# Thread: [SOLVED] Help! Many algebra problems :(

1. ## 2nd attempt

Originally Posted by Elite00
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13. Graph the following piecewise function and give the domain, range, and zeros.
f (x ) = {2x + 3, x<4
{x -1, 4 < x < 9

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Hello,

you have $f(x)=\left\{\begin{array}{l}2x+3,x<4\\x-1,4

I've attached the graph of this function.

From the grap you can see: The domain is $d=(-\infty,4)\ \cup\ (4,9)$. According to the wording of this problem the 4 doesn't belong to the domain.

The range is $r=(-\infty,11)$

There is one zeroat $x=-\frac{3}{2}$

EB

2. Originally Posted by Elite00
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15. A lab technician is trying to make an 8% solution by combining a 2% and 12% solution. If he is trying to make a 150mL solution, how many mL of each solution should be mixed?
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Hello,

let x be the amount of the 2%-solution
let y be the amount of the 12%-solution

Then you get 2 simultanous equations:
A: x+y=150
B: 0.02*x+0.12*y=0.08*150=12

Multiply the equation B by 50 and calculate 50*B-A:

5y=450. Thus y = 90. Plug in this result into A. You'll get x = 60.

EB

3. Originally Posted by Elite00
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20. Solve the following system of equations: { -3x + 6y – 4z = 8
{x + 2z – 4y = -3
{8y=z

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Hello,

Plug in 8y for z in equation A and B.

$\begin{array}{r}A': -3x - 26y = 8\\B': x + 12y = -3 \end{array}$

Multiply B' by 3 and add A'+3*B'

10y = -1. Thus y = -1/10. Calculate z = -4/z and finally x = -9/5

EB

4. Originally Posted by Elite00
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21. Solve 4(x+5)^2 = 20. express the answer exactly and if necessary, rounded to the nearest tenth.

22. Factor 2x^2 – 10x – 48 =0 and solve for x

23. Find the exact values of x in the following quadratic equation. 2x^2 – 5x -12 = 0

24. Simplify the following complex number: 3 + i / 2 – 3i

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Hello,

to 21.)
$4(x+5)^2 = 20\Longleftrightarrow(x+5)^2=5\Longleftrightarrow x+5=\sqrt{5}\vee x+5=-\sqrt{5}$
Thus the exact solutions are: $x=-5+\sqrt{5}\ \vee \ x=-5-\sqrt{5}$
Approximate resultes: x = -2.8 or x = -7.3

to 22.)
$2x^2 - 10x - 48 =0 \Longleftrightarrow 2(x^2-5x-24)=0 \Longleftrightarrow 2(x-8)(x+3)=0$
A produkt is zero if one of the factors is zero. Thus you'll get the solutions: x = -3 or x = 8

to 23.)
Use the formula you know to solve this equation: $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

$x=\frac{5\pm\sqrt{25+96}}{4}$
$x=\frac{5\pm 11}{4}$. Thus x = 4 or x = -3/2

to 24.) Multiply denominator and numerator with the conjugated(?) value of the denominator:
$\frac{3+i}{2-3i} \cdot \frac{2+3i}{2+3i}=\frac{6+9i+2i+3i^2}{4-(-9)}=\frac{3+11i}{13}$

EB

5. Originally Posted by Elite00
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16. Graph, shade the feasible region and label the corner points for the following system of inequalities: {y < -2x -2
{y > 1/2x
{x > -3

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Hello,

have a look at the attached diagram.

EB

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