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aivan3 [116]
1 year ago
8

Explain how you can determine the number of real number solutions of a system of equations in which one equation is linear and t

he other is quadratic–without graphing the system of equations.
Mathematics
1 answer:
Assoli18 [71]1 year ago
3 0

Answer:

To determine the number of real number solutions of as system of equations in which one equation is linear and the other is quadratic

1) Given that there are two variables, x and y as an example, we make y the subject of the equation of the linear equation and substitute the the expression for y in x into the quadratic equation

We simplify and check the number of real roots with the quadratic formula, x = \dfrac{-b\pm \sqrt{b^{2}-4\cdot a\cdot c}}{2\cdot a} for quadratic equations  the form 0 = a·x² - b·x + c

Where b² > 4·a·c there are two possible solutions and when b² = 4·a·c equation there is only one solution.

Step-by-step explanation:

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There are 1751 members under the age of 25.

7 0
1 year ago
A one story home has a total square footage of 1,200. If 900sq ft of the home has a basement, and 300 sq ft has a slab foundatio
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Answer:

it has 1200

Step-by-step explanation:

because 900+300=1200

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1 year ago
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If h:k = 2:5, x:y=3:4 and 2h+x:k+2y = 1:2, find the ratio h-x: k-y.​
fenix001 [56]

Given:

h:k=2:5,x:y=3:4,2h+x:k+2y=1:2

To find:

h-x:k-y

Solution:

We have, h:k=2:5 and x:y=3:4.

Let the values of h and k are 2a and 5a respectively.

Let the values of x and y are 3b and 4b respectively.

We have,

2h+x:k+2y=1:2

It can be written as

\dfrac{2h+x}{k+2y}=\dfrac{1}{2}

\dfrac{2(2a)+(3b)}{5a+2(4b)}=\dfrac{1}{2}

\dfrac{4a+3b}{5a+8b}=\dfrac{1}{2}

2(4a+3b)=1(5a+8b)

On further simplification, we get

8a+6b=5a+8b

8a-5a=8b-6b

3a=2b

a=\dfrac{2b}{3}

Now,

h-x:k-y=\dfrac{h-x}{k-y}

h-x:k-y=\dfrac{2-3b}{5a-4b}

h-x:k-y=\dfrac{2\times \dfrac{2b}{3}-3b}{5\times \dfrac{2b}{3}-4b}

h-x:k-y=\dfrac{\dfrac{4b}{3}-3b}{\dfrac{10b}{3}-4b}

On further simplification, we get

h-x:k-y=\dfrac{\dfrac{4b-9b}{3}}{\dfrac{10b-12b}{3}}

h-x:k-y=\dfrac{-5b}{-2b}

h-x:k-y=\dfrac{5}{2}

h-x:k-y=5:2

Therefore, the ratio h-x:k-y is 5:2.

8 0
1 year ago
Find the value of cosAcos2Acos3A...........cos998Acos999A where A=2π/1999
Lady bird [3.3K]
Hello,

Here is the demonstration in the book Person Guide to Mathematic by Khattar Dinesh.

Let's assume
P=cos(a)*cos(2a)*cos(3a)*....*cos(998a)*cos(999a)
Q=sin(a)*sin(2a)*sin(3a)*....*sin(998a)*sin(999a)

As sin x *cos x=sin (2x) /2

P*Q=1/2*sin(2a)*1/2sin(4a)*1/2*sin(6a)*....
         *1/2* sin(2*998a)*1/2*sin(2*999a) (there are 999 factors)
= 1/(2^999) * sin(2a)*sin(4a)*...
     *sin(998a)*sin(1000a)*sin(1002a)*....*sin(1996a)*sin(1998a)
 as sin(x)=-sin(2pi-x) and 2pi=1999a

sin(1000a)=-sin(2pi-1000a)=-sin(1999a-1000a)=-sin(999a)
sin(1002a)=-sin(2pi-1002a)=-sin(1999a-1002a)=-sin(997a)
...
sin(1996a)=-sin(2pi-1996a)=-sin(1999a-1996a)=-sin(3a)
sin(1998a)=-sin(2pi-1998a)=-sin(1999a-1998a)=-sin(a)

So  sin(2a)*sin(4a)*...
     *sin(998a)*sin(1000a)*sin(1002a)*....*sin(1996a)*sin(1998a)
= sin(a)*sin(2a)*sin(3a)*....*sin(998)*sin(999) since there are 500 sign "-".

Thus
P*Q=1/2^999*Q or Q!=0 then
P=1/(2^999)

       








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Answer:

5

Step-by-step explanation:

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