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3 years ago

What is the value of 4(25-5)?

Mathematics

1 answer:

sweet [91]3 years ago

7 0

First you subtract 25 from 5 which equals 20 then you multipliy 4 to 20 and you get 80...so 80 is your solution.

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100, 91, 82, 73 Number pattern expression equations

emmasim [6.3K]

Answer:

The pattern is subtract 9 (OR -9)

Step-by-step explanation:

100 - 91 = 9

91 - 82 = 9

82 - 73 = 9

Glad I could help! Feel free to mark brainliest!

5 0

3 years ago

Read 2 more answers

Mark tosses a coin 25 times. What is the probability that it lands on heads exactly seven times?

mezya [45]

Answer:

1.43%

Step-by-step explanation:

we have that we can solve it by means of binomial, in this case m = 25. Also the probability of head is 50%. we need to know when it falls 7 times, therefore x = 7

the formula is:

P (x) = mCx * p ^ x * q ^ (m-x)

we know that p is 0.5 therefore q is equal to 0.5, replacing:

P (7) = 25C7 * 0.5 ^ 7 * 0.5 ^ (25-7)

P (7) = 25! / (7! * (25-7)!) * 0.5 ^ 7 * 0.5 ^ (25-7)

P = 0.0143

that is, the probability is 1.43%

6 0

3 years ago

a school bought 831 boxes of computer paper for the computer lab. Each box has 59 sheets of paper inside it. how many sheets of

Bess [88]

<span> 831 *59=49,029

hope this helps!</span>

4 0

3 years ago

Read 2 more answers

What is the value of the expression (–5) -3?

fgiga [73]

The answer to (-5)-3 is -8. I may be wrong but I did the work.

8 0

3 years ago

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Prove that if x is an positive real number such that x + x^-1 is an integer, then x^3 + x^-3 is an integer as well.

Shkiper50 [21]

Answer:

By closure property of multiplication and addition of integers,

If $x + \dfrac{1}{x}$ is an integer

∴ $\left ( x + \dfrac{1}{x} \right) ^3 = x^3 + \dfrac{1}{x^3} +3\cdot \left (x + \dfrac{1}{x} \right )$ is an integer

From which we have;

$x^3 + \dfrac{1}{x^3}$ is an integer

Step-by-step explanation:

The given expression for the positive integer is x + x⁻¹

The given expression can be written as follows;

$x + \dfrac{1}{x}$

By finding the given expression raised to the power 3, sing Wolfram Alpha online, we we have;

$\left ( x + \dfrac{1}{x} \right) ^3 = x^3 + \dfrac{1}{x^3} +3\cdot x + \dfrac{3}{x}$

By simplification of the cube of the given integer expressions, we have;

$\left ( x + \dfrac{1}{x} \right) ^3 = x^3 + \dfrac{1}{x^3} +3\cdot \left (x + \dfrac{1}{x} \right )$

Therefore, we have;

$\left ( x + \dfrac{1}{x} \right) ^3 - 3\cdot \left (x + \dfrac{1}{x} \right )= x^3 + \dfrac{1}{x^3}$

By rearranging, we get;

$x^3 + \dfrac{1}{x^3} = \left ( x + \dfrac{1}{x} \right) ^3 - 3\cdot \left (x + \dfrac{1}{x} \right )$

Given that $x + \dfrac{1}{x}$ is an integer, from the closure property, the product of two integers is always an integer, we have;

$\left ( x + \dfrac{1}{x} \right) ^3$ is an integer and $3\cdot \left (x + \dfrac{1}{x} \right )$ is also an integer

Similarly the sum of two integers is always an integer, we have;

$\left ( x + \dfrac{1}{x} \right) ^3 + \left(- 3\cdot \left (x + \dfrac{1}{x} \right ) \right )$ is an integer

$\therefore x^3 + \dfrac{1}{x^3} = \left ( x + \dfrac{1}{x} \right) ^3 - 3\cdot \left (x + \dfrac{1}{x} \right )= \left ( x + \dfrac{1}{x} \right) ^3 + \left(- 3\cdot \left (x + \dfrac{1}{x} \right ) \right )$ is an integer

From which we have;

$x^3 + \dfrac{1}{x^3}$ is an integer.

4 0

3 years ago