Is 0.05 greater than 0.083

Mathematics

1 answer:

Harrizon [31]3 years ago

6 0

Answer:

no 0.05 is smaller then 0.083

Step-by-step explanation:

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a city got 40 inches of rain in 2012 in 2013 it got 44.8 inches what was the percent increase? help fast

Alex777 [14]

Answer:

10.2% increase

Step-by-step explanation:

44.8-40 = 4.8

4.8/40 = 10.2

6 0

3 years ago

A sphere with a radius of 6cm has the same volume as a cylinder with a height of 4.5cm. What is the radius of the cylinder

yawa3891 [41]

Short answer: r = 8
Remark
The easiest way to do this is to solve the sphere's volume in terms of pi. When you do this, you can equate that to the formula for a cylinder and cancel the pi values.

Step One
Find the volume of the sphere.

Givens
r = 6 cm

Formula
V = (4/3) pi r^3

Sub and Solve
V = 4/3 pi * 6^3
V = 288 * pi

Step two
Find the radius of the cylinder

Givens
V = 288* pi cm^3
h = 4.5 cm

Formula
V = pi r^h

Sub and solve
288 pi cm^3 = pi r^2 * 4.5 Divide both sides by pi
288 cm^3 = 4.5 r^2 Divide both sides by 4.5
388 / 4.5 = r^2
64 = r^2 Take the square root of both sides.
r = square root( 64)
r = 8 <<<<< Answer

4 0

3 years ago

Two problems here I need solved! I need every step, so please have that with your answers!!

Free_Kalibri [48]

QUESTION 1

We want to solve,

$\frac{1}{(x-4)}+\frac{x}{(x-2)}=\frac{2}{x^{2}-6x+8}$

We factor the denominator of the fraction on the right hand side to get,

$\frac{1}{(x-4)}+\frac{x}{(x-2)}=\frac{2}{x^{2}-4x - 2x+8}.$

This implies
$\frac{1}{(x-4)}+\frac{x}{(x-2)}=\frac{2}{x(x-4) - 2(x - 4)}.$

$\frac{1}{(x-4)}+\frac{x}{(x-2)}=\frac{2}{(x-4)(x - 2)}$

We multiply through by LCM of
$(x-4)(x - 2)$

$(x - 2) + x(x-4) = 2$

We expand to get,

$x - 2 + {x}^{2} - 4x= 2$

We group like terms and equate everything to zero,

${x}^{2} + x - 4x - 2 - 2 = 0$

We split the middle term,

${x}^{2} + - 3x - 4 = 0$

We factor to get,

${x}^{2} + x - 4x- 4 = 0$

$x(x + 1) - 4(x + 1) = 0$

$(x + 1)(x - 4) = 0$

$x + 1 = 0 \: or \: x - 4 = 0$

$x = - 1 \: or \: x = 4$

But
$x = 4$
is not in the domain of the given equation.

It is an extraneous solution.

$\therefore \: x = - 1$
is the only solution.

QUESTION 2

$\sqrt{x+11} -x=-1$

We add x to both sides,

$\sqrt{x+11} =x-1$

We square both sides,

$x + 11 = (x - 1)^{2}$

We expand to get,

$x + 11 = {x}^{2} - 2x + 1$

This implies,

${x}^{2} - 3x - 10 = 0$

We solve this quadratic equation by factorization,

${x}^{2} - 5x + 2x - 10 = 0$

$x(x - 5) + 2(x - 5) = 0$

$(x + 2)(x - 5) = 0$

$x + 2 = 0 \: or \: x - 5 = 0$

$x = - 2 \: or \: x = 5$

But
$x = - 2$
is an extraneous solution

$\therefore \: x = 5$