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kykrilka [37]
3 years ago
10

Solve the equation by using the quadratic formula. x^2+2x=6​

Mathematics
1 answer:
katrin [286]3 years ago
6 0

Answer:

\Huge \boxed{{x=-1\pm \sqrt{7}}}

Step-by-step explanation:

x² + 2x = 6

Subtract both sides by 6.

x² + 2x - 6 = 0

ax²+bx+c=0

a=1, b=2, and c=-6

We can apply the quadratic formula.

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

Plug in the values.

\displaystyle x=\frac{-2\pm\sqrt{2^2-4(1)(-6)}}{2(1)}

Evaluate.

\displaystyle x=\frac{-2\pm\sqrt{4-(-24)}}{2}

\displaystyle x=\frac{-2\pm\sqrt{28}}{2}

\displaystyle x=\frac{-2\pm 2\sqrt{7}}{2}=-1 \pm \sqrt{7}

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An open rectangular box having a volume of 108 in.3 is to be constructed from a tin sheet. Find the dimensions of such a box if
murzikaleks [220]

Answer:

6 in x 6 in x 3 in.

Step-by-step explanation:

Given

V = xyz = 108   ⇒   z = 108/(xy)

The amount of the material used is

S = xy + 2yz + 2xz

Put value of z from the volume

S = xy + 2y*108/(xy) + 2x*108/(xy) = xy + 216/x + 216/y

Now, we find the relative minimum of the function S(x,y)

First, we find the critical point. Set Sx = 0  and  Sy = 0

and solve this system:

Sx(x,y) = y - (216/x²) = 0

Sy(x,y) = x - (216/y²) = 0

From the first equation we have

y = 216/x²

Put it in the second equation and find x

x - (216/(216/x²)²) = 0

⇒  x*(1 - (x³/216)) = 0

⇒  x₁ = 0   and  x₂ = 6

Now, we can find y as follows

y₁ = 216/(0)²   which is undefined

y₂ = 216/(6)² = 6

Hence, the only critical point of S is (6, 6). Next, we calculate the second ordered derivatives that we need for the second derivative test:

Sxx(x,y) = 432/x³

Sxy(x,y) = 1

Syy(x,y) = 432/y³

Applying the second derivative test

D(6, 6) = Sxx(6, 6)*Syy(6, 6) - S²xy(6, 6) = 2*2 - 1² = 4 -1 = 3 > 0

Sxx(6, 6) = 2 > 0

Since D(6, 6) > 0   and   Sxx(6, 6) > 0   we can conclude that S has a relative minimum at (6, 6).

z coordinate is:

z = 108/(xy) = 108 / (6*6) = 3

Finally, the dimentions of a box are 6 in x 6 in x 3 in.

6 0
3 years ago
A hockey puck has a diameter of 7.6 cm and a height of 3.4 cm. What is the volume of a cylindrical package containing six pucks?
alexandr402 [8]

Answer:

924.96\ cm^3

Step-by-step explanation:

Given that,

The diameter of a hockey puck, d = 7.6 cm

Height, h = 3.4 cm

We need to find the volume of the volume of a cylindrical package containing six pucks.

Volume of 1 puck,

V=\pi r^2 h\\\\=3.14\times (3.8)^2\times 3.4\\\\=154.16\ cm^3

Volume of 6 pucks,

V=6\times 154.16\ cm^3\\\\=924.96\ cm^3

So, the volume of the 6 puck is 924.96\ cm^3.

5 0
3 years ago
In this figure, AB¯¯¯¯¯∥CD¯¯¯¯¯ and m∠1=110°. What is m∠5? Enter your answer in the box
masha68 [24]

Answer:

It's just like it looks, ∠1 = ∠5

Therefore ∠5 = 110 degrees

Step-by-step explanation:

7 0
3 years ago
The function f(x) = 6x + 8 is transformed to function g through a horizontal stretch by a factor of 5. What is the equation of f
Vsevolod [243]

Answer:

g(x)  = \frac{6}{5}x + 8

Step-by-step explanation:

Given

f(x) = 6x + 8

k = 5 --- horizontal stretch factor

Required

Find g(x)

When a function is stretched horizontally by k, the resulting function will be:

Rule: g(x) = f(\frac{1}{k}x)

So, we have:

g(x) = f(\frac{1}{5}x)

Solve for: f(\frac{1}{5}x)

f(\frac{1}{5}x) = 6 * \frac{1}{5}x + 8

f(\frac{1}{5}x) = \frac{6}{5}x + 8

Recall that:

g(x) = f(\frac{1}{5}x)

g(x)  = \frac{6}{5}x + 8

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Answer:

3 4/100 is equal to 3.04 in decimal form

Step-by-step explanation:

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