm{h}+0.9 {h}

Adding, we have,

2 { h } 54 \text { min }=2.9 {h}

Thus, the total time taken by Joshua is 2.9 {h}

To determine the average speed we use the formula,

speed=\frac{distance}{time}

where distance=7250 and time=2.9

Hence, substituting the values we have,

speed=\frac{7250}{2.9}

Dividing, we get,

speed=2500

Thus, the average speed of Joshua during that time is 2500 m/h.

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3 years ago
Use Cramer Rule to solve the following system: 8x−5y=70 and 9x+7y=3
nlexa [21]

Answer:

(x,y) = (5,-6)

Step-by-step explanation:

\underline{\textbf{Determinant of a matrix.}}\\\\\text{For a}~ 2 \times 2 ~ \text{matrix,}\\\\\begin{vmatrix} a_1&a_2\\b_1&b_2 \end{vmatrix} = a_1b_2 - a_2b_1\\\\\\\text{For a}~ 3 \times 3 ~ \text{matrix,}\\\\\begin{vmatrix} a_1&a_2&a_3\\ b_1&b_2&b_3\\ c_1&c_2&c_3 \end{vmatrix} = a_1\begin{vmatrix} b_2&b_3\\c_2&c_3 \end{vmatrix} - a_2 \begin{vmatrix} b_1&b_3\\c_1&c_3 \end{vmatrix}+ a_3 \begin{vmatrix} b_1&b_2\\c_1&c_2 \end{vmatrix}\\\\\\

                     ~~~~~~~~~~~~~~~~~~=a_1(b_2c_3-b_3c_2) -a_2(b_1c_3-b_3c_1) +a_3(b_1c_2-b_2c_1)

\underline{\textbf{Cramer's Rule to solve a system of two equations.}}\\\\\text{Consider the system of two equations:}\\\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~a_1x + b_1 y= c_1\\\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~a_2x +b_2 y = c_2\\\\\text{Here,}\\\\x = \dfrac{D_x}{D}= \dfrac{\begin{vmatrix} c_1&b_1\\c_2&b_2 \end{vmatrix}}{\begin{vmatrix} a_1&b_1\\a_2&b_2 \end{vmatrix}}\\\\\\ y= \dfrac{D_y}{D}= \dfrac{\begin{vmatrix} a_1&c_1\\a_2&c_2 \end{vmatrix}}{\begin{vmatrix} a_1&b_1\\a_2&b_2 \end{vmatrix}}\\\\

\underline{\textbf{Solution:}}\\\\~~~~~~~~~~~~~~~~~~~~~~~8x-5y = 70~~~~~~...(i)\\\\~~~~~~~~~~~~~~~~~~~~~~~9x +7y = 3~~~~~~~...(ii)\\\\\text{Applying Cramer's rule:}\\\\x = \dfrac{D_x}{D}\\\\\\~~=\dfrac{\begin{vmatrix} 70& -5 \\3&7 \end{vmatrix}}{\begin{vmatrix} 8& -5\\ 9& 7\end{vmatrix}}\\\\\\~~=\dfrac{70(7) -(-5)(3)}{(8)(7)-(-5)(9)}\\\\\\~~=\dfrac{490+15}{56+45}\\\\\\~~=\dfrac{505}{101}\\\\\\~~=5

y = \dfrac{D_y}{D}\\\\\\~~=\dfrac{\begin{vmatrix} 8& 70 \\9&3 \end{vmatrix}}{\begin{vmatrix} 8& -5\\ 9& 7\end{vmatrix}}\\\\\\~~=\dfrac{(8)(3) -(70)(9)}{(8)(7)-(-5)(9)}\\\\\\~~=\dfrac{24-630}{56+45}\\\\\\~~=-\dfrac{606}{101}\\\\\\~~=-6

\textbf{Hence, the solution to the system of equation is}~ (x,y) = (5,-6)

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1 year ago
A standard clock has a 1-cm hour hand and a 2-cm minute hand. At 12 pm, they are both pointing in the same direction and the dis
Levart [38]

Answer:

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Step-by-step explanation:

In a standard clock, the angle between every number is 30°, therefore the angle between 12 and 2 will be 30° x 2 = 60°.

Looking at the diagram, to find c we can make use of our cosine formula

c² = a² + b² –2abCos(C°)

a = 2, b = 1 and C° = 60°

Therefore we have:

c² = 2² + 1² –2 x 2 x 1 x cos(60°) =

c² = 4 + 1 – 4 x 0.5 =

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Therefore, the distance between the hands is √(3)cm ≈ 1.73cm.

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Step-by-step explanation:

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8 0
3 years ago
6/100 + 5/100 +80/100 +9/100
xz_007 [3.2K]

♡ The Question ♡

- 6/100 + 5/100 +80/100 +9/100

*୨୧ ┈┈┈┈┈┈┈┈┈┈┈┈ ୨୧*

♡ The Answer ♡

-  1

*୨୧ ┈┈┈┈┈┈┈┈┈┈┈┈ ୨୧*

♡ The Explanation/Step-By-Step ♡

- 6/100 + 5/100 + 80/100 + 9/100

Apply the fraction rule! --> a/c + b/c = a + b over c

6/100 + 5/100 + 80/100 + 9/100 = 6 + 5 + 80 + 9 over 100

Add the numbers: 6 + 5 + 80 + 9 = 100!

= 100/100 Apply the fraction rule! a/a = 1

a/a = 1 = 100/100 = 1

*୨୧ ┈┈┈┈┈┈┈┈┈┈┈┈ ୨୧*

♡ Tips ♡

- No tips provided!

4 0
3 years ago
Read 2 more answers
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