Help me please I need it

The area of the base of the oblique pentagonal pyramid is 50 cm2 and the distance from the apex to the center of the pentagon is

kirza4 [7]

Answer:

$\text{Height is }6\sqrt2cm$

$\text{The volume is }100\sqrt2cm^3$

Step-by-step explanation:

Given the area of the base of the oblique pentagonal pyramid is $50 cm^2$ and the distance from the apex to the center of the pentagon is 6 cm. The measure of ∠ACB is 45°.

we know that

ABC is a right triangle

∠ACB=45°

AC=hypotenuse------> 6 cm

$\sin 45^{\circ}=\frac{AB}{AC}$

$AB=AC\times \sin 45$

$AB=\frac{6}{\sqrt2}$

$AB=3\sqrt2$

$\text{Height is }6\sqrt2cm$

Now, volume of pyramid

$\text{Volume of the pyramid=}\frac{1}{3}\times \text{Area of the base}\times height$

Area of the base=50 cm²

$Height=6\sqrt2 cm$

so, $\text{volume of the pyramid=}\frac{1}{3}\times 50\times 6\sqrt2=100\sqrt2cm^3$

$\text{the volume is }100\sqrt2cm^3$

7 0

3 years ago

Read 2 more answers

Helppp easy quick points

vladimir2022 [97]

Answer should be y=2x+1

3 0

2 years ago

If P and Q are two sets then n( P-Q) means ?

Lapatulllka [165]

$\huge{\mathbb{ \tt{ANSWER}}}$

I shall use the symbol U for union

^ for intersection

~ for complement

The property of set subtraction states that for any sets X and Y, X-Y = X ^ Y~

That is in words, X throw away Y is the same as the intersection of X and the complement of Y

or X and anything NOT in Y

At the SET level, the proof looks like this:

(P-Q) U (Q-P) = « given

(P^Q~) U (Q ^ P~) = « property of set subtraction

(P U Q) ^ (P U P~) ^ (Q~ U Q) ^ (Q~ U P~) = « distributive property

(P u Q) ^ (Q~ U P~) = « Union of any set with it's complement is the universal

set which gets dropped by any intersection of a non-universal set

(P u Q ) ^ (Q ^ P)~ = « De Morgan's Law

(P u Q) - (Q ^ P) = « property of set subtraction

(P u Q) - (P ^ Q) « commutative property

At the ELEMENT level, the proof looks like this:

» Proves left side is subset of right side

If x is in (P-Q) U (Q-P) then x is in (P-Q) OR x is in (Q-P)

If x is in P-Q then x is in P but not in Q.

Then X is in P U Q.

Since x is not in Q, x is NOT in P ^ Q

Hence x is in P U Q but NOT in P^Q, which

means it is an element if (P U Q) - (P ^ Q)

Therefore x is an element of the set defined by the right side, which means (P-Q) U (Q-P) is a subset of (P U Q) - (P ^ Q)

« proves right side is subset of the left side

If x is an element of (P U Q) - (P ^ Q) then

x is in (P U Q) but NOT in (P^Q)

Since x is in P U Q, x is in P OR x is in Q.

Since x is NOT in (P^Q) then x is NOT in BOTH P and Q.

Hence x is in either P or Q but not both.

If x is in P, then x is not in Q, so x is in P-Q.

Therefore x is in (P-Q) U (Q-P).if x is in Q, then x is not in P, so x is in Q-P.Therefore x is in (Q-P) U (P-Q).There x is an element of set defined by the left side, which menas (P u Q)-(P ^Q) is a subset of (P-Q) U (Q-P).

#CarryOnLearning

#LetsEnjoyTheSummer

8 0

3 years ago

M+3/2n-8=-3m m+3 is over 2n-8

marysya [2.9K]

-3
m=____________
8n - 32

4 0

4 years ago

How do you solve for the quotient of (x^-1) - 1 ÷ x - 1?

Ber [7]

$\bf x^{-1}-1\div x-1\implies \implies \cfrac{1}{x}-1\div x-1\implies \cfrac{\frac{1}{x}-1}{~~x-1~~}\implies \cfrac{~~\frac{1-x}{x}~~}{\frac{x-1}{1}} \\\\\\ \cfrac{1-x}{x}\cdot \cfrac{1}{x-1}\implies \cfrac{-(\begin{matrix} x-1 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix})}{x}\cdot \cfrac{1}{\begin{matrix} x-1 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}}\implies -\cfrac{1}{x}\implies -x^{-1}$

3 0

4 years ago