Shelly ordered the following items below for herself and 5 teammates:

Solve the system of equations. x+y=4 y=x^2 - 8x + 16 a) {(-3,7).(-4, 8)} b) [(4,0)} c) {(3,1),(4,0) d) {(3,7). (4.0)} e) none

Marysya12 [62]

Answer: The required solution of the given system is

(x, y) = (3, 1) and (4, 0).

Step-by-step explanation: We are given to solve the following system of equations :

$x+y=4~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)\\\\y=x^2-8x+16~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(ii)$

From equation (i), we have

$x+y=4\\\\\Rightarrow y=4-x~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(iii)$

Substituting the value of y from equation (iii) in equation (ii), we get

$y=x^2-8x+16\\\\\Rightarrow 4-x=x^2-8x+16\\\\\Rightarrow x^2-8x+16-4+x=0\\\\\Rightarrow x^2-7x+12=0\\\\\Rightarrow x^2-4x-3x+12=0\\\\\Rightarrow x(x-4)-3(x-4)=0\\\\\Rightarrow (x-3)(x-4)=0\\\\\Rightarrow x-3=0,~~~~~~~x-4=0\\\\\Rightarrow x=3,~4.$

When, x = 3, then from (iii), we get

$y=4-3=1.$

And, when x = 4, then from (iii), we get

$y=4-4=0.$

Thus, the required solution of the given system is

(x, y) = (3, 1) and (4, 0).

8 0

3 years ago

The simplified expression

Romashka-Z-Leto [24]

Answer:

$5x^2 y^2$

Step-by-step explanation:

We need to use the properties shown below to solve this:

1. $\sqrt[n]{x^a} =x^{\frac{a}{n}}$

2. $\sqrt{x}\sqrt{x} =x$

3. $\sqrt{x} \sqrt{y}=\sqrt{x*y}$

Area of a triangle is given by 1/2 * base * height, so we do that and simplify:

$A=\frac{1}{2}(\sqrt{5x^3} )(2\sqrt{5xy^4} )\\A=\frac{1}{2}(5x^3)^{\frac{1}{2}}*2*(5xy^4)^{\frac{1}{2}}\\A=\sqrt{5}x^{\frac{3}{2}}*\sqrt{5}\sqrt{x} } y^2\\A=\sqrt{5} \sqrt{5}x^{\frac{3}{2}} x^{\frac{1}{2}}y^2\\A=5*x^2y^2\\A=5x^2 y^2$

6 0

3 years ago

Which function is undefined for x = 0? y=3√x-2 y=√x-2 y=3√x+2 y=√x=2

Andre45 [30]

For this case, we must indicate which of the given functions is not defined for $x = 0$

By definition, we know that:

$f (x) = \sqrt {x}$ has a domain from 0 to infinity.

Adding or removing numbers to the variable within the root implies a translation of the function vertically or horizontally. For it to be defined, the term within the root must be positive.

Thus, we observe that:

$y = \sqrt {x-2}$ is not defined, the term inside the root is negative when $x = 0$ .

While $y = \sqrt {x + 2}$ if it is defined for $x = 0.$

$f(x)=\sqrt[3]{x}$ , your domain is given by all real numbers.

Adding or removing numbers to the variable within the root implies a translation of the function vertically or horizontally. In the same way, its domain will be given by the real numbers, independently of the sign of the term inside the root.

So, we have:

$y = \sqrt [3] {x-2}$ with x = 0: $y = \sqrt [3] {- 2}$ is defined.