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3 years ago

If a pentagon has dimensions of 12 in length, 6in height and 8in width, how can I calculate the area of this pentagon?

Mathematics

1 answer:

ANTONII [103]3 years ago

8 0

12•6= 72

If you were finding volume then you would multiply it my 8 too, but you need to find area

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3 years ago

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Which circle is the graph of the equation (x + 4)2 + (y – 5)2 = 4?

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2 years ago

X=6 is a solution to the inequality x−5≤ 10 True or False?

vekshin1

Answer:

$\boxed{\sf{FALSE}}$

$\boxed{\sf{x\leq 15}}$

Step-by-step explanation:

Isolate the term of x from one side of the equation.

First, add by 5 from both sides.

$\sf{x-5+5\leq10+5}$

Solve.

Add the numbers from left to right.

$\sf{10+5=15}$

$\Longrightarrow: \boxed{\sf{x\leq 15}}$

Therefore, the solution is x≤15, which is our answer.

I hope this helps you! Let me know if my answer is wrong or not.

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2 years ago

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Solve these equations. Thanks.

Nutka1998 [239]

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2 years ago

For a function p(x)= x^2-9, what is the inverse function for the domain [0, infinity\

kkurt [141]

An inverse function is a function that reverses another function, so we have a function called:

$p(x)$

Then, the inverse function will be as follows:

$p^{-1}(x)$

Given that $y = p(x)$ , we need to isolate x in terms of y:

$y = x^{2} -9$
∴ $y+9 = x^{2}$

So:

$x=+\sqrt{y+9}$ and $x=-\sqrt{y+9}$

therefore, exchanging variables x and y:

$y=+\sqrt{x+9}$ and $y=-\sqrt{x+9}$
$p^{-1}(x)=+\sqrt{x+9}$ and $p^{-1}(x)=-\sqrt{x+9}$

Which are the figures shown below.

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3 years ago