All categories

2 years ago

What is the area of the regular polygon?

Mathematics

2 answers:

natta225 [31]2 years ago

7 0

Correct answer would be 50m^2 because 100/2=50

cestrela7 [59]2 years ago

5 0

The diagonal is 10, so a side is ${10/\sqrt{2} }$

The area would be 100/2

or 50 m^2.

You might be interested in

If 3x-4+8=12+2x<br> then x=?

vova2212 [387]

Answer:

Step-by-step explanation:

Combine like terms, -4 and 8, 2x and 3x.

After combining the terms, you get x+4=12.

Subtract 4 from both sides and get x=8

5 0

3 years ago

Which equation represents the value of x? x=100−y2−−−−−−−√ x=10+y2 x=10−y x=y2+100−−−−−−−√ Right triangle A B C with angle B as

Nostrana [21]

Answer:

Therefore the required value of x,

$x=\sqrt{(100-y^{2})}$

Step-by-step explanation:

Given:

ΔBC is a Right Angle Triangle at ∠ B = 90°

As ∠ B = 90° , AC will be the Hypotenuse

AC = 10 = Hypotenuse

BC = y = Longer leg ( say )

AB = x = Shorter leg ( say )

To Find :

x = ?

Solution:

In Right Angle Triangle Δ ABC , By Pythagoras Theorem we get

$(\textrm{Hypotenuse})^{2} = (\textrm{Shorter leg})^{2}+(\textrm{Longer leg})^{2}$

Substituting the given values we get

$10^{2}= x^{2}+y^{2} \\\\x^{2}=100-y^{2} \\\\\textrm{square rooting on both the side we get}\\\\x=\sqrt{(100-y^{2})}\\\therefore x=\sqrt{(100-y^{2})}\ \textrm{ which is the required value of x}$

Therefore the required value of x,

$x=\sqrt{(100-y^{2})}$

8 0

3 years ago

Find the equation of the line shown.

ExtremeBDS [4]

Answer: y = -1/4x + 2
explanation: line passes through y axis on 2, so 2 is your y-intercept. rise/run is ur slope, which is -1/4 .

3 0

2 years ago

<img src="https://tex.z-dn.net/?f=15%20x%5E%7B2%7D%20%2B14x-49%3D0" id="TexFormula1" title="15 x^{2} +14x-49=0" alt="15 x^{2} +1

Alex777 [14]

$15x^2+14x-49=0\\\\
Terms:\\
a=15\ \ \ \ b=14\ \ \ \ c=-49\\
Quadratic\ Formula:\\\\
x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\
x=\dfrac{-14\pm\sqrt{196+2,940}}{30}\\\\
x=\dfrac{-14\pm\sqrt{3,136}}{30}\\\\
x=\dfrac{-14\pm56}{30}$

$Roots:\\\\
x'=\dfrac{-14+56}{30}\\\\
x'=\dfrac{42}{30}\\\\
x'=\dfrac{21}{15}\\\\
\boxed{x'=\dfrac{7}{5}}\\\\\\\\
x''=\dfrac{-14-56}{30}\\\\
x''=\dfrac{-70}{30}\\\\
\boxed{x''=\dfrac{-7}{3}}$

8 0

3 years ago

Read 2 more answers

What is the range of f(x)=1/2^x

SCORPION-xisa [38]

Downloaded Photomath it’s going to help you a lot

8 0

3 years ago