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3 years ago

Explain how are the decimals 0.009 and 0.09 related

Mathematics

2 answers:

Anna11 [10]3 years ago

8 0

Because .... Well obviously there's a number nine and it's a part of a number line

Maurinko [17]3 years ago

5 0

0.009 is 1/10 of 0.09

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Please help me do number 3 and number 4

Marina86 [1]

Hello :
let : A(0,3) B(4,8)
<span>the slope is : (YB - YA)/(XB -XA)
</span>(8-3)/(4-0) = 5/4
same method for 4

3 0

3 years ago

What is the initial velocity of the boiling water? H=-16t^2+170t

erastova [34]

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5 0

3 years ago

Passes through (1,12) and has a vertex (10,-4)

agasfer [191]

Answer: 16/81 (x-10)^2 -4

Step-by-step explanation:

To write a vertex equation with just a point and the vertex, you have to figure out the variables.

In vertex form, the equation is y = a (x-h)^2 + k

Your y is 12, x = 1, h = 10, and k = -4

Plug everything into equation

12 = a (1 - 10)^2 -4

12 = a (-9)^2 - 4

12 = 81a - 4

16 = 81a

16/81 = a

Now you know what the 'a' value is.

If you graph 16/81 (x-10)^2 -4 , you will get a point at (1,12) and a vertex of (10,-4)!

I hope this helps!

3 0

3 years ago

A hoop, a uniform solid cylinder, a spherical shell, and a uniform solid sphere are released from rest at the top of an incline.

Serjik [45]

Answer:

Step-by-step explanation:

Given

Hoop, Uniform Solid Cylinder, Spherical shell and a uniform Solid sphere released from Rest from same height

Suppose they have same mass and radius

time Period is given by

$t=\sqrt{\frac{2h}{a}}$ ,where h=height of release

a=acceleration

$a=\frac{g\sin \theta }{1+\frac{I}{mr^2}}$

Where I=moment of inertia

a for hoop

$a=\frac{g\sin \theta }{1+\frac{mr^2}{mr^2}}$

$a=\frac{g\sin \theta }{2}$

a for Uniform solid cylinder

$a=\frac{g\sin \theta }{1+\frac{mr^2}{2mr^2}}$

$a=\frac{2g\sin \theta }{3}$

a for spherical shell

$a=\frac{g\sin \theta }{1+\frac{2mr^2}{3mr^2}}$

$a=\frac{3g\sin \theta }{5}$

a for Uniform Solid

$a=\frac{g\sin \theta }{1+\frac{2mr^2}{5mr^2}}$

$a=\frac{5g\sin \theta }{7}$

time taken will be inversely proportional to the square root of acceleration

$t_1=k\sqrt{2}=1.414k$

$t_2=k\sqrt{\frac{3}{2}}=1.224k$

$t_3=k\sqrt{\frac{5}{3}}=1.2909k$

$t_4=k\sqrt{\frac{7}{5}}=1.183k$

thus first one to reach is Solid Sphere

second is Uniform solid cylinder

third is Spherical Shell

Fourth is hoop

3 0

3 years ago

Which property was used to simply the expression 3c+9+4c=3c+4c+9

34kurt

<span> commutative property</span>

7 0

3 years ago