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3 years ago

6

Explain how you know these figures are similar

2 answers:

sergeinik [125]3 years ago

7 0

Answer:

because of them is bigger than the other one so that means its differ but it has the same length

Step-by-step explanation:

ziro4ka [17]3 years ago

7 0

The figures are the same size

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Square root of -49 in terms of i

natita [175]

Answer:

7i

Step-by-step explanation:

We know that i = √-1

Then solving the given:

√-49 = √49×(-1) = √49 × √-1 = 7×√-1 = 7i

So the answer is:

√-49 = 7i

3 0

3 years ago

A football team loses 3 yards, gains 12 yards, gains 10 yards, and then loses 15 yards. Did the team gain yards or lose yards ov

e-lub [12.9K]

Answer: the answer is 4

Step-by-step explanation:

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3 years ago

WILL CHOOSE BRAINLIEST!!! Quadrilateral ABCD is congruent to quadrilateral RSTU. Without seeing the figures visually, which stat

garri49 [273]

2nd option

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8 0

2 years ago

Read 2 more answers

Given tan theta =9, use trigonometric identities to find the exact value of each of the following:_______

Ludmilka [50]

Answer:

$(a)\ \sec^2(\theta) = 82$

$(b)\ \cot(\theta) = \frac{1}{9}$

$(c)\ \cot(\frac{\pi}{2} - \theta) = 9$

$(d)\ \csc^2(\theta) = \frac{82}{81}$

Step-by-step explanation:

Given

$\tan(\theta) = 9$

Required

Solve (a) to (d)

Using tan formula, we have:

$\tan(\theta) = \frac{Opposite}{Adjacent}$

This gives:

$\frac{Opposite}{Adjacent} = 9$

Rewrite as:

$\frac{Opposite}{Adjacent} = \frac{9}{1}$

Using a unit ratio;

$Opposite = 9; Adjacent = 1$

Using Pythagoras theorem, we have:

$Hypotenuse^2 = Opposite^2 + Adjacent^2$

$Hypotenuse^2 = 9^2 + 1^2$

$Hypotenuse^2 = 81 + 1$

$Hypotenuse^2 = 82$

Take square roots of both sides

$Hypotenuse =\sqrt{82}$

So, we have:

$Opposite = 9; Adjacent = 1$

$Hypotenuse =\sqrt{82}$

Solving (a):

$\sec^2(\theta)$

This is calculated as:

$\sec^2(\theta) = (\sec(\theta))^2$

$\sec^2(\theta) = (\frac{1}{\cos(\theta)})^2$

Where:

$\cos(\theta) = \frac{Adjacent}{Hypotenuse}$

$\cos(\theta) = \frac{1}{\sqrt{82}}$

So:

$\sec^2(\theta) = (\frac{1}{\cos(\theta)})^2$

$\sec^2(\theta) = (\frac{1}{\frac{1}{\sqrt{82}}})^2$

$\sec^2(\theta) = (\sqrt{82})^2$

$\sec^2(\theta) = 82$

Solving (b):

$\cot(\theta)$

This is calculated as:

$\cot(\theta) = \frac{1}{\tan(\theta)}$

Where:

$\tan(\theta) = 9$ ---- given

So:

$\cot(\theta) = \frac{1}{\tan(\theta)}$

$\cot(\theta) = \frac{1}{9}$

Solving (c):

$\cot(\frac{\pi}{2} - \theta)$

In trigonometry:

$\cot(\frac{\pi}{2} - \theta) = \tan(\theta)$

Hence:

$\cot(\frac{\pi}{2} - \theta) = 9$

Solving (d):

$\csc^2(\theta)$

This is calculated as:

$\csc^2(\theta) = (\csc(\theta))^2$

$\csc^2(\theta) = (\frac{1}{\sin(\theta)})^2$

Where:

$\sin(\theta) = \frac{Opposite}{Hypotenuse}$

$\sin(\theta) = \frac{9}{\sqrt{82}}$

So:

$\csc^2(\theta) = (\frac{1}{\frac{9}{\sqrt{82}}})^2$

$\csc^2(\theta) = (\frac{\sqrt{82}}{9})^2$

$\csc^2(\theta) = \frac{82}{81}$

4 0

3 years ago

Find the midpoint of the segment with the given endpoints.<br><br><br> M(–2, 1) and N(–3, 2)

mafiozo [28]

$\bf ~~~~~~~~~~~~\textit{middle point of 2 points }
\\\\
M(\stackrel{x_1}{-2}~,~\stackrel{y_1}{1})\qquad 
N(\stackrel{x_2}{-3}~,~\stackrel{y_2}{2})
\qquad
% coordinates of midpoint 
\left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right)
\\\\\\
\left( \cfrac{-3-2}{2}~~,~~\cfrac{2+1}{2} \right)\implies \left(-\frac{5}{2}~,~\frac{3}{2} \right)$

5 0

3 years ago

Read 2 more answers