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Preeti Dabral 2 years, 9 months ago
{tex}Put x+y=t \begin{aligned} & \frac{d y}{d x}=\frac{d t}{d x}-1 \\ & \frac{d t}{d x}-1=\operatorname{sect} \\ & \frac{d t}{d x}=\operatorname{sect}+1 \\ & \int \frac{d t}{\operatorname{sect}+1}=\int d x \\ & \int \frac{\operatorname{costdt}}{1+\operatorname{cost}} d t=x \\ & \int \frac{1+\operatorname{costdt}-1}{1+\operatorname{cost}} d t=x \\ & \int\left(d t-\frac{1}{1+\cos t}\right) d t=x \end{aligned} {/tex}
{tex}\begin{aligned} & \mathrm{t}-\frac{1}{2} \int \sec ^2 \frac{\mathrm{t}}{2} \mathrm{dt}=\mathrm{x} \\ & \mathrm{t}-\tan \frac{\mathrm{t}}{2}=\mathrm{x}+\mathrm{C} \\ & \mathrm{x}+\mathrm{y}-\tan \left(\frac{\mathrm{x}+\mathrm{y}}{2}\right)=\mathrm{x}+\mathrm{C} \\ & \mathrm{y}-\tan \left(\frac{\mathrm{x}+\mathrm{y}}{2}\right)=C \end{aligned}{/tex}
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Preeti Dabral 2 years, 9 months ago
Basically, integration is a way of uniting the part to find a whole. It is the inverse operation of differentiation. Thus the basic integration formula is ∫ f'(x) dx = f(x) + C.
Posted by Siddharth Yadav 2 years, 9 months ago
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Preeti Dabral 2 years, 9 months ago
Given,
There are some dietary requirements of the staple food of the adolescent students of some schools.
To Find,
(1) 2 food items including one pulse and one cereal.
(2)The minimum cost of one pulse and one cereal.
(3) The formulated version of the corresponding linear programming problem.
(4)The problem graph,
Solution,
We can solve this mathematical linear programming problem using the method.
At first let's assume that the 2 food items be, Wheat and arhar dal or pigeon peas.
The rate of Wheat is Rs. 20 per kg and the rate of Arhar dal is Rs. 100 per kg.
So, we need to minimize the 100x+20y.
We know protein in arhar dal is 220gm / kg and in wheat 100gm/kg.
Carbohydrate in arhar dal is 630gm / kg and in wheat 760gm/kg.
Suppose the dietary requirements of adolescent students are as below,
- The requirement for protein is 60gm
- Carbohydrate is 1500gm.
So we can formulate our corresponding linear programming problem as follows,
Zminimize=100x+20y
Constrains are-
{tex}\begin{aligned} & 220 x+100 y \geq 60 \\ & 630 x+760 y \geq 1500 \text { and } \mathrm{x}, \mathrm{y} \geq 0 \end{aligned}{/tex}
Hence, the answers are as follows- The two food items including one pulse and one cereal are Wheat and Arhar dal/ pigeon peas. The arhar dal is Rs.100/kg and wheat is Rs. 20/kg. The linear programming problem is -Zminimize=100x+20y, Constrains are- {tex}\begin{aligned} & 220 x+100 y \geq 60 \\ & 630 x+760 y \geq 1500 \text { and } \mathrm{x}, \mathrm{y} \geq 0 \end{aligned}{/tex} and x,y{tex}\geq 0{/tex}
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Preeti Dabral 2 years, 9 months ago
In geometry, a locus is a set of all points, whose location satisfies or is determined by one or more specified conditions. In other words, the set of the points that satisfy some property is often called the locus of a point satisfying this property.
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Sofil Dangi 2 years, 8 months ago
5Thank You