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Shtirlitz [24]
3 years ago
12

2.27 is a terminating decimal true or false??

Mathematics
2 answers:
IRINA_888 [86]3 years ago
5 0

Answer:

true

Step-by-step explanation:

alina1380 [7]3 years ago
5 0

Answer:

The correct answer is True.

Step-by-step explanation:

  A terminating decimal, as the name implies, is a decimal that has an end.

The characteristic of these numbers is that their termination is with finite numbers of digits, in this case they are two numbers (27).

For a number to classify as terminating decimal it must also fulfill the function that can be transformed into a fraction and the number 2.27 fulfills it, being its fraction: \frac{227}{100}

Given this information, we can say that the answer is true.

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What is the opposite of the integer -4 plss thx :)
qwelly [4]

Answer:

+4

Step-by-step explanation:

any opposite of a natural number is created with a "minus"

- (-4) = +4

7 0
3 years ago
Read 2 more answers
Which expression is equivalent to (7x−5)−(3x−2)?
Alchen [17]

Answer:

D. 4x-3

Step-by-step explanation:

First you subtract the 7x and 3x to get 4x.

Then you subtract the -5 and the -2 to get -3

Therfore, your answer is D

4 0
3 years ago
I need help on all of #1
grandymaker [24]
I think it's fractions
6 0
3 years ago
Solve for x. 5(x 1) = 4(x 8)
tresset_1 [31]
Asssuming blanks are plus

5(x+1)=4(x+8)

distribute
a(b+c)=ab+ac

5(x+1)=5x+5

4(x+8)=4x+32

5x+5=4x+32
minus 4x from both sides
x+5=32
minus 5 from both sides
x=27
5 0
3 years ago
The intensity I of light varies inversely as the square of the distance D from the source. If the intensity of illumination on a
zhenek [66]

Answer:

The intensity on a screen 70 ft from the light is 1.728 foot candle.

Step-by-step explanation:

Given that,

The magnitude of  intensity I of light varies inversely as the square of the magnitude of distance D from the source.

That is

I\propto \frac{1}{D^2}

Then,

\frac{I_1}{I_2}=\frac{D_2^2}{D_1^2}

Given that,

The magnitude of intensity of illumination on a screen 56 ft from a light is 2.7 foot-candle.

Here,

I_1=2.7 foot-candle, D_1= 56 ft

I_2=?, D_2= 70 ft.

\frac{I_1}{I_2}=\frac{D_2^2}{D_1^2}

\Rightarrow \frac{2.7}{I_2}=\frac{70^2}{56^2}

\Rightarrow \frac{I_2}{2.7}=\frac{56^2}{70^2}

\Rightarrow {I_2}=\frac{56^2\times 2.7}{70^2}

\Rightarrow {I_2}=1.728 foot-candle

The intensity on a screen 70 ft from the light is 1.728 foot candle.

3 0
4 years ago
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