ESTIMATINGplease show all working out please!!thanks!!

Mathematics

1 answer:

GaryK [48]3 years ago

5 0

1) 304 would round to 300, 9.96 would round to 10, and 0.51 would round to 0.5, so your final expression would be (300 × 10)/0.5. 300 × 10 = 3000, and 3000 ÷ 0.5 = 6000, which is your answer.

2) 6.7 would round to 7, 192 would round to 200, and 0.051 would round to 0.05. (7 × 200)/0.05. 7 × 200 = 1400, and 1400 ÷ 0.05 = 28,000, which is your answer.

I hope this helps!

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The trip from Carville to Nikpath requires $4\frac 12$ hours when traveling at an average speed of 70 miles per hour. How many h

KonstantinChe [14]

From the given speed and time, the distance of the trip can

be determined.

Correct response;

Number of hours the trip will take when traveling 60 mph is $\underline{5\frac{1}{4} \ hours}$

<h3>How is the time taken at a given speed calculated?</h3>

Given;

Time required for the trip from Carville to Nikpath = $4\frac{1}{2}$ hours

Average speed for the trip = 70 mph

Required;

The duration of the trip when traveling 60 mph

Solution;

Distance = Velocity × Time

Distance of trip = $\mathbf{4\frac{1}{2}}$ hour × 70 miles/hour = 315 miles

Duration of trip when traveling 60 mph is therefore;

$Time = \dfrac{315 \, miles}{60 \ miles/hour} = 5.25 \ hours = \mathbf{ 5\frac{1}{4} \ hours}$

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Learn more about speed and time calculations here;

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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:

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

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

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

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

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