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2 years ago

Find the domain & range of the relation {(-5,13), (6,18), (2,-7), (1,4), (-3,-1)}

Mathematics

1 answer:

Neporo4naja [7]2 years ago

5 0

Answer:

Domain: -5,6,2,1,-3

Range: 13,18,-7,4,-1

Step-by-step explanation:

Domain is the independent value/ x / input

Range is the dependent value/ y / output

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3 years ago

211 minus 87.4 decimal subtraction

Pavlova-9 [17]

<u>211 - 87.40</u> equals 123.60.

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3 years ago

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ABCD is a rhombus, where m∠AED = 5x – 10. Use the properties of a rhombus to determine the value of x. Question 10 options: A) x

fomenos

Answer:

A) x = 20

Step-by-step explanation:

ABCD is a rhombus, AC and BD are diagonals which intersect each other at point E.

Since, diagonals of a rhombus are perpendicular bisector.

$\therefore \: m \angle AED = 90 \degree \\ \therefore \: (5x - 10) \degree = 90 \degree \\ \therefore \: 5x - 10 = 90 \\ \therefore \: 5x = 90 + 10 \\ \therefore \: 5x = 100 \\ \\ \therefore \: x = \frac{100}{5} \\ \\ \huge \red{ \boxed{ \therefore \: x =20}}$

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3 years ago

Find the exact value of cos(sin^-1(-5/13))

son4ous [18]

bearing in mind that the hypotenuse is never negative, since it's just a distance unit, so if an angle has a sine ratio of -(5/13) the negative must be the numerator, namely -5/13.

$\bf cos\left[ sin^{-1}\left( -\cfrac{5}{13} \right) \right] \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{then we can say that}~\hfill }{sin^{-1}\left( -\cfrac{5}{13} \right)\implies \theta }\qquad \qquad \stackrel{\textit{therefore then}~\hfill }{sin(\theta )=\cfrac{\stackrel{opposite}{-5}}{\stackrel{hypotenuse}{13}}}\impliedby \textit{let's find the \underline{adjacent}}$

$\bf \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies \pm\sqrt{c^2-b^2}=a \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ \pm\sqrt{13^2-(-5)^2}=a\implies \pm\sqrt{144}=a\implies \pm 12=a \\\\[-0.35em] ~\dotfill\\\\ cos\left[ sin^{-1}\left( -\cfrac{5}{13} \right) \right]\implies cos(\theta )=\cfrac{\stackrel{adjacent}{\pm 12}}{13}$

le's bear in mind that the sine is negative on both the III and IV Quadrants, so both angles are feasible for this sine and therefore, for the III Quadrant we'd have a negative cosine, and for the IV Quadrant we'd have a positive cosine.

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3 years ago

What are the correct steps for solving the following equation: 5x - = 21

ololo11 [35]

Answer:

x = -21/5 = -4.200

Step-by-step explanation:

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3 years ago

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