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scoundrel [369]
3 years ago
11

Mr. Wells runs a telecommunications company. While going through the company's project records, he found that there were 8 engin

eers under every team leader and 5 project managers for every 50 engineers. Mr. Wells currently employs 15 team leaders.
How many project managers are employed by Mr. Wells? A.
16 project managers
B.
12 project managers
C.
32 project managers
D.
24 project managers

Mathematics
1 answer:
Allisa [31]3 years ago
7 0

There are 12 project managers that are employed by Mr. Wells

<h3>Further explanation</h3>

Solving linear equation mean calculating the unknown variable from the equation.

Let the linear equation : y = mx + c

If we draw the above equation on Cartesian Coordinates , it will be a straight line with :

<em>m → gradient of the line</em>

<em>( 0 , c ) → y - intercept</em>

Gradient of the line could also be calculated from two arbitrary points on line ( x₁ , y₁ ) and ( x₂ , y₂ ) with the formula :

\large {\boxed{m = \frac{y_2 - y_1}{x_2 - x_1}}}

If point ( x₁ , y₁ ) is on the line with gradient m , the equation of the line will be :

\large {\boxed{y - y_1 = m ( x - x_1 )}}

<em>Let us tackle the problem!</em>

\texttt{ }

This problem is about Directly Proportional.

<em>There were 8 engineers under every team leader.</em>

\texttt{1 Team Leader} \rightarrow \texttt{8 engineers}

\texttt{15 Team Leader} \rightarrow 15 \times \texttt{8 engineers} = \boxed{\texttt{120 engineers}}

\texttt{ }

<em>There were 5 project managers for every 50 engineers.</em>

\texttt{50 engineers} \rightarrow \texttt{5 project managers}

\texttt{120 engineers} \rightarrow (120 \div 50) \times \texttt{5 project managers} = \boxed{\texttt{12 project managers}}

\texttt{ }

<h3>Learn more</h3>
  • Infinite Number of Solutions : brainly.com/question/5450548
  • System of Equations : brainly.com/question/1995493
  • System of Linear equations : brainly.com/question/3291576

<h3>Answer details</h3>

Grade: High School

Subject: Mathematics

Chapter: Linear Equations

Keywords: Linear , Equations , 1 , Variable , Line , Gradient , Point

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Anna [14]

Answer:

\large\boxed{1\dfrac{1}{3}\ u^2}

Step-by-step explanation:

Let's sketch graphs of functions f(x) and g(x) on one coordinate system (attachment).

Let's calculate the common points:

x^2=\sqrt{x}\qquad\text{square of both sides}\\\\(x^2)^2=\left(\sqrt{x}\right)^2\\\\x^4=x\qquad\text{subtract}\ x\ \text{from both sides}\\\\x^4-x=0\qquad\text{distribute}\\\\x(x^3-1)=0\iff x=0\ \vee\ x^3-1=0\\\\x^3-1=0\qquad\text{add 1 to both sides}\\\\x^3=1\to x=\sqrt[3]1\to x=1

The area to be calculated is the area in the interval [0, 1] bounded by the graph g(x) and the axis x minus the area bounded by the graph f(x) and the axis x.

We have integrals:

\int\limits_{0}^1(\sqrt{x})dx-\int\limits_{0}^1(x^2)dx=(*)\\\\\int(\sqrt{x})dx=\int\left(x^\frac{1}{2}\right)dx=\dfrac{2}{3}x^\frac{3}{2}=\dfrac{2x\sqrt{x}}{3}\\\\\int(x^2)dx=\dfrac{1}{3}x^3\\\\(*)=\left(\dfrac{2x\sqrt{x}}{2}\right]^1_0-\left(\dfrac{1}{3}x^3\right]^1_0=\dfrac{2(1)\sqrt{1}}{2}-\dfrac{2(0)\sqrt{0}}{2}-\left(\dfrac{1}{3}(1)^3-\dfrac{1}{3}(0)^3\right)\\\\=\dfrac{2(1)(1)}{2}-\dfrac{2(0)(0)}{2}-\dfrac{1}{3}(1)}+\dfrac{1}{3}(0)=2-0-\dfrac{1}{3}+0=1\dfrac{1}{3}

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