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3 years ago

6

I need help on 24 please

1 answer:

dusya [7]3 years ago

4 0

Where's the table. you need it to answer the questions.

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What is the Area of the circle in pi units if the diameter is 6 cm?

sergejj [24]

The area of the circle in pi units when the diameter of this circle is 6 centimeters is 9π centimeters.

<h3>What is the area of the circle?</h3>

The area of the circle is the space occupied by it. It is the product of pi and square of its diameter divided by 4. The area of the circle can be given as,

$A=\pi\dfrac{d^2}{4}$

Here (d) is the diameter of the circle. The diameter of the circle is 6 cm.

$d=6$

Put this value in the above formula to find the area of the circle as,

$A=\pi\dfrac{6^2}{4}\\A=\pi\dfrac{36}{4}\\A=\pi\times9\\A=9\pi\rm\; cm$

Thus, the area of the circle in pi units when the diameter of this circle is 6 centimeters is 9π centimeters.

Learn more about the area of the circle here;

brainly.com/question/402655

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6 0

2 years ago

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Please help me solve this problem ASAP

Setler [38]

Find the height of it with pythagoras
Root(100-36)=8
Volume = 1/3 pi r^2 h
1/3(36pi x 8)
=96pi
about 301.59ft^3

3 0

2 years ago

Solve the following equations.<br> log2(x^2 − 16) − log^2(x − 4) = 1

Alenkasestr [34]

Answer:

$x=\frac{4*(2+e)}{e-2}$

Step-by-step explanation:

Let's rewrite the left side keeping in mind the next propierties:

$log(\frac{1}{x} )=-log(x)$

$log(x*y)=log(x)+log(y)$

Therefore:

$log(2*(x^{2} -16))+log(\frac{1}{(x-4)^{2} })=1\\ log(\frac{2*(x^{2} -16)}{(x-4)^{2}})=1$

Now, cancel logarithms by taking exp of both sides:

$e^{log(\frac{2*(x^{2} -16)}{(x-4)^{2}})} =e^{1} \\\frac{2*(x^{2} -16)}{(x-4)^{2}}=e$

Multiply both sides by $(x-4)^{2}$ and using distributive propierty:

$2x^{2} -32=16e-8ex+ex^{2}$

Substract $16e-8ex+ex^{2}$ from both sides and factoring:

$-(x-4)*(-8-4e-2x+ex)=0$

Multiply both sides by -1:

$(x-4)*(-8-4e-2x+ex)=0$

Split into two equations:

$x-4=0\hspace{3}or\hspace{3}-8-4e-2x+ex=0$

Solving for $x-4=0$

Add 4 to both sides:

$x=4$

Solving for $-8-4e-2x+ex=0$

Collect in terms of x and add $4e+8$ to both sides:

$x(e-2)=4e+8$

Divide both sides by e-2:

$x=\frac{4*(2+e)}{e-2}$

The solutions are:

$x=4\hspace{3}or\hspace{3}x=\frac{4*(2+e)}{e-2}$

If we evaluate x=4 in the original equation:

$log(0)-log(0)=1$

This is an absurd because log (x) is undefined for $x\leq 0$

If we evaluate $x=\frac{4*(2+e)}{e-2}$ in the original equation:

$log(2*((\frac{4e+8}{e-2})^2-16))-log((\frac{4e+8}{e-2}-4)^2)=1$

Which is correct, therefore the solution is:

$x=\frac{4*(2+e)}{e-2}$

6 0

3 years ago

What is half of 215 ?

bogdanovich [222]

Answer:

107.5

Step-by-step explanation:

6 0

2 years ago

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Find the elapsed time.<br> from 9:16 a.m. to 1:44 p.m.

Masteriza [31]

Answer:

5 hours and 28 minutes

Step-by-step explanation:

8 0

2 years ago

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