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3 years ago

Solve the proportion below. 5/x = 30/72 a.12 b.9 c.6 d.30

Mathematics

2 answers:

AnnyKZ [126]3 years ago

3 0

Answer:

A. 12

Step-by-step explanation:

Cross multiply:

30x = 72(5)

x = 360/30

x = 12

Stels [109]3 years ago

3 0

Hey there! :)

Answer:

x = 12.

Step-by-step explanation:

Set the two proportions equal to each other:

$\frac{5}{x} = \frac{30}{72}$

Cross multiply to solve for x:

5 · 72 = 30 · x

360 = 30x

Divide both sides by 30:

360/30 = 30x/30

12 = x.

Therefore, the value of 'x' in this proportion is 12.

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How many eggs in 3 and a half dozen?

MariettaO [177]

42.
As
1 dozen is 12 eggs, 3 dozens of the eggs is 12×3=36.
half of a dozens is 6 so
36+6 =42

6 0

3 years ago

Read 2 more answers

In a classroom 12 students are wearing glasses, while 6 are not. What percent of the students are wearing glasses? Explainnn

ASHA 777 [7]

Answer:

66,666666666666666666666666666666666667% :))

Step-by-step explanation:

total number of students in 1 class: 12 + 6 = 18 students

percentage of student wearing glasses are:

$\frac{12}{18} \times 100 = 66.666666667\%$

Done :))

4 0

3 years ago

How do you compare 36.782 or 37.762

Ber [7]

Answer:

36.782 < 37.762

6 0

3 years ago

WHY IS THIS GETTING DELETED?? I NEED HELP

aliina [53]

If this exact question is repeatedly deleted, it's probably because of the ambiguity of the given equation. I see two likely interpretations, for instance:

$\dfrac{(5\times5)^k}{5^{-8}} = 5^3$

$\dfrac{5\times 5^k}{5^{-8}} = 5^3$

If the first one is what you intended, then

$\dfrac{(5\times5)^k}{5^{-8}} = \dfrac{(5^2)^k}{5^{-8}} = \dfrac{5^{2k}}{5^{-8}} = 5^{2k-(-8)} = 5^{2k+8} = 5^3$

and it follows that

2k + 8 = 3 ==> 2k = -5 ==> k = -5/2

If you meant the second one, then

$\dfrac{5\times 5^k}{5^{-8}} = \dfrac{5^1\times5^k}{5^{-8}} = \dfrac{5^{k+1}}{5^{-8}} = 5^{k+1-(-8)} = 5^{k+9} = 5^3$

which would give

k + 9 = 3 ==> k = -6

And for all I know, you might have meant some other alternative... When you can, you should include a picture of your problem.

3 0

3 years ago

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