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3 years ago

2x+3y=12 find four solutions

Mathematics

1 answer:

solong [7]3 years ago

5 0

Answer:

1. x=3,y=2

2.x=0,y=4

3.x=-3,y=6

4.x=6,y=0

Step-by-step explanation:

x=3, y=2

2(3)+3(2)=6+6=12

x=0, y=4

2(0)+3(4)=0+12=12

x=-3, y=6

2(-3)+3(6)=-6+18=12

x=6,y=0

2(6)+3(0)=12+0=12. Hope this helps!

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The base and the height of a triangle are multiplied by 5/4. Which of the following describes the

lions [1.4K]

Answer:

The option that describe the effect of the change is;

The perimeter is multiplied by 5/4

Step-by-step explanation:

The parameters from the question are;

The (scale) factor by which the base and the height of the triangle are multiplied = 5/4

By the constant proportion of their sides, the triangle before and the triangle obtained after the multiplication are similar

Let 'a' and 'b' represent the length of the base and the height of the triangle respectively, we have;

The area of the triangle, A₁ = 1/2 × a × b = a·b/2

With the application of the scale factor of 5/4, we have;

The area of the scaled triangle, A₂ = 1/2 × (5/4) × a × (5/4) × b = (5/4)² × a·b/2

Therefore, the scale factor of the area = (5/4)²

The scale factor of the perimeter = √(The scale facto of area)

∴ The scale factor of the perimeter = √(5/4)² = (5/4)

The scale factor of the perimeter = (5/4)

Therefore, the perimeter of the scaled triangle is obtained by multiplying the perimeter of the initial triangle by (5/4).

8 0

3 years ago

Read 2 more answers

Hi, can someone please help, and possibly explain the answer please.

mr_godi [17]

Answer:

$x=\frac{-(-2)\±\sqrt{(-2)^2-4(3)(0)} }{2(3)}$

Step-by-step explanation:

Quadratic formula: $x=\frac{-b\±\sqrt{b^2-4ac} }{2a}$ when the equation is $0=ax^2+bx+c$

The given equation is $1=-2x+3x^2+1$ . Let's first arrange this so its format looks like $y=ax^2+bx+c$ :

$1=-2x+3x^2+1$

$1=3x^2-2x+1$

Subtract 1 from both sides of the equation

$1-1=3x^2-2x+1-1\\0=3x^2-2x+0$

Now, we can easily identify 3 as a, -2 as b and 0 as c. Plug these into the quadratic formula:

$x=\frac{-b\±\sqrt{b^2-4ac} }{2a}\\x=\frac{-(-2)\±\sqrt{(-2)^2-4(3)(0)} }{2(3)}$

I hope this helps!

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crimeas [40]

Answer:

6/25

Step-by-step explanation:

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2x+3y=12 find four solutions​

2x+3y=12 find four solutions