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Alborosie
3 years ago
15

if you are writing equivalent expression for 2^3 · 2^4, how many times would you write 2 as a factor?

Mathematics
1 answer:
Liono4ka [1.6K]3 years ago
6 0

Answer:

7 times

Step-by-step explanation:

In order to find the answer to this question you have to remember that when dealing with exponents you need to multiply the number to itself as many times as the exponent. In this case we have 2^3 we then need to multiply two by itself three times because three is it's exponent.

2^3 \times2^4

2^3=2\times2\times2

2^4=2\times2\times2\times2

2\times2\times2\times2\times2\times2\times2=

3+4=7

=7

Hope this helps.

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3 years ago
Which graph could be used to show the situation described?
vredina [299]

Answer:

Option (2) [Graph on the top right]

Step-by-step explanation:

To find the temperature decrease by 0.5 degrees every hour we will find the slopes of the lines given in every option.

Option (1) [Top left graph]

Slope = \frac{\text{Rise}}{\text{Run}}

          = \frac{-20}{2}

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Option (2) [Top right]

Slope = \frac{-2}{4}

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Option (3) [Bottom left]

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Option (4) [Bottom right]

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4 0
3 years ago
On a test victor asked to find the square root of 0.83. Wich of the following is closest to the square root of 0.83?
SVETLANKA909090 [29]

Answer:

I do not know what choices you have, but 0.9110433579144299 is the square root of 0.83

Step-by-step explanation:

Step 1:

Divide the number (0.83) by 2 to get the first guess for the square root .

First guess = 0.83/2 = 0.415.

Step 2:

Divide 0.83 by the previous result. d = 0.83/0.415 = 2.

Average this value (d) with that of step 1: (2 + 0.415)/2 = 1.2075 (new guess).

Error = new guess - previous value = 0.415 - 1.2075 = 0.7925.

0.7925 > 0.001. As error > accuracy, we repeat this step again.

Step 3:

Divide 0.83 by the previous result. d = 0.83/1.2075 = 0.6873706004.

Average this value (d) with that of step 2: (0.6873706004 + 1.2075)/2 = 0.9474353002 (new guess).

Error = new guess - previous value = 1.2075 - 0.9474353002 = 0.2600646998.

0.2600646998 > 0.001. As error > accuracy, we repeat this step again.

Step 4:

Divide 0.83 by the previous result. d = 0.83/0.9474353002 = 0.8760492667.

Average this value (d) with that of step 3: (0.8760492667 + 0.9474353002)/2 = 0.9117422835 (new guess).

Error = new guess - previous value = 0.9474353002 - 0.9117422835 = 0.0356930167.

0.0356930167 > 0.001. As error > accuracy, we repeat this step again.

Step 5:

Divide 0.83 by the previous result. d = 0.83/0.9117422835 = 0.9103449681.

Average this value (d) with that of step 4: (0.9103449681 + 0.9117422835)/2 = 0.9110436258 (new guess).

Error = new guess - previous value = 0.9117422835 - 0.9110436258 = 0.0006986577.

0.0006986577 <= 0.001. As error <= accuracy, we stop the iterations and use 0.9110436258 as the square root.

So, we can say that the square root of 0.83 is 0.911 with an error smaller than 0.001 (in fact the error is 0.0006986577). this means that the first 3 decimal places are correct. Just to compare, the returned value by using the javascript function 'Math.sqrt(0.83)' is 0.9110433579144299.

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