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3 years ago

Walter had 2/4 yard of string. He used 3/6 of it on a school project. How much string did he use?

Mathematics

2 answers:

andrezito [222]3 years ago

8 0

1/4 is the correct answer

frozen [14]3 years ago

3 0

2/4 reduces to 1/2

3/6 reduces to 1/2

1/2 * 1/2 = 1/4

he used 1/4 yard of string

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Which number should follow in this series: 64/64 9/12 6/12

zheka24 [161]

Answer: $\frac{1}{4}$

Step-by-step explanation:

You can reduce the fractions, then:

$\frac{64}{64}=1$

$\frac{9}{12}=\frac{3}{4}$

$\frac{6}{12}=\frac{3}{6}=\frac{1}{2}$

Rewrite them as following:

$1,\frac{3}{4},\frac{1}{2}$

If you subtract the first number and the second number, you obtain:

$1-\frac{3}{4}=\frac{1}{4}$

If you subtract the second number and the second number, you obtain:

$\frac{3}{4}-\frac{1}{2}=\frac{1}{4}$

Therefore, you must subtract $\frac{1}{2}$ and $\frac{1}{4}$ to obtain the number asked. Then, this is:

$\frac{1}{2}-\frac{1}{4}=\frac{1}{4}$

6 0

3 years ago

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Jim is saving money to buy a game. So far he saved $36, which is three-fourths of the total cost of the game. How much does the

Andru [333]

Answer:

$48

Step-by-step explanation:

Because he has 3/4 you would divide 36 by 3 whick is 12 then multiply by 4 to get the total of $48

4 0

3 years ago

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Given the functionf ( x ) = x^2 + 7 x + 10/ x^2 + 9 x + 20

vladimir1956 [14]

<em>x = -4 is a vertical asymptote for the function.</em>

<h2>Explanation:</h2>

The graph of $y=f(x)$ is a vertical has an asymptote at $x=a$ if at least one of the following statements is true:

$1) \ \underset{x\rightarrow a^{-}}{lim}f(x)=\infty\\ \\ 2) \ \underset{x\rightarrow a^{-}}{lim}f(x)=-\infty \\ \\ 3) \ \underset{x\rightarrow a^{+}}{lim}f(x)=\infty \\ \\ 4) \ \underset{x\rightarrow a^{+}}{lim}f(x)=\infty$

The function is:

$f(x)=\frac{x^2+7x+10}{x^2+9x+20}$

First of all, let't factor out:

$f(x)=\frac{x^2+5x+2x+10}{x^2+5x+4x+20} \\ \\ f(x)=\frac{x(x+5)+2(x+5)}{x(x+5)+4(x+5)} \\ \\ f(x)=\frac{(x+5)(x+2)}{(x+5)(x+4)} \\ \\ f(x)=\frac{(x+2)}{(x+4)}, \ x\neq 5$

From here:

$\bullet \ When \ x \ approaches \ -4 \ on \ the \ right: \\ \\ \underset{x\rightarrow -4^{+}}{lim}\frac{(x+2)}{(x+4)}=? \\ \\ \underset{x\rightarrow -4^{+}}{lim}\frac{(-4^{+}+2)}{(-4^{+}+4)} \\ \\ \\ The \ numerator \ is \ negative \ and \ the \ denominator \\ is \ a \ small \ positive \ number. \ So: \\ \\ \underset{x\rightarrow -4^{+}}{lim}\frac{(x+2)}{(x+4)}=-\infty$

$\bullet \ When \ x \ approaches \ -4 \ on \ the \ left: \\ \\ \underset{x\rightarrow -4^{-}}{lim}\frac{(x+2)}{(x+4)}=? \\ \\ \underset{x\rightarrow -4^{-}}{lim}\frac{(-4^{-}+2)}{(-4^{-}+4)} \\ \\ \\ The \ numerator \ is \ a \ negative \ and \ the \ denominator \\ is \ a \ small \ negative \ number \ too. \ So: \\ \\ \underset{x\rightarrow -4^{-}}{lim}\frac{(x+2)}{(x+4)}=+\infty$