Will enyone help me answer this question please

What should you substitute for y in the bottom equation to solve the system by the substitution method?

Contact [7]

Answer: y= -x-5

Step 1: Determine which equation to change

Since we will be using the substitution method, we will need to substitute y from the bottom equation with information from the top equation. This will be the equation we change.

Step 2: Change the equation

Let’s rewrite the equation we will be working with.

3x+3y= -15

Our goal will be to get y alone on the left side. To start, let’s subtract 3x from both sides, leaving only 3y on the left side.

3x+3y= -15
3y= -3x-15

Now we need to get y completely alone by eliminating the coefficient of 3. Let’s do this by dividing each term by 3.

3y= -3x-15
y= -x-5

This is your answer. Hope this helps! Comment below for more questions.

3 0

3 years ago

4/6 = x/9 <br>what x is? Can solve it

drek231 [11]

So for this, multiply both sides by 9, and your answer will be:

$\frac{36}{6}=x\\\\6=x$

8 0

3 years ago

Read 2 more answers

Let $f(x) = x^2$ and $g(x) = \sqrt{x}$. Find the area bounded by $f(x)$ and $g(x).$

Anna [14]

Answer:

$\large\boxed{1\dfrac{1}{3}\ u^2}$

Step-by-step explanation:

Let's sketch graphs of functions f(x) and g(x) on one coordinate system (attachment).

Let's calculate the common points:

$x^2=\sqrt{x}\qquad\text{square of both sides}\\\\(x^2)^2=\left(\sqrt{x}\right)^2\\\\x^4=x\qquad\text{subtract}\ x\ \text{from both sides}\\\\x^4-x=0\qquad\text{distribute}\\\\x(x^3-1)=0\iff x=0\ \vee\ x^3-1=0\\\\x^3-1=0\qquad\text{add 1 to both sides}\\\\x^3=1\to x=\sqrt[3]1\to x=1$

The area to be calculated is the area in the interval [0, 1] bounded by the graph g(x) and the axis x minus the area bounded by the graph f(x) and the axis x.

We have integrals:

$\int\limits_{0}^1(\sqrt{x})dx-\int\limits_{0}^1(x^2)dx=(*)\\\\\int(\sqrt{x})dx=\int\left(x^\frac{1}{2}\right)dx=\dfrac{2}{3}x^\frac{3}{2}=\dfrac{2x\sqrt{x}}{3}\\\\\int(x^2)dx=\dfrac{1}{3}x^3\\\\(*)=\left(\dfrac{2x\sqrt{x}}{2}\right]^1_0-\left(\dfrac{1}{3}x^3\right]^1_0=\dfrac{2(1)\sqrt{1}}{2}-\dfrac{2(0)\sqrt{0}}{2}-\left(\dfrac{1}{3}(1)^3-\dfrac{1}{3}(0)^3\right)\\\\=\dfrac{2(1)(1)}{2}-\dfrac{2(0)(0)}{2}-\dfrac{1}{3}(1)}+\dfrac{1}{3}(0)=2-0-\dfrac{1}{3}+0=1\dfrac{1}{3}$