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shtirl [24]
3 years ago
6

A set of problems on Algebra, Geometry, and Combinatorics consists of 52, 70, and 88 problems. If 2 more problems from each cate

gory were added to the set, what is the largest number of tests that can be made containing an equal amount of problems from each category?
Mathematics
2 answers:
Talja [164]3 years ago
6 0
Anwser: 36
explanation: i take geometry so ik
MatroZZZ [7]3 years ago
4 0

Answer:

The largest number of tests that can be made is 36

Step-by-step explanation:

Let us list the information

  • Algebra consists of 52 problems
  • Geometry consists of 70 problems
  • Combinatorics consists of 88 problems
  • 2 more problems from each category were added to the set

That means each category will increase by 2

∴ Algebra = 52 + 2 = 54

∴ Geometry = 70 + 2 = 72

∴ Combinatorics = 88 + 2 = 90

We need to find the largest number of test that can be made containing an equal amount of problem

∵ 54 = 6 × 9

∵ 72 = 6 × 12

∵ 90 = 6 × 15

6 is the number of problem in each test because we need an equal amount of problems in each category and the other numbers are the number of tests in each category

∴ The largest number of Algebra tests is 9

∴ The largest number of Geometry tests is 12

∴ The largest number of Combinatorics test is 15

Add them to find the largest number of tests that can be made

∵ 9 + 12 + 15 = 36

∴ The largest number of tests that can be made is 36

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Solve the equation <img src="https://tex.z-dn.net/?f=%2835%20x%5E%7B4%7D%20y%2B14%20x%5E%7B5%7D%20y-2%20y%5E%7B3%7D-4x%20y%5E%7B
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\underbrace{35x^4y+14x^5y-2y^3-4xy^3}_M\,\mathrm dx+\underbrace{7x^5-6xy^2}_N\,\mathrm dy=0

M_y=35x^4+14x^5-6y^2-12xy^2
N_x=35x^4-6y^2

\dfrac{N_x-M_y}N=\dfrac{-14x^5+12xy^2}{7x^5-6xy^2}=-2

This suggests an integrating factor depending on x only is possible, and given by

\mu(x)=\exp\left(-\displaystyle\int\frac{N_x-M_y}N\,\mathrm dx\right)=e^{2x}

Distributing across the ODE, we end up with

\underbrace{(35x^4y+14x^5y-2y^3-4xy^3)e^{2x}}_{M^*}\,\mathrm dx+\underbrace{(7x^5-6xy^2)e^{2x}}_{N^*}\,\mathrm dy=0

The equation is now exact, with

{M^*}_y={N^*}_x=(35x^4+14x^5-6y^2-12xy^2)e^{2x}

Now we find the solution:

F_x=M^*
F=\displaystyle\int(35x^4+14x^5-2y^3-4xy^3)e^{2x}\,\mathrm dx
F=(7x^5y-2xy^3)e^{2x}+f(y)

F_y=N^*
(7x^5-6xy^2)e^{2x}+f'(y)=(7x^5-6xy^2)e^{2x}
f'(y)=0
\implies f(y)=C

The general solution is then

F(x,y)=(7x^5y-2xy^3)e^{2x}=C
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