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Bingel [31]
3 years ago
11

Alan bought 20 yards of fencing. he used 5.9 yards to surround one flower garden and 10.3 yards to surround another garden.write

the amount remaining as a fraction in simplest form.
Mathematics
2 answers:
marusya05 [52]3 years ago
8 0
The answer for this question is that the remaining amount fencing left after Alan has completed his garden will be 3 4/5 yards. This can be worked out, by first determining that the overall amount of fencing Alan used was 5.9 + 10.3, to create a total of 16.2. The remaining amount of fence can therefore be calculated by subtracting 16.2 from 20 to get 3.8 yeards. As a fraction this would then by written as 3 8/10, which can then be simplified down to get your answer of 3 4/5.
Paladinen [302]3 years ago
4 0

<u>step 1</u>

<u>Determine the total amount of fencing used by Alan</u>

5.9+10.3=16.2\ yards

<u>step 2</u>

<u>Determine the remaining amount of fence</u>

Subtract 16.2 from 20

20-16.2=3.8\ yards

<u>step 3</u>

<u>convert the amount remaining as a fraction in simplest form</u>

3.8= 3+0.8 =3+\frac{8}{10} = 3\frac{4}{5} \ yards

therefore

<u>the answer is</u>

3\frac{4}{5} \ yards


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If 20% of a school was made up of out-of-district students last year, but the number of out-of-district students increased by 50
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<h2>30% is enrolled out-of-district.</h2>

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4 0
3 years ago
72÷x=3.6÷0.015 what is the answer?
lapo4ka [179]

Answer:

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Step-by-step explanation:

D( x )

x = 0

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4 0
3 years ago
Read 2 more answers
Remember to show work and explain. Use the math font.
MrMuchimi

Answer:

\large\boxed{1.\ f^{-1}(x)=4\log(x\sqrt[4]2)}\\\\\boxed{2.\ f^{-1}(x)=\log(x^5+5)}\\\\\boxed{3.\ f^{-1}(x)=\sqrt{4^{x-1}}}

Step-by-step explanation:

\log_ab=c\iff a^c=b\\\\n\log_ab=\log_ab^n\\\\a^{\log_ab}=b\\\\\log_aa^n=n\\\\\log_{10}a=\log a\\=============================

1.\\y=\left(\dfrac{5^x}{2}\right)^\frac{1}{4}\\\\\text{Exchange x and y. Solve for y:}\\\\\left(\dfrac{5^y}{2}\right)^\frac{1}{4}=x\qquad\text{use}\ \left(\dfrac{a}{b}\right)^n=\dfrac{a^n}{b^n}\\\\\dfrac{(5^y)^\frac{1}{4}}{2^\frac{1}{4}}=x\qquad\text{multiply both sides by }\ 2^\frac{1}{4}\\\\\left(5^y\right)^\frac{1}{4}=2^\frac{1}{4}x\qquad\text{use}\ (a^n)^m=a^{nm}\\\\5^{\frac{1}{4}y}=2^\frac{1}{4}x\qquad\log_5\ \text{of both sides}

\log_55^{\frac{1}{4}y}=\log_5\left(2^\frac{1}{4}x\right)\qquad\text{use}\ a^\frac{1}{n}=\sqrt[n]{a}\\\\\dfrac{1}{4}y=\log(x\sqrt[4]2)\qquad\text{multiply both sides by 4}\\\\y=4\log(x\sqrt[4]2)

--------------------------\\2.\\y=(10^x-5)^\frac{1}{5}\\\\\text{Exchange x and y. Solve for y:}\\\\(10^y-5)^\frac{1}{5}=x\qquad\text{5 power of both sides}\\\\\bigg[(10^y-5)^\frac{1}{5}\bigg]^5=x^5\qquad\text{use}\ (a^n)^m=a^{nm}\\\\(10^y-5)^{\frac{1}{5}\cdot5}=x^5\\\\10^y-5=x^5\qquad\text{add 5 to both sides}\\\\10^y=x^5+5\qquad\log\ \text{of both sides}\\\\\log10^y=\log(x^5+5)\Rightarrow y=\log(x^5+5)

--------------------------\\3.\\y=\log_4(4x^2)\\\\\text{Exchange x and y. Solve for y:}\\\\\log_4(4y^2)=x\Rightarrow4^{\log_4(4y^2)}=4^x\\\\4y^2=4^x\qquad\text{divide both sides by 4}\\\\y^2=\dfrac{4^x}{4}\qquad\text{use}\ \dfrac{a^n}{a^m}=a^{n-m}\\\\y^2=4^{x-1}\Rightarrow y=\sqrt{4^{x-1}}

6 0
3 years ago
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