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2 years ago

Pls help in this maths problem

Mathematics

1 answer:

tester [92]2 years ago

5 0

Answer:

SORRY I don't know second number . I did only first .

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Suppose theta is an angle in the standard position whose terminal side is in quadrant 4 and cot theta = -6/7. find the exact val

zimovet [89]

First off, let's notice that the angle is in the IV Quadrant, where sine is negative and the cosine is positive, likewise the opposite and adjacent angles respectively.

Also let's bear in mind that the hypotenuse is never negative, since it's simply just a radius unit.

$\bf cot(\theta )=\cfrac{\stackrel{adjacent}{6}}{\stackrel{opposite}{-7}}\qquad \impliedby \textit{let's find the \underline{hypotenuse}} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies c=\sqrt{a^2+b^2} \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ c=\sqrt{6^2+(-7)^2}\implies c=\sqrt{36+49}\implies c=\sqrt{85} \\\\[-0.35em] ~\dotfill$

$\bf tan(\theta)=\cfrac{\stackrel{opposite}{-7}}{\stackrel{adjacent}{6}} ~\hfill csc(\theta)=\cfrac{\stackrel{hypotenuse}{\sqrt{85}}}{\stackrel{opposite}{-7}} ~\hfill sec(\theta)=\cfrac{\stackrel{hypotenuse}{\sqrt{85}}}{\stackrel{adjacent}{6}} \\\\\\ sin(\theta)=\cfrac{\stackrel{opposite}{-7}}{\stackrel{hypotenuse}{\sqrt{85}}}\implies \stackrel{\textit{and rationalizing the denominator}}{sin(\theta)=\cfrac{-7}{\sqrt{85}}\cdot \cfrac{\sqrt{85}}{\sqrt{85}}\implies sin(\theta)=-\cfrac{7\sqrt{85}}{85}}$

$\bf cos(\theta)=\cfrac{\stackrel{adjacent}{6}}{\stackrel{hypotenuse}{\sqrt{85}}}\implies \stackrel{\textit{and rationalizing the denominator}}{cos(\theta)=\cfrac{6}{\sqrt{85}}\cdot \cfrac{\sqrt{85}}{\sqrt{85}}\implies cos(\theta)=\cfrac{6\sqrt{85}}{85}}$

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2 years ago

A prism is filled with 60 cubes with 1/2 unite side length

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2 years ago

Which number is the smallest? 5.6123 5.6245 5.6023 5.6985

alexandr402 [8]

5.6023, hope this helps!

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2 years ago

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Sidana [21]

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Step-by-step explanation:

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2 years ago

The area of the base of a rectangular prisim four and three fourths and the height of two and one third what is the volume

VikaD [51]

Data:
$A_{base} = 4 \frac{3}{4} = \frac{4*4+3}{4} = \frac{16+3}{4} \to A_{base} = \frac{19}{4}$

$h (height) = 2 \frac{1}{3} = \frac{3*2+1}{3} = \frac{6+1}{3} \to h (height) = \frac{7}{3}$
$v (volume) = ?$

Formula: $V = A_{b}*h$

Solving:
$V = A_{b}*h$
$V = \frac{19}{4} * \frac{7}{3}$
$V = \frac{133}{12}$
$V = 11.08333....$
$\boxed{\boxed{V \approx 11.08}}\end{array}}\qquad\quad\checkmark$

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3 years ago