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3 years ago

7

Find the value of given expression

ula1" title="(55 - 54) {}^{2} " alt="(55 - 54) {}^{2} " align="absmiddle" class="latex-formula">

2 answers:

yaroslaw [1]3 years ago

7 0

Answer:

your answer will be 1

Step-by-step explanation:

......

LiRa [457]3 years ago

4 0

Answer:

(55-54)²=1²=1 is your answer

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Simplify (2√5+3√7)^2

Anna35 [415]

The answer is: 83 + 12√35

(a + b)² = a² + 2ab + b²

In (2√5 + 3√7)², a = 2√5, b = 3√7

Just substitute a and b:
(a + b)² = a² + 2ab + b² = (2√5)² + 2 * 2√5 * 3√7 + (3√7)² =
= 2²√5² + 2 * 2 * 3 * √5 * √7 + 3²√7² =
= 4 * 5 + 12 * √(5*7) + 9 * 7 =
= 20 + 12 *√35 + 63 =
= 83 + 12√35

3 0

3 years ago

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Rewrite 3.47 • 10^5 in standard form

sesenic [268]

The answer is 347000

5 0

4 years ago

Two factors whose product is 84 are 12 and _[blank]_.

djyliett [7]

Answer: 7

Step-by-step explanation:

To solve this problem you must apply the proccedure shown below:

1. You know that a multiplication has the following form:

a*b=c

Where a and b are the factors and c is the product.

2. You know one of the factor, then you can find the second one as following:

12*b=84

b=84/12

b=7

Therefore the answer is 7.

7 0

4 years ago

What is 7/3 times 2/3 multiplied in simplest form

Sav [38]

Multiplying fractions is simple. You just multiply across so 7/3 x 2/3 is 14/9 and that is the simplest form.

4 0

3 years ago

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Assume you have such a watch. if a minimum of 18.0% of the original tritium is needed to read the dial in dark places, for how m

laila [671]

Given that the half life of 3H is 12.3 years.

The amount of substance left of a radioactive substance with half life of $t_{ \frac{1}{2} }$ after t years is given by

$N(t)=N_0\left(\frac{1}{2} \right)^{ \frac{t}{t_{\frac{1}{2}} }$

Therefore, the number of years it will take for 18% of the original tritinum to remain is given by

$18 \% = 100 \% \left(\frac{1}{2} \right)^{ \frac{t}{12.3}} \\ \\ \Rightarrow\left(\frac{1}{2} \right)^{ \frac{t}{12.3}} =0.18 \\ \\ \Rightarrow\frac{t}{12.3}\ln\left(\frac{1}{2} \right)=\ln0.18 \\ \\ \Rightarrow\frac{t}{12.3}= \frac{\ln0.18}{\ln\left(\frac{1}{2} \right)} = \frac{-1.715}{-0.6931} =2.474\\ \\ \Rightarrow t=2.474(12.3)=30.4$

Therefore, the number of <span>years that the time could be read at night is 30.4 years.</span>

5 0

3 years ago