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2 years ago

How do you write 4.279E-4 in standard form?

Mathematics

2 answers:

vampirchik [111]2 years ago

7 0

Answer:

0.0004279 in decimal form

Step-by-step explanation:

alex41 [277]2 years ago

3 0

Answer:

4279/10000000

Step-by-step explanation:

Multiply and divide by 10 for every number after the decimal point. There are 7 digits to the right of the decimal point, therefore multiply and divide by 10000000.

'Hope this helps :)'

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I don’t get the problem and I’ve been trying to solve it

Black_prince [1.1K]

I'm gonna use c for circle, t for triangle, and s for square
c+t+s=13
t+s=8
c+8=13
c=5
c+t=7
t=2
s=6

circle is 5, triangle is 2, and square is 6

8 0

3 years ago

I need help with math look at the picture and help me please with all the parts math isn’t my strong suit

Zepler [3.9K]

Answer:

which form is that

Step-by-step explanation:

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3 years ago

26. Define a relation ∼ ∼ on R 2 R2 by stating that ( a , b ) ∼ ( c , d ) (a,b)∼(c,d) if and only if a 2 + b 2 ≤ c 2 + d 2 . a2+

Tresset [83]

Answer:

~ is reflexive.

~ is asymmetric.

~ is transitive.

Step-by-step explanation:

~ is reflexive:

i.e., to prove $\forall (a, b) \in \mathbb{R}^2$ , $(a, b) R(a, b)$ .

That is, every element in the domain is related to itself.

The given relation is $\sim: (a,b) \sim (c, d) \iff a^2 + b^2 \leq c^2 + d^2$

Reflexive:

$(a, b) \sim (a, b)$ since $a^2 + b^2 = a^2 + b^2$

This is true for any pair of numbers in $\mathbb{R}^2$ . So, $\sim$ is reflexive.

Symmetry:

$\sim$ is symmetry iff whenever $(a, b) \sim (c, d)$ then $(c, d) \sim (a, b)$ .

Consider the following counter - example.

Let (a, b) = (2, 3) and (c, d) = (6, 3)

$a^2 + b^2 = 2^2 + 3^2 = 4 + 9 = 13$

$c^2 + d^2 = 6^2 + 3^2 = 36 + 9 = 42$

Hence, $(a, b) \sim (c, d)$ since $a^2 + b^2 \leq c^2 + d^2$

Note that $c^2 + d^2 \nleq a^2 + b^2$

Hence, the given relation is not symmetric.

Transitive:

$\sim$ is transitive iff whenever $(a, b) \sim (c, d) \hspace{2mm} \& \hspace{2mm} (c, d) \sim (e, f)$ then $(a, b) \sim (e, f)$

To prove transitivity let us assume $(a, b) \sim (c, d)$ and $(c, d) \sim (e, f)$ .

We have to show $(a, b) \sim (e, f)$

Since $(a, b) \sim (c, d)$ we have: $a^2 + b^2 \leq c^2 + d^2$

Since $(c, d) \sim (e, f)$ we have: $c^2 + d^2 \leq e^2 + f^2$

Combining both the inequalities we get:

$a^2 + b^2 \leq c^2 + d^2 \leq e^2 + f^2$

Therefore, we get: $a^2 + b^2 \leq e^2 + f^2$

Therefore, $\sim$ is transitive.

Hence, proved.

3 0

3 years ago

Simplify -5-√-44<br><br> i have no idea

Tatiana [17]

Answer:

Undefined

Step-by-step explanation:

The square root of a negative number does not exist in the set of real numbers so it would be Undefined

6 0

3 years ago

Which equations represent inverse variation? Check all that apply.

Bogdan [553]

For this case, what you should know is that the equations that represent an inverse variation are those that could not form a straight line, for example.

We have then:
Equation 1:
pv = 13
Rewriting:
p = 13 / v
P and v are represent an inverse variation.
Equation 2:
z = (2 / x)
z and x are represent an inverse variation.
Answer:
equations represented inverse variation are:
pv = 13
z = (2 / x)

4 0

3 years ago

Read 2 more answers