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3 years ago

Express 4 250 000 000 000 in standard form

Mathematics

1 answer:

AlladinOne [14]3 years ago

7 0

hi <3

standard form is written in the form a x 10^n

where a is a number between 1 and 10

in this case, you would get 4.25 x 10^12

hope this helps :)

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What is a general pattern in data

Andrew [12]

Answer:

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Step-by-step explanation:

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1 year ago

Find the TWO integers whos product is -12 and whose sum is 1<br>

ahrayia [7]

Answer:

$\rm Numbers = 4 \ and \ -3.$

Step-by-step explanation:

Given :-

The sum of two numbers is 1 .

The product of the nos . is 12 .

And we need to find out the numbers. So let us take ,

First number be x

Second number be 1-x .

According to first condition :-

$\rm\implies 1st \ number * 2nd \ number= -12\\\\\rm\implies x(1-x)=-12\\\\\rm\implies x - x^2=-12\\\\\rm\implies x^2-x-12=0\\\\\rm\implies x^2-4x+3x-12=0\\\\\rm\implies x(x-4)+3(x-4)=0\\\\\rm\implies (x-4)(x+3)=0\\\\\rm\implies\boxed{\red{\rm x = 4 , -3 }}$

Hence the numbers are 4 and -3

8 0

3 years ago

Read 2 more answers

Determine all real values of p such that the set of all linear combination of u = (3, p) and v = (1, 2) is all of R2. Justify yo

Rama09 [41]

Answer:

p ∈ IR - {6}

Step-by-step explanation:

The set of all linear combination of two vectors ''u'' and ''v'' that belong to R2

is all R2 ⇔

$u\neq 0_{R2}$

$v\neq 0_{R2}$

And also u and v must be linearly independent.

In order to achieve the final condition, we can make a matrix that belongs to $R^{2x2}$ using the vectors ''u'' and ''v'' to form its columns, and next calculate the determinant. Finally, we will need that this determinant must be different to zero.

Let's make the matrix :

$A=\left[\begin{array}{cc}3&1&p&2\end{array}\right]$

We used the first vector ''u'' as the first column of the matrix A

We used the second vector ''v'' as the second column of the matrix A

The determinant of the matrix ''A'' is

$Det(A)=6-p$

We need this determinant to be different to zero

$6-p\neq 0$

$p\neq 6$

The only restriction in order to the set of all linear combination of ''u'' and ''v'' to be R2 is that $p\neq 6$

We can write : p ∈ IR - {6}

Notice that is $p=6$ ⇒

$u=(3,6)$

$v=(1,2)$

If we write $3v=3(1,2)=(3,6)=u$ , the vectors ''u'' and ''v'' wouldn't be linearly independent and therefore the set of all linear combination of ''u'' and ''b'' wouldn't be R2.