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4 years ago

Which number between 20 and 28 are prime

Mathematics

2 answers:

ollegr [7]4 years ago

8 0

The number is 23, that is the only number prime between 20 and 28

ad-work [718]4 years ago

4 0

Prime numbers basically numbers that dont have any numbers that will be multiplied to get to it. So, it would be:
23
-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-
All the other numbers in between can be divided by something and not become a decimal. Other than 23, it can be divided by anything without becoming a decimal.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
~Hope this helped :)

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Solve the math problem

Korvikt [17]

Answer:

3=-8y+5

Step-by-step explanation:

Gather your like terms, - 3y and - 5y which in turn in - 8y. You cant plus by 5 as it is not a like term. IF we were finding y we would rearrange and bring +5 to the other side and so on...

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3 years ago

Adrian's annual income changes every year because of the following three factors: On average, his salary is 1.2 times the previo

olya-2409 [2.1K]

In this question, Adrian initially has $52,000 and his salary is 1.2 times the previous year's salary. This means the income should look like this:52000*1.2^n

30% of his income is budgeted for rent, then the equation should look like this: 30% * income= 30%*(52000*1.2^n)

Adrian's income increases by $2,300 each year as a result of gifts will make this equation:2300n

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7 0

3 years ago

Insert a digit to make numbers that are divisible by 24 if it is possible: 83...8

AfilCa [17]

Answer:

Step-by-step explanation:

8328÷24=347

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3 0

3 years ago

The graphs below have the same shape. What is the equation of the red

anyanavicka [17]

Answer:

Option B is correct .

Step-by-step explanation:

According to Question , both the graph have same shape . If we look at the the first graph it cuts x - axis at (0 , 2) and ( 0 , -2) . Hence x = 2 and -2 are the zeroes of the equation .

And ,the given function is ,

$\implies f(x) = 4 - x^2 \\\\\implies f(x) = 4-x^2=0 \\\\\implies 2^2-x^2=0\\\\\implies (2-x)(2+x) = 0 \\\\\boxed{\red{\bf \implies x = 2 , (-2) }}$

Hence ,we can can see that x = 2 and (-2) are the zeroes of graph. 

This implies that if we know the zeroes , we can frame the Equation.

On looking at second parabola , it's clear that cuts x - axis at ( 1, 0 ) and (-1,0). So , 1 and -1 are the zeroes of the quadratic equation . Let the function be g(x) . Here , a and ß are the zeroes.

$\implies g(x) = (x-\alpha)(x-\beta) \\\\\implies g(x) = (1+x)(1-x) \\\\\boxed{\pink{\bf \implies g(x) = 1 - x^2}}$

Hence option B is correct .

8 0

3 years ago

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