Plz What is 0.04 is 1/10 of

Mathematics

2 answers:

Mnenie [13.5K]2 years ago

8 0

If 0.04 is 1/10 of the answer, multiply 0.04 by 10 to find the answer:

0.04 x 10 = 0.4

netineya [11]2 years ago

7 0

0.04 is 1/10 of :

0.04 x 10 = 0,4

Ans = 0,4

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69) Half-way to the center of a planet of uniform density, your weight compared to that at the surface would be :

mel-nik [20]

Answer:

One- half

Step-by-step explanation:

Since the density is uniform

Mass of the sphere = [(4/3) π r^3] d

where d is the uniform density.

Mass of the sphere = [(4/3) π d] r^3 = k r^3

where k = [(4/3) π d] is a constant

Weight = mg = G m M / r^2 = G m [k r^3] /r^2 = G m k r

Using Gauss’ law for gravitation,

Half way to the center of a planet the weight is only due to the inner sphere and the outer sphere does not contribute to his weight,

Inside his weight is mg’ = (G m k r) /2 = mg/2

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7 0

2 years ago

Bradley and Kelly are out flying kites at a park one afternoon. A model of Bradley and Kelly's kites are shown below on the coor

Alex787 [66]

The correct answer is:

C. They are similar because the corresponding sides of kites KELY and BRAD all have the relationship 2:1.

Using the distance formula,

$d=\sqrt{(y_2-y_1)^2+(x_2-x_1)^2}$

the lengths of the sides of BRAD are:

$\text{BR}=\sqrt{(7-6)^2+(3-4)^2}=\sqrt{1^2+(-1)^2}=\sqrt{1+1}=\sqrt{2}\\\\\text{RA}=\sqrt{(6-3)^2+(4-3)^2}=\sqrt{3^2+1^2}=\sqrt{9+1}=\sqrt{10}\\\\\text{AD}=\sqrt{(3-6)^2+(3-2)^2}=\sqrt{(-3)^2+1^2}=\sqrt{9+1}=\sqrt{10}\\\\\text{DB}=\sqrt{(6-7)^2+(2-3)^2}=\sqrt{(-1)^2+(-1)^2}=\sqrt{1+1}=\sqrt{2}$

The lengths of the sides of KELY are:

$\text{KE}=\sqrt{(2-0)^2+(11-9)^2}=\sqrt{2^2+2^2}=\sqrt{4+4}=\sqrt{8}=2\sqrt{2}\\\\\text{EL}=\sqrt{(0-2)^2+(9-3)^2}=\sqrt{(-2)^2+6^2}=\sqrt{40}=2\sqrt{10}\\\\\text{LY}=\sqrt{(2-4)^2+(3-9)^2}=\sqrt{(-2)^2+(-6)^2}=\sqrt{40}=2\sqrt{10}\\\\\text{YK}=\sqrt{(4-2)^2+(9-11)^2}=\sqrt{2^2+(-2)^2}=\sqrt{8}=2\sqrt{2}$