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2 years ago

14

A cable company had 132 subscribers. The ratio of regular subscribers to premium subscribers was 3:8. How many regular subscribe

rs did they have?

2 answers:

svp [43]2 years ago

7 0

Answer:

49.5

Step-by-step explanation:

vampirchik [111]2 years ago

3 0

Answer: 49.5 ~50

Step-by-step explanation: 132/8=16.5 16.5(3)=49.5

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Fawn plants 2/3 of the garden with vegetables. Her son plants the remainder of the garden. He decides to use 1/2 of his space to

Furkat [3]

1/6th because he used half of a whole part. You just multiply the fraction by 2 and count how many parts are flowers in the whole garden afterward :)

Drawing it out helps too! Good luck!

4 0

2 years ago

sam had 70 erasers and rulers. after giving away 1/3 of his erasers and 10 rulers, he had an equal number of erasers and rulers

Radda [10]

Answer:

36 erasers

Step-by-step explanation:

Let number of erasers be e

let number of rulers be r

We can write:

e + r = 70

and

After giving away, he has

Erasers: 2/3e

Rulers: r - 10

These two are equal, so we can write and solve:

2/3e = r - 10

2/3e + 10 = r

Putting this in initial equation, we have:

e + (2/3e + 10) = 70

5/3e + 10 = 70

5/3 e = 60

e = 36

And rulers is:

r = 2/3(36) + 10 = 34

Hence, he had 36 erasers in the beginning

4 0

3 years ago

X + 9 <4x plz awser

Setler [38]

Answer:

x>3

Step-by-step explanation:

7 0

2 years ago

Evaluate log1∕6 36. please help

Sholpan [36]

$\log_{\tfrac 16} 36 \\\\\\=\dfrac{\ln 36}{ \ln \left(\tfrac 16\right)}\\\\\\=\dfrac{ \ln 6^2}{\ln (6^{-1})}\\\\\\=\dfrac{2 \ln 6}{-1 \ln 6}\\\\= - 2$

4 0

2 years ago

Consider the expression. (StartFraction 2 m Superscript negative 1 baseline n Superscript 5 Baseline Over 3 m Superscript 0 Base

suter [353]

Answer:

The correct option is option (3) 4 ÷ 25.

Step-by-step explanation:

The expression in terms of <em>m</em> and <em>n</em> is:

$F(m,n)=[\frac{2m^{-1}n^{5}}{3m^{0}n^{4}}]^{2}$

Exponent rule of division:

$a^{x}\div a^{y}=a^{x-y}$

Compute the value of the expression for <em>m</em> = -5 and <em>n</em> = 3 as follows:

$F(m,n)=[\frac{2m^{-1}n^{5}}{3m^{0}n^{4}}]^{2}$

$F(-5,3)=[\frac{2\csdot (-5)^{-1}\cdot (3)^{5}}{3\cdot (-5)^{0}\cdot (3)^{4}}]^{2}$

$=\{\frac{2}{3}\times [(-5)^{-1-0}\times (3)^{5-4}}]\}^{2}\\\\=\{\frac{2}{3}\times \frac{-1}{5}\times 3\}^{2}\\\\=\{-\frac{2}{5}\}^{2}\\\\=\frac{4}{25}$

Thus, the correct option is option (3) 4 ÷ 25.

8 0

3 years ago

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