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pochemuha
3 years ago
8

Explain why the ratio 2 to 5 is different from the ratio 5 to 2 if they both represent the ratio of cats to dogs.

Mathematics
1 answer:
Fynjy0 [20]3 years ago
7 0
The ratio 2 to 5 means there are 2 cats and 5 dogs but the 5 to 2 ratio means there are 5 cats and 2 dogs.
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substances that are used up in a reaction

Step-by-step explanation:

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5. Classify each triangle by its angle and sides
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5 is D

6 is C

Step-by-step explanation:

Define:

Equilateral - All sides and angles are congruent

Isosceles: Two congruent sides and two congruent angles

Scalene: No congruent sides or angles

Right triangle: A triangle with a right angle ( note that a right angle has a measure of 90 degrees )

obtuse triangle: A triangle with an obtuse angle ( obtuse angles have a measure of more than 90 degrees. )

Acute triangle : A triangle with an acute angle ( an acute angle has a measure of less than 90 degrees )

Answer:

the triangle shown in # 5 has a right angle and 2 congruent sides and angles

Therefore the triangle in #5 is a right isosceles

The triangle shown in #6 has an angle with a measure that is more than 90 degrees and have no congruent sides or angles therefore the triangle in #6 is an obtuse scalene.

3 0
3 years ago
Find the equation of a line that is perpendicular to y = 3x – 5 and passes through the point (1, -3).
Svetradugi [14.3K]

keeping in mind that perpendicular lines have negative reciprocal slopes, let's check for the slope of the equation above

\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}\qquad \qquad y = \stackrel{\stackrel{m}{\downarrow }}{3}x-5

well then therefore

\stackrel{\textit{perpendicular lines have \underline{negative reciprocal} slopes}} {\stackrel{slope}{3\implies \cfrac{3}{1}} ~\hfill \stackrel{reciprocal}{\cfrac{1}{3}} ~\hfill \stackrel{negative~reciprocal}{-\cfrac{1}{3}}}

so we're really looking for the equation of a line with slope of -1/3 and that passes through (1, -3 )

(\stackrel{x_1}{1}~,~\stackrel{y_1}{-3})\qquad \qquad \stackrel{slope}{m}\implies -\cfrac{1}{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}{(-3)}=\stackrel{m}{-\cfrac{1}{3}}(x-\stackrel{x_1}{1})\implies y+3=-\cfrac{1}{3}x+\cfrac{1}{3} \\\\\\ y=-\cfrac{1}{3}x+\cfrac{1}{3}-3\implies y=-\cfrac{1}{3}x-\cfrac{8}{3}

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