Can someone help with this question

Mathematics

1 answer:

NikAS [45]2 years ago

4 0

1st : Find the volume of the rectangular block:
12 × 10 × 20 = 2400 cm^3.

2nd : Find the volume of a cylinder
$\pi {r}^{2} h$
Where r is radius & h is height.
so:
$\pi \times 2.2 {}^{2} \times 5.0$
= 76.02654222 cm^3

Lastly, divide the volume of the rectangular block by the volume of the cylinder
2400 ÷ 76.026 = 31.5679226

You can't round it up so the answer is F) 31

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In simplest radical form, what are the solutions to the quadratic equation 6 = x2 – 10x?

vichka [17]

Answer:

$x = 5+\sqrt{31}\,\, and\,\, x=5-\sqrt{31}$

Step-by-step explanation:

We need to solve the quadratic equation

6 = x^2 -10x

Rearranging we get,

x^2-10x-6=0

Using quadratic formula to solve the quadratic equation

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

a= 1, b =-10 and c=6

Putting values in the quadratic formula

$x=\frac{-(-10)\pm\sqrt{(-10)^2-4(1)(-6)}}{2(1)}\\x=\frac{10\pm\sqrt{100+24}}{2}\\x=\frac{10\pm\sqrt{124}}{2}\\x=\frac{10\pm\sqrt{2*2*31}}{2}\\x=\frac{10\pm\sqrt{2^2*31}}{2}\\x=\frac{10\pm2\sqrt{31}}{2}\\x = 5\pm\sqrt{31}$

So, $x=5+\sqrt{31}\,\, and\,\, x=5-\sqrt{31}$

3 0

3 years ago

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Solve the following system of equations <br> x^2+2y^2=59<br> 2x^2+y^2=43

aivan3 [116]

Answer: x = (-3,3), y = (-5,5)

Step-by-step explanation:

Add both of the equations together, for $3x^{2}+3y^{2} = 102$ . Now we can divide both sides by 3, getting $x^2+y^2=34$ . we subtract the first equation from our equation we just got, getting $y^2=25$ y= (5,-5). once we plug that in, we get $50+x^2 = 59$ , $x^2 = 9$ x = (-3,3)