Please help !! the answer isn’t 1:9.75 x 10^7 like i thought !

Mathematics

1 answer:

postnew [5]10 months ago

8 0

The required scale model will be 1:6.24 10^9

<h3>Finding scale factors</h3>

A scale factor is the ratio of the length of a side of one figure to the length of the corresponding side of the other figure.

Given the following parameters

Distance of sun to Jupiter = 7.8 * 10^8 km

If the distance of sun to Jupiter is 8m on his scale, hence the value of n is calculated as;

n = 8 * 7.8 *10^8

n = 62.4 * 1068

n = 6.24 10^9

Hence the required scale model will be 1:6.24 10^9

Learn more on scale model here: brainly.com/question/24105256

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What is the area of ABC below? If necessary, round your answer to two decimal places.

irina [24]

Our goal is to find the area
area=1/2 times base times height
we know the height is 7.9 but what is the base?

use the pythagorean theorem twice
alrighty
so
remember for legs length a and b and hytponuse c in a right triangle
a²+b²=c²

we need AD and DC

so
AD²+7.9²=9.4²
AD=√25.95

DC²+7.9²=23.2²
DC=√475.83

so
AD+DC=base=(√25.95)+(√475.83)
area=1/2bh
area=1/2((√25.95)+(√475.83))(7.9)
area≈106.28518654591426812803776879893
round to 2 decimal places
area≈106.29 square units

6 0

2 years ago

The slope-intercept form of the equation of a line that passes through point (-3, 8) is y = -2/3x + 6. What is the point-

Elena-2011 [213]

$y = \stackrel{\stackrel{m}{\downarrow }}{-\cfrac{2}{3}}x+6\qquad \impliedby \begin{array}{|c|ll} \cline{1-1} slope-intercept~form\\ \cline{1-1} \\ y=\underset{y-intercept}{\stackrel{slope\qquad }{\stackrel{\downarrow }{m}x+\underset{\uparrow }{b}}} \\\\ \cline{1-1} \end{array}$

so hmmm then we know the slope of that line is -2/3, so we're really looking for the point-slope form of a line with a slope of -2/3 and that passes through (-3 , 8)

$(\stackrel{x_1}{-3}~,~\stackrel{y_1}{8})\hspace{10em} \stackrel{slope}{m} ~=~ -\cfrac{2}{3} \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{8}=\stackrel{m}{-\cfrac{2}{3}}(x-\stackrel{x_1}{(-3)})\implies y-8=-\cfrac{2}{3}(x+3)$