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3 years ago

Write the fractional equivalent (in reduced form) to each number.

2 answers:

7 0

0.3 = 3/10

0.125 = 1/8

0.16 = 4/25

0.1 = 1/10

0.6 = 3/5

0.2 = 1/5

0.75 = 3/4

Hope this helped!
(please mark me as brainliest)

LenaWriter [7]3 years ago

6 0

Here you go:

1) 0.3 = 3/10
2) 0.125 = 1/8
3) 0.16 = 16/100 = 8/50 = 4/25
4) 0.1 = 1/10
5) 0.6 = 6/10 = 3/5
6) 0.2 = 2/10 = 1/5
7) 0.75 = 75/100 = 3/4

Please mark as Brainliest, I want to rank up :D

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2x - 7y = 14

x intercept is when y=0

2x = 14

x = 14/2 = 7

Answer: x intercept is 7

y intercept is when x=0

-7y = 14

y = -2

Answer: y intercept is -2

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3 years ago

Read 2 more answers

The sum of a number w and 3 is less than or equal to ten.

Zinaida [17]

<h2>Answer:</h2>

In this problem we must translate English words into Algebraic Expressions. So we have the following English words:

<em>The sum of a number w and 3 is less than or equal to ten</em>

<em />

So let's do that using two steps:

<h3>1. The sum of a number w and 3:</h3>

Here we have two numbers w and 3. So we must makes the sum these numbers, therefore:

$w+3$

<h3>2. The sum of a number w and 3 is less than or equal to ten:</h3>

The symbol for the English words less than or equal to is ≤ hence we can write these words as:

$\boxed{w+3\leq 10}$

As you can see, the algebraic expression is an inequality.

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4 years ago

A tank contains 250 liters of fluid in which 40 grams of salt is dissolved. Brine containing 1 gram of salt per liter is then pu

VashaNatasha [74]

Answer:

$A(t)=250-210e^{-\frac{t}{50}}$

Step-by-step explanation:

We are given that

Volume,V=250 L

Mass of salt=m=40 g

Brine containing 1 g per liter

Pumped into the tank at the rate=5 L/min

We have to find the number A(t) of grams of salt in the tank at time t.

Rate of change of salt in the tank

$\frac{dA}{dt}=$ Rate in-Rate out

Rate in=5 L/min

Rate out= $\frac{A}{250}\times 5=\frac{A}{50}$ L/min

$\frac{dA}{dt}=5-\frac{A}{50}=\frac{250-A}{50}$

$\int \frac{50dA}{250-A}=\int dt$

$-50ln(250-A)=t+C$

Using the formula

$\int \frac{dx}{x}=ln x$

A=40 and t=0

$-50ln(250-40)=0+C$

$-50ln(210)=C$

Substitute the value

$-50ln(250-A)=t-50ln(210)$

$-50ln(250-A)+50ln(210)=t$

$50ln\frac{210}{250-A}=t$

$t=50ln\frac{210}{250-A}$

$\frac{t}{50}=ln\frac{210}{250-A}$

$\frac{210}{250-A}=e^{\frac{t}{50}$

$\frac{250-A}{210}=e^{-\frac{t}{50}}$

$250-A=210e^{-\frac{t}{50}}$

$A(t)=250-210e^{-\frac{t}{50}}$

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