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3 years ago

3. [02.03] Solve for x: |x + 2| + 16 = 14 (2 points) x = −32 and x = −4 x = −4 and x = 0 x = 0 and x = 28 No solution

Mathematics

2 answers:

MArishka [77]3 years ago

6 0

Answer:

The correct option is D) No solution.

Step-by-step explanation:

Consider the provided equation.

$|x + 2|+16=14$

We need to solve the above equation for x.

Subtract 16 from both the sides.

$|x + 2| + 16-16 = 14-16$

$|x + 2| = -2$

The absolute value of a real number "a" is the non-negative value of "a" without regard to its sign. It cannot be a negative number.

Thus the equation has no solution.

Hence, the correct option is D) No solution.

Svetlanka [38]3 years ago

4 0

The answer is x= -4 and x= 0, simply subtract 16 on both sides and solve it with x-2 = -2 and x+2= -2

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The graph of the quadratic function y = ax^2 + bx + c is given. Find a + b + c.

andrey2020 [161]

Answer:

$\frac{145}{24}\approx 6$ ***

Answers may vary depending on the person reading the graph.

It is hard to tell what they think crosses nicely.

For example, I decided that the graph crossed nicely at the following points:

$(0,3)$

$(2,8)$

$(6,5)$

Step-by-step explanation:

I see that the graph crosses at $(0,3)$ , $(2,8)$ , and $(6,5)$ .

The equation $y=ax^2+bx+c$ tells us the $y$ -intercept is $c$ since when $x=0$ we have $y=a(0)^2+b(0)+c=0+0+c=c$ .

The $y$ -intercept for our graph is $3$ . Therefore, $c=3$ .

So far we have the equation:

$y=ax^2+bx+3$ .

Let's enter the other points creating a system of linear equations to solve:

$8=a(2)^2+b(2)+3$

$5=a(6)^2+b(6)+3$

Let's simplify:

$8=4a+2b+3$

$5=36a+6b+3$

Let's subtract 3 on both sides:

$5=4a+2b$

$2=36a+6b$

I choose to solve the system by elimination.

Let's multiply the top equation by -3:

$-15=-12a-6b$

$2=36a+6b$

Now adding the equations results in:

$-13=24a$

Divide both sides by 24:

$\frac{-13}{24}=a$

Now using one of the equations we can find $b$ :

$5=4a+2b$ with $a=\frac{-13}{24}$

$5=4\frac{-13}{24}+2b$

$5=\frac{-13}{6}+2b$

Add 13/6 on both sides:

$5+\frac{13}{6}=2b$

$\frac{30+13}{6}=2b$

$\frac{43}{6}=2b$

Divide both sides by 2:

$\frac{43}{2(6)}=b$

$\frac{43}{12}=b$

So we have the equation:

$y=\frac{-13}{24}x^2+\frac{43}{12}+3$

Let's evaluate $a+b+c$ now:

$\frac{-13}{24}+\frac{43}{12}+3$

$\frac{-13+86+72}{24}$

$\frac{73+72}{24}$

$\frac{145}{24}$

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