All categories

2 years ago

Reflect P (-5,10) over the x-axis. What are the coordinates of the reflected point? What quadrant is the reflected point in?

Mathematics

1 answer:

erastova [34]2 years ago

8 0

Answer:

<h3>I need the answer 2 QwQ</h3><h3>....</h3>

You might be interested in

Prove that:- (1-sinA)(1+sinA)÷(1+cosA)(1-cosA)=cot^2A

astraxan [27]

Answer:

see explanation

Step-by-step explanation:

Using the trigonometric identity

sin²x + cos²x = 1 , then

sin²x = 1 - cos²x and cos²x = 1 - sin²x

Consider the left side

$\frac{(1-sinA)(1 + sinA)}{(1+cosA)(1-cosA)}$ ← expand numerator/denominator using FOIL

= $\frac{1-sin^2A}{1-cos^2A}$

= $\frac{1-(1-cos^2A)}{1-(1-sin^2A)}$

= $\frac{1-1+cos^2A}{1-1+sin^2A}$

= $\frac{cos^2A}{sin^2A}$

= cot²A = right side , thus proven

6 0

2 years ago

Differentiate Square numbers and cube numbers

pashok25 [27]

Answer:

square numbers are numbers raised up to the power of two in essence the number multiplied by it self

6 0

2 years ago

7. Triangles CDE and RST are similar triangles. What is

andreyandreev [35.5K]

Answer:

CD = 25 ft

Step-by-step explanation:

$\triangle CDE \sim \triangle RST$

$\therefore \frac{CD}{20}=\frac{40}{32}$

(Corresponding sides of similar triangles are in proportion)

$\therefore \frac{CD}{20}=\frac{5}{4}$

$\therefore CD = \frac{5}{\cancel 4}\times \cancel {20} \:\:\blue{\bold{5}}$

$\therefore CD = 5\times 5$

$\therefore CD = 25\: ft$

6 0

2 years ago

Read 2 more answers

How much is 3x4x5 plz help

Vikentia [17]

3×4×5 is 60 I think it help u

3 0

2 years ago

Read 2 more answers

Find the six trigonometric function values for angle ∅ where its adjacent side is -9 and its hypotenuse is 41. (Theta is located

arsen [322]

Check the picture below.

$\bf \textit{using the pythagorean theorem}
\\\\
c^2=a^2+b^2\implies \sqrt{c^2-a^2}=b
\qquad 
\begin{cases}
c=hypotenuse\\
a=adjacent\\
b=opposite\\
\end{cases}
\\\\\\
\sqrt{41^2-(-9)^2}=b\implies \sqrt{1681-81}=b\\\\\\ \sqrt{1600}=b\implies 40=b\\\\
-------------------------------$

$\bf sin(\theta )=\cfrac{\stackrel{opposite}{40}}{\stackrel{hypotenuse}{41}}\qquad~~ cos(\theta )=\cfrac{\stackrel{adjacent}{-9}}{\stackrel{hypotenuse}{41}}\qquad~~ tan(\theta )=\cfrac{\stackrel{opposite}{40}}{\stackrel{adjacent}{-9}}
\\\\\\
csc(\theta )=\cfrac{\stackrel{hypotenuse}{41}}{\stackrel{opposite}{40}}\qquad ~~sec(\theta )=\cfrac{\stackrel{hypotenuse}{41}}{\stackrel{adjacent}{-9}}\qquad ~~cot(\theta )=\cfrac{\stackrel{adjacent}{-9}}{\stackrel{opposite}{40}}$

3 0

2 years ago