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3 years ago

How many times larger is 3 x 1015 than 6 x 1010

Mathematics

2 answers:

Deffense [45]3 years ago

8 0

it is 3,000 times larger! that is the difference.

Klio2033 [76]3 years ago

7 0

(3 x 10^15) / (6 x 10^10)

= (3/6) x 10^(15 - 10)

= 0.5 x 10^5

= 5 x 10^4

= 50,000

Answer

50,000 times larger

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greg invested $20 and qudrupled his money in five years.sabrena also invested $20 and after five years the amount was eqaul to 2

const2013 [10]

Answer:

The money of Greg after five years is $80

The money of Sabrena after five years is $160,000

So,The money of Sabrena is more than Greg after five years .

Step-by-step explanation:

Given as :

Greg invested $20 and quadrupled his money in five years

Let the quadrupled money after five years = $x

∵ quadrupled means the money becomes 4 times of previous

So, x = $20 × 4

i.e x = $80

So, The money of Greg after five years is $80

Again

Sabrena invested $20 and after five years money becomes 20 to the fourth power

Let the money after 20 to the fourth power = $y

So, y = $20^{4}$

i.e y = 160,000

So, The money of Sabrena after five years is $160,000

∵ Sabrena money is more than Greg money

Hence, The money of Sabrena is more after five years . Answer

6 0

3 years ago

B: Alex is 12 years older than George.

UkoKoshka [18]

Answer:

The ratio of George age to Carl's age is 1:12.

Step-by-step explanation:

Let the age of George be 'g'.

Let the age of Alex be 'a'.

Also Let the age of Carl be 'c'.

Given:

The sum of their ages is 68.

So equation can be framed as;

$g+a+c=68 \ \ \ \ equation 1$

Also Given:

Alex is 12 years older than George.

So equation can be framed as;

$a =g+12 \ \ \ \ equation \ 2$

Now Given:

Carl is three times older than Alex.

$c=3a$

But $a =g+12$

So we get;

$c = 3(g+12) \\\\c= 3g+36 \ \ \ \ equation \ 3$

Now Substituting equation 2 and equation 3 in equation 1 we get;

$g+a+c=68\\\\g+g+12+3g+36=68\\\\5g+48=68$

Subtracting both side by 48 using subtraction property of equality we get;

$5g+48-48=68-48\\\\5g=20$

Now Dividing both side by 5 using Division property of equality we get;

$\frac{5g}{5}=\frac{20}{5}\\\\g =4$

Hence George age $g = 4 \ years$

Now Alex age $a=g+12 = 4+12 =16\ years$

Also Carl's age $c=3g+36=3\times 4+36 =12+36 =48\ years$

Now we need to find the ratio of George age to Carl's age.

$\frac{g}{c}=\frac{4}{48} = \frac{1}{12}$

Hence the ratio of George age to Carl's age is 1:12.

3 0

3 years ago

Prove that: (b²-c²/a)CosA+(c²-a²/b)CosB+(a²-b²/c)CosC = 0

IRISSAK [1]

Prove that:

$\:\:\sf\:\:\left(\dfrac{b^2-c^2}{a}\right)\cos A+\left(\dfrac{c^2-a^2}{b}\right)\cos B +\left(\dfrac{a^2-b^2}{c}\right)\cos C=0$

Proof: 

We know that, by Law of Cosines,

$\sf \cos A=\dfrac{b^2+c^2-a^2}{2bc}$
$\sf \cos B=\dfrac{c^2+a^2-b^2}{2ca}$
$\sf \cos C=\dfrac{a^2+b^2-c^2}{2ab}$

Taking LHS

$\left(\dfrac{b^2-c^2}{a}\right)\cos A+\left(\dfrac{c^2-a^2}{b}\right)\cos B +\left(\dfrac{a^2-b^2}{c}\right)\cos C$

Substituting the value of cos A, cos B and cos C,

$\longmapsto\left(\dfrac{b^2-c^2}{a}\right)\left(\dfrac{b^2+c^2-a^2}{2bc}\right)+\left(\dfrac{c^2-a^2}{b}\right)\left(\dfrac{c^2+a^2-b^2}{2ca}\right)+\left(\dfrac{a^2-b^2}{c}\right)\left(\dfrac{a^2+b^2-c^2}{2ab}\right)$

$\longmapsto\left(\dfrac{(b^2-c^2)(b^2+c^2-a^2)}{2abc}\right)+\left(\dfrac{(c^2-a^2)(c^2+a^2-b^2)}{2abc}\right)+\left(\dfrac{(a^2-b^2)(a^2+b^2-c^2)}{2abc}\right)$

$\longmapsto\left(\dfrac{(b^2-c^2)(b^2+c^2)-(b^2-c^2)(a^2)}{2abc}\right)+\left(\dfrac{(c^2-a^2)(c^2+a^2)-(c^2-a^2)(b^2)}{2abc}\right)+\left(\dfrac{(a^2-b^2)(a^2+b^2)-(a^2-b^2)(c^2)}{2abc}\right)$

$\longmapsto\left(\dfrac{(b^4-c^4)-(a^2b^2-a^2c^2)}{2abc}\right)+\left(\dfrac{(c^4-a^4)-(b^2c^2-a^2b^2)}{2abc}\right)+\left(\dfrac{(a^4-b^4)-(a^2c^2-b^2c^2)}{2abc}\right)$

$\longmapsto\dfrac{b^4-c^4-a^2b^2+a^2c^2}{2abc}+\dfrac{c^4-a^4-b^2c^2+a^2b^2}{2abc}+\dfrac{a^4-b^4-a^2c^2+b^2c^2}{2abc}$

On combining the fractions,

$\longmapsto\dfrac{(b^4-c^4-a^2b^2+a^2c^2)+(c^4-a^4-b^2c^2+a^2b^2)+(a^4-b^4-a^2c^2+b^2c^2)}{2abc}$

$\longmapsto\dfrac{b^4-c^4-a^2b^2+a^2c^2+c^4-a^4-b^2c^2+a^2b^2+a^4-b^4-a^2c^2+b^2c^2}{2abc}$

Regrouping the terms,

$\longmapsto\dfrac{(a^4-a^4)+(b^4-b^4)+(c^4-c^4)+(a^2b^2-a^2b^2)+(b^2c^2-b^2c^2)+(a^2c^2-a^2c^2)}{2abc}$

$\longmapsto\dfrac{(0)+(0)+(0)+(0)+(0)+(0)}{2abc}$

$\longmapsto\dfrac{0}{2abc}$

$\longmapsto\bf 0=RHS$

LHS = RHS proved.

7 0

2 years ago

Help me on math please really need help

Zanzabum

If the height was 10, then the volume of the cube would be 1000 because you find volume you must multiply the length*width*height and the value of those three is 10.
Now since volume is 1000 and the volume of a 2in cube is 8 (again, lwh=V) you can divide 1000 by 8 and you would get 125. So that means 125 2in cubes can fit inside the bigger cube.

If the volume of this cube were 750in^3 and you had to find the height, you would use the Volume formula again:
l*w*h=V
10*10*h=750
20h=750
((divide both sides of the equation by 20 to find the value of h))
h=37.5

If the surface area of the cube were 680in^2 then you would use the surface area formula to find the value of h:
(2(lw))+(2(lh))+(2(wh))=A
(2(10*10))+(2(10h))+(2(10h))=680
200+20h+20h=680
subtract 200 from both sides of the equation:
40h=480
divide both sides by 40 to get the value of h:
h=12

5 0

3 years ago

Find the measure of the angle indicated in bold

vova2212 [387]

Answer:

60 for both

Step-by-step explanation:

Since these angles are parallel, they are equal to each other :

6x = 5x + 10

Subtracting 5x from both sides gives us :

x = 10

Substituting this value back into both angles gives us

6x = 60

5x+10 = 60

8 0

2 years ago