3 years ago

What is 18,300,000 rounded to the nearest tenth

Mathematics

1 answer:

max2010maxim [7]3 years ago

3 0

13,000,000 rounded to the nearest tenth is 18,000,000

You might be interested in

What number belongs in the blank space below

Setler79 [48]

The number that belongs in the blank space is answer choice A. 6

6 0

3 years ago

Wht is the answer for 2/3+2/9

garik1379 [7]

2/3 = 6/9 So,

6/9 + 2/9 = 8/9

4 0

3 years ago

Read 2 more answers

Evaluate 3-8/2 +3x5

wolverine [178]

<h2>Answer:</h2><h2>3-8/2+3×5</h2><h2>5/2</h2><h2>2.5 + 3×5</h2><h2>2.5+15</h2><h2>= 17.5*</h2>

3 0

3 years ago

Read 2 more answers

Solve for x triangles

wel

Answer:

115

Step-by-step explanation:

180-50=130

130/2=65

180-65=115

4 0

2 years ago

Find the modulus of the complex number 6-2i

Papessa [141]

Answer:

The modulus of the complex number 6-2i is:

$|z|\:=2\sqrt{10}$

Step-by-step explanation:

Given the number

$6-2i$

We know that

$z = x + iy$

where x and y are real and $\sqrt{-1}=i$

The modulus or absolute value of z is:

$|z|\:=\sqrt{x^2+y^2}$

Therefore, the modulus of $6-2i$ will be:

$z=6-2i$

$z=6+(-2)i$

$|z|\:=\sqrt{x^2+y^2}$

$|z|\:=\sqrt{6^2+\left(-2\right)^2}$

$=\sqrt{6^2+2^2}$

$=\sqrt{36+4}$

$=\sqrt{40}$

$=\sqrt{2^2}\sqrt{2\cdot \:5}$

$=2\sqrt{2\cdot \:5}$

$=2\sqrt{10}$

Therefore, the modulus of the complex number 6-2i is:

$|z|\:=2\sqrt{10}$

3 0

2 years ago