i am lousy at math~ plz help mi solve!

**xiaoz** · June 6th 2006, 08:02 PM

[IMG]D:\Documents and Settings\fang\My Documents\Scan0002.jpg[/IMG]

OPS~ how to put picture?!?!?!?!

**CaptainBlack** · June 6th 2006, 08:54 PM

Originally Posted by xiaoz

[IMG]D:\Documents and Settings\fang\My Documents\Scan0002.jpg[/IMG]

OPS~ how to put picture?!?!?!?!

When creating a post scroll down past the text editor window,
you will see a button marked manage attachments, click on
it and it will allow you to upload a jpg attachment.

RonL

**xiaoz** · June 6th 2006, 09:07 PM

kk. thxz .. here is questions i dunno.. i am confuse at math tis yr!!

**xiaoz** · June 6th 2006, 09:08 PM

they say canot use calculator.. i dunno how to do the working.. my math fail badly too.. HELP!

**CaptainBlack** · June 6th 2006, 09:21 PM

1.a Evaluate:

$\sqrt{1\frac{11}{25}}$

first change what is under the root symbol from a mixed number to an
improper fraction:

$ \sqrt{1\frac{11}{25}}=\sqrt{\frac{36}{25}}=\frac{ \sqrt {36}}{\sqrt{25}} $

You should now be able to complete this without a calculator as
both $36$ and $25$ are perfect squares.

RonL

**xiaoz** · June 6th 2006, 09:49 PM

o.. i find out the answer liao... is 6 over 5.. thx!!

1b.. explain plz..

**CaptainBlack** · June 6th 2006, 11:19 PM

1.b Evaluate $\sqrt{2120}$ without a calculator given that

$\sqrt{2.12}=1.456$ and $\sqrt{21.2}=4.004$ .

We want to write $2120$ as a product of a square and one of
the number we are given the square root of. Now:

$2120=1000 \times 2.12$ ,

and

$2120=100 \times 21.2$ .

$1000$ is not the square of an integer, but $100=10^2$ . So we can write:

$ \sqrt{2120}=\sqrt{100 \times 21.2}=\sqrt{100}\times \sqrt{21.2} $ .

RonL

**CaptainBlack** · June 6th 2006, 11:32 PM

1.c Evaluate without a calculator:

$ \left(-\frac{1}{216}\right)^{-2/3} $

Now:

$ \left(-\frac{1}{216}\right)^{-2/3}=\frac{1}{(-\frac{1}{216})^{2/3}} $ $ =\frac{1}{((-\frac{1}{216})^{1/3})^2}=\frac{1}{((-\frac{1}{(216)^{1/3}}))^2} $

Now this can be evaluated from the innermost brackets outward, as 216 is
a perfect cube (of a small number which you can find by trial and error).

RonL

**Soroban** · June 7th 2006, 03:41 AM

Hello,xiaoz!

Here's #6 . . .

6(a) Solve the inequality: $5 \;< \;6x - 13 \;< \;7$

We have: . $5 \;< \; 6x - 13 \;< \;7$

Add 13 to each side: . $18 \:< \:6x \:< \:20$

Divide by 6: . $3 \;< \;x \;< \;\frac{10}{3}$

(b) Find a fraction $x$ such that: $\frac{7}{11}\;<\:x\:<\:\frac{8}{11}$

There are several ways to solve this problem.

(1) To find a number between two given numbers, average the numbers.
. . . $x\;=\;\frac{\frac{7}{11} + \frac{8}{11}}{2} \;=\;\frac{15}{22}$

(2) Convert to decimals: $\begin{array}{cc}\frac{7}{11} = 0.63\hdots \\ \frac{8}{11} = 0.72\cdots\end{array}$
Then let $x$ be any decimal between $0.64$ and $0.72$

For example: . $x\:=\:0.65\:=\:\frac{65}{100} \:= \:\frac{13}{20}$

(c) If $x^{2y} = 5$ , find the value of $2x^{6y} - 4$

Change the first expression into the second, step by step.

We have: . $x^{2y}\:=\:5$

Cube both sides: . $(x^{2y})^3\:=\:5^3\quad\Rightarrow\quad x^{6y}\:=\:125$

Multiply by 2: . $2x^{6y}\;=\;250$

Subtract 4: . $2x^{6y} - 4 \:= \:246$

**xiaoz** · June 7th 2006, 05:02 AM

thxz.. 1b i undersyand.. 1c i dunno still... 6c the last part.. the mutiply by 2 de.. got the power still can mutiply?? how about question 2a and b.. explaint plz.. thxz

**Soroban** · June 7th 2006, 06:37 AM

Hello, xiaoz!

Let me give 1(c) a try . . .

1(c) Simplify: $\left(-\frac{1}{216}\right)^{-\frac{2}{3}}$

Did you know that a negative exponent on a fraction flips the fraction?
. . $\left(\frac{a}{b}\right)^{-n}\;=\;\left(\frac{b}{a}\right)^n$

We have: $\left(-\frac{1}{216}\right)^{-\frac{2}{3}}\:=\: (-216)^{\frac{2}{3}}$

We know that: $-216 = (-6)^3$

So we have: $[(-6)^3]^{\frac{2}{3}}\;=\;6^{(3\cdot\frac{2}{3})} \;=\;6^2\;=\;36$

**xiaoz** · June 7th 2006, 09:40 AM

woow..i did not know about tt formula.. hahax.. thxz to u.. i understand liao..
other question.. plz help.. explain to mi..

**Ranger SVO** · June 7th 2006, 06:42 PM

http://www.mathhelpforum.com/math-he...ntid=630&stc=1

**xiaoz** · June 7th 2006, 09:25 PM

kk.. thxz all...now plz help mi wif question 2 , 3 , 4 , 5

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