What is the exact volume of the cylinder?

IRISSAK [1]

Answer:

5.8 m

Step-by-step explanation:

Using the Pythagorean Theorem (a² + b² = c²) you get the equation a² + 1.5² = 6² (or 1.5² + b² = 6²). Then you solve from there and rounding to the first decimal place you get 5.8

8 0

3 years ago

Q3: Identify the graph of the equation and write and equation of the translated or rotated graph in general form. (Picture Provi

natta225 [31]

Answer:

b. circle; $2(x')^2+2(y')^2-5x'-5\sqrt{3}y'-6 =0$

Step-by-step explanation:

The given conic has equation;

$x^2-5x+y^2=3$

We complete the square to obtain;

$(x-\frac{5}{2})^2+(y-0)^2=\frac{37}{4}$

This is a circle with center;

$(\frac{5}{2},0)$

This implies that;

$x=\frac{5}{2},y=0$

When the circle is rotated through an angle of $\theta=\frac{\pi}{3}$ ,

The new center is obtained using;

$x'=x\cos(\theta)+y\sin(\theta)$ and $y'=-x\sin(\theta)+y\cos(\theta)$

We plug in the given angle with x and y values to get;

$x'=(\frac{5}{2})\cos(\frac{\pi}{3})+(0)\sin(\frac{\pi}{3})$ and $y'=--(\frac{5}{2})\sin(\frac{\pi}{3})+(0)\cos(\frac{\pi}{3})$

This gives us;

$x'=\frac{5}{4} ,y'=\frac{5\sqrt{3} }{4}$

The equation of the rotated circle is;

$(x'-\frac{5}{4})^2+(y'-\frac{5\sqrt{3} }{4})^2=\frac{37}{4}$

Expand;

$(x')^2+(y')^2-\frac{5\sqrt{3} }{2}y'-\frac{5}{2}x'+\frac{25}{4} =\frac{37}{4}$

Multiply through by 4; to get

$4(x')^2+4(y')^2-10\sqrt{3}y'-10x'+25 =37$

Write in general form;

$4(x')^2+4(y')^2-10x'-10\sqrt{3}y'-12 =0$

Divide through by 2.

$2(x')^2+2(y')^2-5x'-5\sqrt{3}y'-6 =0$

8 0

3 years ago

**PLEASE HELP ASAP** Terri is analyzing a circle, y^2 + x^2 = 36, and a linear function g(x). Will they intersect?

Alborosie

We analyze the linear function to find that it has equation $y=x$ . Drawing the line, we find that it will intersect the circle at negative and positive x-coordinates.

We can prove this by substituting $y=x$ into the given equation:

$x^2+x^2=36$

$2x^2=36$

$x^2 = 18$

$x = \pm 3\sqrt{2}$

Hence, $x = 3\sqrt{2}$ or $-\sqrt{2}$ , so there are both positive and negative values for $x$ .

5 0

3 years ago

Full working out please ^^ for the first question , ty

sp2606 [1]

Answer:

E

Step-by-step explanation:

Given

y = 2x² - kx + 3

with a = 2, b = - k and c = 3

Since the curve touches the x- axis at one place then the roots are real and equal and the discriminant for this condition is

b² - 4ac = 0, that is