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3 years ago

What is the value of 4 in the decimal 4.706

Mathematics

1 answer:

juin [17]3 years ago

6 0

Answer:

whole number

Step-by-step explanation:

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Which of the following shows the polynomial below written in descending order?

Vladimir79 [104]

Answer:

A. 4x¹² + 9x⁷ + 3x³ -x

Step-by-step explanation:

Hi!

==================================================================

To write a polynomial in descending order, we write the terms with higher degrees, or exponents, first.

3x³ + 9x⁷ -x + 4x¹²

4x¹² has the highest degree, so it is written as the first term.

⇒4x¹²

9x⁷ has the next highest degree, so it is written next.

⇒4x¹² + 9x⁷

3x³ has the next highest degree, so it is written next.

⇒4x¹² + 9x⁷ + 3x³

-x has the lowest degree, so it is written last.

⇒4x¹² + 9x⁷ + 3x³ -x

4x¹² + 9x⁷ + 3x³ -x

==================================================================

Hope I Helped, Feel free to ask any questions to clarify :)

Have a great day!

-Aadi x

8 0

3 years ago

In ΔDEF,{EF}\cong{DE} EF ≅ DE and m∠E = 104°. Find m∠D.

Alexeev081 [22]

The 3 inside angles need to equal 180.

Subtract E from 180 to find the sum of the other two angles:

180 - 104 = 76

Now because EF ≅ DE , angle D and F would be the same so divide 76 by 2

Angle D = 38

7 0

3 years ago

A locus of points equidistant from a fixed point is a

kap26 [50]

Answer:

B. bisector of an angle.

5 0

3 years ago

3/7 rounded to the nearest thousand

iogann1982 [59]

3/7= 0.42857143
rounded to the neared thousandth, it is 0.429

6 0

3 years ago

Read 2 more answers

Which statement describes if there is an extraneous solution to the equation √x-3 = x-5? A. there are no solutions to the equati

Brrunno [24]

Remember that an extraneous solution of an equation, is the solution that emerges from solving the equation but is not a valid solution.

Lets solve our equation to find out what is the extraneous solution:
 $\sqrt{x-3} =x-5$
$(\sqrt{x-3})^2 =(x-5)^2$
$x-3=x^2-10x+25$
$x^2-11x+28=0$
$(x-4)(x-7)=0$
$x-4=0$ and $x-7=0$
$x=4$ and $x=7$

So, the solutions of our equation are $x=4$ and $x=7$ . Lets replace each solution in our original equation to check if they are valid solutions:
- For $x=7$
$\sqrt{x-3} =x-5$
$\sqrt{7-3} =7-5$
$\sqrt{4} =2$
$2=2$
We can conclude that 7 is a valid solution of the equation.

- For $x=4$
$\sqrt{x-3} =x-5$
$\sqrt{4-3} =4-5$
$\sqrt{1} =1$
$1 \neq 1$
We can conclude that 4 is not a valid solution of the equation; therefore, 4 is a extraneous solution.

We can conclude that the correct answer is: D. the extraneous solution is x = 4

7 0

2 years ago