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3 years ago

What is the name given to these numbers 1,3,6,10,15

Mathematics

1 answer:

ZanzabumX [31]3 years ago

8 0

Answer:

Other than one, products to prime numbers.

Step-by-step explanation:

1=1x1 (But 1 isn't prime, so it's the exception.)

3=1x3 (1 isn't prime, but it'll pass.)

6=2x3 (Prime and prime)

10=2x5(Prime and prime)

15=3x5(Prime and Prime)

I also guess you could call these NOT prime numbers, because you can't call these composite, since one is neither prime or composite.

DUBS, Signing off, GETTING 5 BRAINLIEST CHALLENGE!

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Find a polynomial function of degree 3 with 2, i, -i as zeros.

sergiy2304 [10]

Answer:

$p(x)= x^3-2x^2+x-2$

Step-by-step explanation:

Here we are given that a polynomial has zeros as 2 , i and -i . We need to find out the cubic polynomial . In general we know that if $\alpha , \ \beta \ \& \ \gamma$ are the zeros of the cubic polynomial , then ,

$\sf \longrightarrow p(x)= (x -\alpha )(x-\beta)(x-\gamma)$

Here in place of the Greek letters , substitute 2,i and -i , we get ,

$\sf\longrightarrow p(x)= (x -2 )(x-i)(x+i)$

Now multiply (x-i) and (x+i ) using the identity (a+b)(a-b)=a² - b² , we have ,

$\sf \longrightarrow p(x)= (x-2)\{ x^2 - (i)^2\}$

Simplify using i = √-1 ,

$\sf \longrightarrow p(x)= (x-2)( x^2 + 1 )$

Multiply by distribution ,

$\sf \longrightarrow p(x)= x(x^2+1) -2(x^2+1)$

Simplify by opening the brackets ,

$\sf\longrightarrow p(x)= x^3+x-2x^2-2$

Rearrange ,

$\sf\longrightarrow \underline{\boxed{\blue{\sf p(x)= x^3-2x^2+x-2}}}$

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2 years ago

5^-1 = blank * 5^4 find the missing factor

skad [1K]

Answer:

1/3125 or, 5^-5

Step-by-step explanation:

5^-1 = blank * 5^4

1/5/625= blank or, 5^-1/5^4=blank

blank=1/3125 5^-5=blank

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3 years ago

Solve: 3x+5=14<br><br> Find the value of x

Levart [38]

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