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2 years ago

6

A sphere is inscribed in a cylinder. Use complete sentences and geometric formulas to compare the surface area of the sphere and

the lateral area of the cylinder.

1 answer:

crimeas [40]2 years ago

8 0

Ke y Fredy no se pasa del modo y el Fredy que no tiene el nombre y el que le gusta el ratón y el Brian de que no se puede hacer nada por el la tuba de Brian o si no no te pasó la netta y no se pasan de Brian o no son muy buenos o no se puede hacer nada porque no se

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Solve for x in the equation 2xsquared-5+1=3.

gizmo_the_mogwai [7]

Answer:

x = 1.87

Step-by-step explanation:

2x^2 - 5 + 1 = 3

2x^2 = 3 + 5 - 1

2x^2 = 7

x^2 = 7/2

x = √3.5

x = 1.87082.....

x = 1.87 ( 3 significant figures )

4 0

3 years ago

I honestly need help with these

Brilliant_brown [7]

9. The curve passes through the point (-1, -3), which means

$-3 = a(-1) + \dfrac b{-1} \implies a + b = 3$

Compute the derivative.

$y = ax + \dfrac bx \implies \dfrac{dy}{dx} = a - \dfrac b{x^2}$

At the given point, the gradient is -7 so that

$-7 = a - \dfrac b{(-1)^2} \implies a-b = -7$

Eliminating $b$ , we find

$(a+b) + (a-b) = 3+(-7) \implies 2a = -4 \implies \boxed{a=-2}$

Solve for $b$ .

$a+b=3 \implies b=3-a \implies \boxed{b = 5}$

10. Compute the derivative.

$y = \dfrac{x^3}3 - \dfrac{5x^2}2 + 6x - 1 \implies \dfrac{dy}{dx} = x^2 - 5x + 6$

Solve for $x$ when the gradient is 2.

$x^2 - 5x + 6 = 2$

$x^2 - 5x + 4 = 0$

$(x - 1) (x - 4) = 0$

$\implies x=1 \text{ or } x=4$

Evaluate $y$ at each of these.

$\boxed{x=1} \implies y = \dfrac{1^3}3 - \dfrac{5\cdot1^2}2 + 6\cdot1 - 1 = \boxed{y = \dfrac{17}6}$

$\boxed{x = 4} \implies y = \dfrac{4^3}3 - \dfrac{5\cdot4^2}2 + 6\cdot4 - 1 \implies \boxed{y = \dfrac{13}3}$

11. a. Solve for $x$ where both curves meet.

$\dfrac{x^3}3 - 2x^2 - 8x + 5 = x + 5$

$\dfrac{x^3}3 - 2x^2 - 9x = 0$

$\dfrac x3 (x^2 - 6x - 27) = 0$

$\dfrac x3 (x - 9) (x + 3) = 0$

$\implies x = 0 \text{ or }x = 9 \text{ or } x = -3$

Evaluate $y$ at each of these.

$A:~~~~ \boxed{x=0} \implies y=0+5 \implies \boxed{y=5}$

$B:~~~~ \boxed{x=9} \implies y=9+5 \implies \boxed{y=14}$

$C:~~~~ \boxed{x=-3} \implies y=-3+5 \implies \boxed{y=2}$

11. b. Compute the derivative for the curve.

$y = \dfrac{x^3}3 - 2x^2 - 8x + 5 \implies \dfrac{dy}{dx} = x^2 - 4x - 8$

Evaluate the derivative at the $x$ -coordinates of A, B, and C.

$A: ~~~~ x=0 \implies \dfrac{dy}{dx} = 0^2-4\cdot0-8 \implies \boxed{\dfrac{dy}{dx} = -8}$

$B:~~~~ x=9 \implies \dfrac{dy}{dx} = 9^2-4\cdot9-8 \implies \boxed{\dfrac{dy}{dx} = 37}$

$C:~~~~ x=-3 \implies \dfrac{dy}{dx} = (-3)^2-4\cdot(-3)-8 \implies \boxed{\dfrac{dy}{dx} = 13}$

12. a. Compute the derivative.

$y = 4x^3 + 3x^2 - 6x - 1 \implies \boxed{\dfrac{dy}{dx} = 12x^2 + 6x - 6}$

12. b. By completing the square, we have

$12x^2 + 6x - 6 = 12 \left(x^2 + \dfrac x2\right) - 6 \\\\ ~~~~~~~~ = 12 \left(x^2 + \dfrac x2 + \dfrac1{4^2}\right) - 6 - \dfrac{12}{4^2} \\\\ ~~~~~~~~ = 12 \left(x + \dfrac14\right)^2 - \dfrac{27}4$

so that

$\dfrac{dy}{dx} = 12 \left(x + \dfrac14\right)^2 - \dfrac{27}4 \ge 0 \\\\ ~~~~ \implies 12 \left(x + \dfrac14\right)^2 \ge \dfrac{27}4 \\\\ ~~~~ \implies \left(x + \dfrac14\right)^2 \ge \dfrac{27}{48} = \dfrac9{16} \\\\ ~~~~ \implies \left|x + \dfrac14\right| \ge \sqrt{\dfrac9{16}} = \dfrac34 \\\\ ~~~~ \implies x+\dfrac14 \ge \dfrac34 \text{ or } -\left(x+\dfrac14\right) \ge \dfrac34 \\\\ ~~~~ \implies \boxed{x \ge \dfrac12 \text{ or } x \le -1}$

13. a. Compute the derivative.

$y = x^3 + x^2 - 16x - 16 \implies \boxed{\dfrac{dy}{dx} = 3x^2 - 2x - 16}$

13. b. Complete the square.

$3x^2 - 2x - 16 = 3 \left(x^2 - \dfrac{2x}3\right) - 16 \\\\ ~~~~~~~~ = 3 \left(x^2 - \dfrac{2x}3 + \dfrac1{3^2}\right) - 16 - \dfrac13 \\\\ ~~~~~~~~ = 3 \left(x - \dfrac13\right)^2 - \dfrac{49}3$

Then

$\dfrac{dy}{dx} = 3 \left(x - \dfrac13\right)^2 - \dfrac{49}3 \le 0 \\\\ ~~~~ \implies 3 \left(x - \dfrac13\right)^2 \le \dfrac{49}3 \\\\ ~~~~ \implies \left(x - \dfrac13\right)^2 \le \dfrac{49}9 \\\\ ~~~~ \implies \left|x - \dfrac13\right| \le \sqrt{\dfrac{49}9} = \dfrac73 \\\\ ~~~~ \implies x - \dfrac13 \le \dfrac73 \text{ or } -\left(x-\dfrac13\right) \le \dfrac73 \\\\ ~~~~ \implies \boxed{x \le 2 \text{ or } x \ge \dfrac83}$

5 0

2 years ago

Angle A is twice the size of Angle B.Their sum is 87 degrees.Find x

Nataliya [291]

A=2b
a+b=87
a=4x-2
a=87-2x+1
4x-2=88-2x
6x=90
x=15
b=29
a=58

6 0

3 years ago

Which of the following is the graph of y= 2x2+3/4?

liberstina [14]

Y= (2×2)- 4 +3/4 = 12/4 or 3 that I'd the best of my

5 0

3 years ago

Eight students share 12 mini oatmeal muffins equally and 6 students share 15 mini apple muffins equally Carmine is in both group

Lana71 [14]

Group 1:
divide number of muffins by number of students

=8 ÷ 12
=8/12
reduce by 4
=2/3 muffins each in group 1

Group 2:
=6 ÷ 15
=6/15
reduce by 3
=2/5 muffins each in group 2

Add fractions together:
= 2/3 + 2/5
need common denominator
multiply 2/3 by 5 and 2/5 by 3 for common denominator of 15

=(2/3 * 5/5) + (2/5 * 3/3)
=(2*5/3*5) + (2*3/5*3)
= 10/15 + 6/15
= 16/15
= 1 1/15 muffin

ANSWER: Carmine gets 1 1/15 muffins because she was in both groups. She received 2/3 in one group, 2/5 in another group and added together she received 1 1/15 muffins.

Hope this helps! :)

4 0

4 years ago