All categories

2 years ago

How many times greater is 9 times 10 to the 5 power than 8 times 10 to the 2 power

Mathematics

1 answer:

Schach [20]2 years ago

6 0

Answer:

1125 times bigger

Step-by-step explanation:

You might be interested in

Functions Question ! Please help with this one question!

Rus_ich [418]

G(a/2)= 2(a/2)-3
=2a/2-3
then just subtract 2a/2 by 3/1 by finding the common denominator and same for the next one.

8 0

3 years ago

Dominic has $35.42 in his bank account. He earns $18 for mowing lawns. He spends $1.89 on Gatorade and then deposits the rest in

Pavel [41]

answer

yes he has enough

Step-by-step explanation:

35.42 + 18 -1.89 +35

he has 87.53 this is enough for the skateboard

4 0

2 years ago

Read 2 more answers

Josiah and his children went into a grocery store and he bought $15 worth of peaches and mangos. Each peach costs $0.50 and each

Westkost [7]

Step•BY•Step Explanation~|

 ♡♡♡♡♡♡♡♡♡♡♡♡♡ 

y=1.50

x=0.50

1.50

1.59

______+

3.00

0.50

_____+

3.50

<h2>Answer:♡~</h2>

3.50

HOPE IT HELPSS

8 0

1 year ago

Consider the following region R and the vector field F. a. Compute the two-dimensional curl of the vector field. b. Evaluate bo

Shalnov [3]

Looks like we're given

$\vec F(x,y)=\langle-x,-y\rangle$

which in three dimensions could be expressed as

$\vec F(x,y)=\langle-x,-y,0\rangle$

and this has curl

$\mathrm{curl}\vec F=\langle0_y-(-y)_z,-(0_x-(-x)_z),(-y)_x-(-x)_y\rangle=\langle0,0,0\rangle$

which confirms the two-dimensional curl is 0.

It also looks like the region $R$ is the disk $x^2+y^2\le5$ . Green's theorem says the integral of $\vec F$ along the boundary of $R$ is equal to the integral of the two-dimensional curl of $\vec F$ over the interior of $R$ :

$\displaystyle\int_{\partial R}\vec F\cdot\mathrm d\vec r=\iint_R\mathrm{curl}\vec F\,\mathrm dA$

which we know to be 0, since the curl itself is 0. To verify this, we can parameterize the boundary of $R$ by

$\vec r(t)=\langle\sqrt5\cos t,\sqrt5\sin t\rangle\implies\vec r'(t)=\langle-\sqrt5\sin t,\sqrt5\cos t\rangle$

$\implies\mathrm d\vec r=\vec r'(t)\,\mathrm dt=\sqrt5\langle-\sin t,\cos t\rangle\,\mathrm dt$

with $0\le t\le2\pi$ . Then

$\displaystyle\int_{\partial R}\vec F\cdot\mathrm d\vec r=\int_0^{2\pi}\langle-\sqrt5\cos t,-\sqrt5\sin t\rangle\cdot\langle-\sqrt5\sin t,\sqrt5\cos t\rangle\,\mathrm dt$

$=\displaystyle5\int_0^{2\pi}(\sin t\cos t-\sin t\cos t)\,\mathrm dt=0$

7 0

2 years ago

What is the solution to the equation 3(y-2)=2y-8 A.-2/5 B.-14 C. -2 D. -2 4/5 I NEED-HELP PLeAse

AlladinOne [14]

3(y - 2) = 2y - 8

3y - 6 = 2y - 8

3y - 2y = -8 + 6

y = -2

Answer: C

4 0

2 years ago

How many times greater is 9 times 10 to the 5 power than 8 times 10 to the 2 power​

How many times greater is 9 times 10 to the 5 power than 8 times 10 to the 2 power