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4 years ago

The__________ method is a way of solving a system of equations using addition or subtraction to eliminate a variable.

Mathematics

1 answer:

Mandarinka [93]4 years ago

8 0

The anwer is like term

You might be interested in

Find all real solutions to the equation (x² − 6x +3)(2x² − 4x − 7) = 0.

Jet001 [13]

Answer:

x = 3 + √6 ; x = 3 - √6 ; $x = \frac{2+3\sqrt{2}}{2}$ ; $x = \frac{2-(3)\sqrt{2}}{2}$

Step-by-step explanation:

Relation given in the question:

(x² − 6x +3)(2x² − 4x − 7) = 0

Now,

for the above relation to be true the following condition must be followed:

Either (x² − 6x +3) = 0 ............(1)

(2x² − 4x − 7) = 0 ..........(2)

now considering the equation (1)

(x² − 6x +3) = 0

the roots can be found out as:

$x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}$

for the equation ax² + bx + c = 0

thus,

the roots are

$x = \frac{-(-6)\pm\sqrt{(-6)^2-4\times1\times(3)}}{2\times(1)}$

$x = \frac{6\pm\sqrt{36-12}}{2}$

$x = \frac{6+\sqrt{24}}{2}$ and, x = $x = \frac{6-\sqrt{24}}{2}$

$x = \frac{6+2\sqrt{6}}{2}$ and, x = $x = \frac{6-2\sqrt{6}}{2}$

x = 3 + √6 and x = 3 - √6

similarly for (2x² − 4x − 7) = 0.

we have

the roots are

$x = \frac{-(-4)\pm\sqrt{(-4)^2-4\times2\times(-7)}}{2\times(2)}$

$x = \frac{4\pm\sqrt{16+56}}{4}$

$x = \frac{4+\sqrt{72}}{4}$ and, x = $x = \frac{4-\sqrt{72}}{4}$

$x = \frac{4+\sqrt{2^2\times3^2\times2}}{2}$ and, x = $x = \frac{4-\sqrt{2^2\times3^2\times2}}{4}$

$x = \frac{4+(2\times3)\sqrt{2}}{2}$ and, x = $x = \frac{4-(2\times3)\sqrt{2}}{4}$

$x = \frac{2+3\sqrt{2}}{2}$ and, $x = \frac{2-(3)\sqrt{2}}{2}$

Hence, the possible roots are

x = 3 + √6 ; x = 3 - √6 ; $x = \frac{2+3\sqrt{2}}{2}$ ; $x = \frac{2-(3)\sqrt{2}}{2}$

7 0

3 years ago

5x^{4}-7x^{3}-5x^{2}+5x+1-0

den301095 [7]

Answer:

5x^4 -7x^3-5x^2+5x+1

Step-by-step explanation:

Hey!

You would just add all the numbers up, and by making sure that you combine all like terms. In this case it is just 1-0 = 1 and so the solution would be the same as your question but with a 1 and you would get rid of the -0.

Hope this helps!

7 0

3 years ago

If f(x)=3x^3 then what is the area enclosed by the graph of the function, the horizontal axis, and vertical lines at x=2 and x=4

Ber [7]

Answer:

Area: 180 units2 (units 2 is because since the are no specific unit given but every area should have a unit of measurement)

Step-by-step explanation:

The area enclosed by the graph of the function, the horizontal axis, and vertical lines is the integral of the function between thos two points (x=2 and x=4)

So , let's solve the integral of f(x)

Area = $\int\limits^2_4 3{x}^3 \, dx = 3*x^4/4$ +C