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3 years ago

Can somebody plz help answer these questions correctly (only if u know how to do this) thanks :D!

Mathematics

1 answer:

nignag [31]3 years ago

4 0

14. Range=22 Mean= 31.2
15. Range=33 Mean= 88.2
16. Range= 81 Mean= 161.3
To find range you take the smallest number and the largest number and subtract them. To find mean you add all of the totals together and then divide by how many numbers are given.

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X^4-100=0 solve the equation by factoring

timurjin [86]

$x^4-100=0 \\
(x^2)^2-10^2=0 \\
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x^2-10=0 \ \lor \ x^2+10=0 \\
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3 years ago

Pls help asap ill give brainlist if it lets me

zaharov [31]

The mistake is

3/5/5/3 =-3/5 this is the mistake

but the answer is

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6 0

3 years ago

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I WILL GIVE 20 POINTS TO THOSE WHO ANSWER THIS QUESTION RIGHT NOOOO SCAMS AND EXPLAIN WHY THAT IS THE ANSWER

Step2247 [10]

Answer:

$A = 297\pi$

Step-by-step explanation:

To solve this problem we need to be familiar with the formula for the surface area of a cone:

$A = \pi r(r + \sqrt{h^2+r^2})$

We are given the length of a side and the diameter, to calculate the radius divide the diameter in half:

$r = \frac{d}{2}\\r = \frac{18}{2}\\r = 9 cm$

To calculate the height of the cone, we must use the Pythagorean Theorem:

$C^2 = A^2 + B^2$

We can treat the side length as the hypotenuse $C$ , the radius as the base $A$ , and solve for height $B$ . Set the expression up like this:

$C^2 = A^2 + B^2\\24^2 = 9^2 + B^2\\B^2 = 24^2 - 9^2\\B = \sqrt{24^2 - 9^2}\\B = \sqrt{576 - 81}\\B = \sqrt{495}\\B \approx 22.25$

Now we can plug into our original formula:

$A = \pi r(r + \sqrt{h^2+r^2})\\A = \pi 9(9+\sqrt{\sqrt{495}^2+9^2}\\A = \pi 9(9+\sqrt{495 + 81}\\A = \pi 9(9+\sqrt{576})\\A = \pi 9(9 + 24)\\A = \pi 9(33)\\A = 297\pi$

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2 years ago

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rusak2 [61]

26.

You simply read the table on this one

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3 years ago

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What is the length of the diagonal if a 10 cm by 15cm rectangle ?

mojhsa [17]

In other words, if a rectangle is 10 cm by 15 cm, what is the length of the diagonal? Use the Pythagorean Theorem. a^2+b^2=c^2, where c is the measure of the hypotenuse.

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4 years ago

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