All categories

1 year ago

If x² + x - 6=0, then x=

Mathematics

1 answer:

alexgriva [62]1 year ago

5 0

Answer:

x = - 3 or x = 2

Step-by-step explanation:

Solve the quadratic equation by factorizasion

x² + x - 6 = 0
x² - 2x + 3x - 6 = 0
x(x - 2) + 3(x - 2) = 0
(x + 3)(x - 2) = 0
x + 3 = 0 or x - 2 = 0
x = - 3 or x = 2

You might be interested in

A right triangle has a leg measuring 52 m and a hypotenuse measuring 63 m. find the length of the other leg.

tangare [24]

By Pythagoras Theorem,
the length of the other leg= (63^2-52^2)^1/2
hope this helps :)

4 0

3 years ago

Find the surface area of the solid generated by revolving the region bounded by the graphs of y = x2, y = 0, x = 0, and x = 2 ab

Nikitich [7]

Answer:

See explanation

Step-by-step explanation:

The surface area of the solid generated by revolving the region bounded by the graphs can be calculated using formula

$SA=2\pi \int\limits^a_b f(x)\sqrt{1+f'^2(x)} \, dx$

If $f(x)=x^2,$ then

$f'(x)=2x$

and

$b=0\\ \\a=2$

Therefore,

$SA=2\pi \int\limits^2_0 x^2\sqrt{1+(2x)^2} \, dx=2\pi \int\limits^2_0 x^2\sqrt{1+4x^2} \, dx$

Apply substitution

$x=\dfrac{1}{2}\tan u\\ \\dx=\dfrac{1}{2}\cdot \dfrac{1}{\cos ^2 u}du$

Then

$SA=2\pi \int\limits^2_0 x^2\sqrt{1+4x^2} \, dx=2\pi \int\limits^{\arctan(4)}_0 \dfrac{1}{4}\tan^2u\sqrt{1+\tan^2u} \, \dfrac{1}{2}\dfrac{1}{\cos^2u}du=\\ \\=\dfrac{\pi}{4}\int\limits^{\arctan(4)}_0 \tan^2u\sec^3udu=\dfrac{\pi}{4}\int\limits^{\arctan(4)}_0(\sec^3u+\sec^5u)du$

Now

$\int\limits^{\arctan(4)}_0 \sec^3udu=2\sqrt{17}+\dfrac{1}{2}\ln (4+\sqrt{17})\\ \\ \int\limits^{\arctan(4)}_0 \sec^5udu=\dfrac{1}{8}(-(2\sqrt{17}+\dfrac{1}{2}\ln(4+\sqrt{17})))+17\sqrt{17}+\dfrac{3}{4}(2\sqrt{17}+\dfrac{1}{2}\ln (4+\sqrt{17}))$

Hence,

$SA=\pi \dfrac{-\ln(4+\sqrt{17})+132\sqrt{17}}{32}$

3 0

2 years ago

Amelia rented a DVD and it was due to be returned on 26 Nov. She actually returned it to the shop on 12 Dec. The rental shop app

svet-max [94.6K]

Answer:

144 (pounds? euros? idk what currency it is)

Step-by-step explanation:

there are 16 day between the date. 16 x 9 = 144

6 1

3 years ago

Read 2 more answers

Hep asappppppppppppp

schepotkina [342]

Answer:

(6 , -1)

Step-by-step explanation:

2x+ y = 11 ----------------(I)

y = 11 - 2x --------------------(II)

$\frac{1}{2}x - 5y = 8$

Multiply the whole equation by 2

$2*\frac{1}{2}x -2*5y = 2*8$

x - 10y = 16 --------------------(III)

Substitute y = 11- 2x in equation (III)

x - 10(11 - 2x) = 16

x - 110 + 20x = 16

21x - 110 = 16

21x = 16 +110

21x = 126

x = 126/21

x = 6

Plugin x = 6 in equation (II)

y = 11 - 2*6

y = 11 - 12

y = -1

5 0

2 years ago

which ordered pairs in the form (x, y) are solutions to the equation 3x−4y=21 ? Select each correct answer. (−1, −6) (−3, 3) (11

Valentin [98]

Answer:

see explanation

Step-by-step explanation:

To determine which ordered pairs are solutions to the equation

Substitute the x and y values into the left side of the equation and if equal to the right side then they are a solution.

(- 1, - 6)

3(- 1) - 4(- 6) = - 3 + 24 = 21 = right side ← thus a solution

(- 3, 3)

3(- 3) - 4(3) = - 9 - 12 = - 21 ≠ 21 ← not a solution

(11, 3)

3(11) - 4(3) = 33 - 12 = 21 = right side ← thus a solution

(7, 0)

3(7) - 4(0) = 21 - 0 = 21 = right side ← thus a solution

The ordered pairs (- 1, - 6), (11, 3), (7, 0) are solutions to the equation

3 0

3 years ago

Read 2 more answers