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4 years ago

The table shows where sixth grade students at Sharonton

Mathematics

1 answer:

e-lub [12.9K]4 years ago

3 0

Did you add the picture of the table? I do not see it :(

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The world's oceans hold roughly 1.4 x 109 cubic kilometers of water. A typical bucket holds roughly 20,000 cubic centimeters of

MissTica

Answer:

It'll take a bucket loads of $7 * 10^{20}$ to empty world's ocean

Step-by-step explanation:

Given

$World\ Ocean: 1.4 * 10^9 km^3$

$Bucket: 20,000 cm^3$

$1 km^3 = 10^{15} cm^3$

To get the number of bucket loads, we simply divide the size of the ocean by the size of the bucket;

$Number = \frac{Ocean\ Size}{Bucket\ Size}$

$Number = \frac{1.4 * 10^9 km^3}{20,000cm^3}$

Expand the numerator

$Number = \frac{1.4 * 10^9 * 1km^3}{20,000cm^3}$

Recall that $1 km^3 = 10^{15} cm^3$

So, the expression becomes

$Number = \frac{1.4 * 10^9 * 10^{15} cm^3}{20,000cm^3}$

Simplify the numerator using laws of indices

$Number = \frac{1.4 * 10^{9+15} cm^3}{20,000cm^3}$

$Number = \frac{1.4 * 10^{25} cm^3}{20,000cm^3}$

Expand the denominator

$Number = \frac{1.4 * 10^{25} cm^3}{2 * 10,000cm^3}$

$Number = \frac{1.4 * 10^{25} cm^3}{2 * 10^4\ cm^3}$

Split fraction

$Number = \frac{1.4}{2} * \frac{10^{25} cm^3}{10^4\ cm^3}$

$Number = 0.7 * \frac{10^{25}}{10^4\ }$

Simplify fraction using laws of indices

$Number = 0.7 * 10^{25-4}$

$Number = 0.7 * 10^{21}$

$Number = 7 * 10^{20}$

Hence, it'll take a bucket loads of $7 * 10^{20}$ to empty world's ocean

4 0

3 years ago

The length of a rectangle is 2 m greater than the width. The area is 143m^2. Find the length and the width.

iogann1982 [59]

Answer:

L = 13 m W = 11 m

Step-by-step explanation:

L = W + 2

area = L x W

143 = (W+2) * W

143 = W^2 + 2w

W^2 + 2W - 143 = 0

Use Quadratic Formula (a = 1 b = 2 c = - 143)

to find W = 11 m then L = 13

8 0

2 years ago

The human resources manager at a company records the length, in hours, of one shift at work, X. He creates the

kiruha [24]

Answer:

.84

Step-by-step explanation:

8 0

4 years ago

If the differential equation t 2y 00 − 2y 0 + (3 + t)y = 0 has y1(t) and y2(t) as a fundamental set of solutions and if W[y1, y2

podryga [215]

In the ODE, solve for $y''$ :

$t^2y''-2y'+(3+t)y=0\implies y''=\dfrac{2y'+(3+t)y}{t^2}$

The Wronskian is then

$W=\begin{vmatrix}y_1&y_2\\{y_1}'&{y_2}'\end{vmatrix}=y_1{y_2}'-{y_1}'y_2$

Differentiating the Wronskian gives

$W'=({y_1}'{y_2}'+y_1{y_2}'')-({y_1}''y_2+{y_1}'{y_2}')=y_1{y_2}''-{y_1}''y_2$

Substitute $y_1,y_2$ into the equation for $y''$ , then substitute ${y_1}'',{y_2}''$ into $W'$ :

$W'=y_1\dfrac{2{y_2}'+(3+t)y_2}{t^2}-y_2\dfrac{2{y_1}'+(3+t)y_1}{t^2}$

$\implies W'=\dfrac{2W}{t^2}$

which is another separable ODE; we have

$\dfrac{\mathrm dW}W=\dfrac2{t^2}\,\mathrm dt\implies \ln|W|=-\dfrac2t+C\implies W=Ce^{-2/t}$

Given that $W(y_1,y_2)(2)=3$ , we find

$3=Ce^{-2/2}\implies C=3e$

so that

$W(y_1,y_2)(t)=3e^{1-2/t}$

and so

$W(y_1,y_2)(6)=\boxed{3e^{2/3}}$

8 0

3 years ago

Let <img src="https://tex.z-dn.net/?f=i" id="TexFormula1" title="i" alt="i" align="absmiddle" class="latex-formula"> be the imag

VLD [36.1K]

$Hey~freckledspots!\\----------------------$

$We~will~solve~for~i^{425}!$

$Rule~of~exponent: a^{b + c} = a^ba^c\\Apply:~i^{425}~=~i^{424}i\\ \\Rule~of~exponent: a^{bc} = (a^{b})^c\\Apply: i^{424} = i(i^2)^{212} \\\\Rule~of~imaginary~number: i^2 = -1\\Apply: i(i^2)^{212} = -1^{212}i\\\\Rule~of~exponent~if~n~is~even: -a^n = a^n\\Apply: -1^{212}i = 1^{212}i\\\\Simplify: 1^{212}i = 1i\\Multiply: 1i * 1 = i\\----------------------\\$

$Now~let's~solve~1^{14}!\\\\Rule~of~exponent: a^{b + c} = a^ba^c\\Apply: i^{14} = (i^2)^7\\\\Rule~of~imaginary~number: i^2 = -1\\Apply: (i^2)^7 = -1^7\\\\Rule~of~exponent~if~n~is~odd: (-a)^n = -a^n\\Apply: -1^7 = -1^7\\\\Simplify: -1^7 = -1\\----------------------\\Now,~we~have: i-1+i^{-14}+i^{44}\\----------------------$

$Now~lets~solve~i^{-14}\\\\Rule~of~exponent: a^{-b} = \frac{1}{a^b} \\Apply: i^{-14} = \frac{1}{i^{14}} \\\\Rule~of~exponent: a^{bc} = (a^b)^c\\Apply: \frac{1}{i^{14}} = \frac{1}{(i^2)^7}\\ \\Rule~of~imagianry~number: i^2 = -1\\Apply: \frac{1}{(i^2)^7} = \frac{1}{-1^7} \\\\Simplify: \frac{1}{-1^7} = \frac{1}{-1} \\\\Rule~of~fractions: \frac{a}{-b} = -\frac{a}{b} \\Apply: \frac{1}{-1} = -\frac{1}{1} = -1\\----------------------\\Now,~we~have: i-1-1+i^44\\----------------------$

$Now~let's~solve~i^{44}!\\\\Rule~of~exponent: a^{bc} = (a^b)^c\\Apply: i^{44} = (i^2)^{22}\\\\Rule~of~imaginary~numbers: i^2 = -1\\Apply: (i^2)^{22} = -1^{22}\\\\Rule~of~exponent~if~n~is~even: (-a)^n = a^n\\Apply: -1^{22} = 1^{22}\\\\Simplify: 1^{22} = 1\\----------------------\\Now,~we~have~i-1-1+1\\----------------------$

$Now~let's~simplify~the~expression!\\\\= i-1-1+1 \\= 1 + i -2\\= -1+i\\----------------------$

$Answer:\\\large\boxed{-1+i}\\----------------------$

$Hope~This~Helped!~Good~Luck!$

8 0

4 years ago

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