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3 years ago

Solve the system below.

1 answer:

6 0

To solve this system, let y = x^2 - 4 = -4. Then y = x^2. This is true for all x. There is no single numerical solution. y = x^2 is a function with infinitely many inputs/outputs.

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I need help with the area of this

Anarel [89]

Answer:

14 inches

Step-by-step explanation:

If you use the formula shown and calculate 1/2 (B x H) you will get 14 in. The base multiplied by the height is 28, and when divided in half, it is 14 in.

7 0

3 years ago

What is the letter a? 6a+14=-7-a. a=?

grin007 [14]

6a+14=-7-a

you simplify the equation to the form, which is simple to understand
6a+14=-7-a

you move all terms containing a to the left and all other terms to the right.
+6a+1a=-7-14

you need to simplify left and right side of the equation.
+7a=-21

you divide both sides of the equation by 7 to get a.
a=-3

5 0

3 years ago

Read 2 more answers

Simplify the following expression. 2x^2 + 7x+ 3 / x^2 - 9

podryga [215]

Answer:

(x3 - 3x2 - 6x - 3) • (2x - 1)

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Step-by-step explanation:

6 0

3 years ago

Can you please help me out and when you do it please the division house

Zolol [24]

Answer:

115.75

Step-by-step explanation:

463/4 friends = 115.75

5 0

3 years ago

Read 2 more answers

Identify the x-intercepts of the function below f(x)=x^2+12x+24

damaskus [11]

ANSWER:

x-intercepts of $\mathrm{x}^{2}+12 \mathrm{x}+24=0 \text { are }(-6+2 \sqrt{3}),(-6-2 \sqrt{3})$

SOLUTION:

Given, $f(x)=x^{2}+12 x+24$ -- eqn 1

x-intercepts of the function are the points where function touches the x-axis, which means they are zeroes of the function.

Now, let us find the zeroes using quadratic formula for f(x) = 0.

$X=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$

Here, for (1) a = 1, b= 12 and c = 24

$X=\frac{-(12) \pm \sqrt{(12)^{2}-4 \times 1 \times 24}}{2 \times 1}$

$\begin{array}{l}{X=\frac{-12 \pm \sqrt{144-96}}{2}} \\\\ {X=\frac{-12 \pm \sqrt{48}}{2}} \\\\ {X=\frac{-12 \pm \sqrt{16 \times 3}}{2}} \\\\ {X=\frac{-12 \pm 4 \sqrt{3}}{2}} \\ {X=\frac{2(-6+2 \sqrt{3})}{2}, \frac{2(-6-2 \sqrt{3})}{2}} \\\\ {X=(-6+2 \sqrt{3}),(-6-2 \sqrt{3})}\end{array}$

Hence the x-intercepts of $\mathrm{x}^{2}+12 \mathrm{x}+24=0 \text { are }(-6+2 \sqrt{3}),(-6-2 \sqrt{3})$

8 0

3 years ago