Find the volume of the cone. 13 cm r= 9 cm V = [?] cm3 Round to the nearest tenth.

In ΔPQR, find the measure of ∡P.

Artyom0805 [142]

Answer:

30.4°

Step-by-step explanation:

< P = arc tan QR/PQ

= arc tan 33.8/57.6

= arc tan 0.5868

= 30.4°

8 0

3 years ago

HELP will give brainliest if all correct

ladessa [460]

1. Definition of Parallelogram

2. Angle 2 plus Angle 3 is 180=Def of Supple

3.Angle Congruence Postulate

4. Angle 1 and Angle 3 equal to 180

5 0

3 years ago

Consider the graph of the line y = .5x- 4 and the point

Katena32 [7]

Answer:

slope of parallel line and perpendicular line are 5 and -1/5 espectively

equation of parallel and perpendicular line are y = 5x + 22 $y= \frac{-1}{5} x+\frac{6}{5}$ respectively

Step-by-step explanation:

y = 5x - 4 is in the form

y = mx + c

where m is the slope of the line and c is the y intercept of thr line

therefore slope of the line = 5 and y intercept = -4

when an another line is parallel to the given line then the slope of both the lines are equal

therefore the slope the parallel line = 5

equation of a line passing through a given point $(x_{1} ,y_{1})$ with slope m is given by $y-y_{1} = m(x-x_{1} )$

given $(x_{1} ,y_{1})$ = (-4,2)

therefore equation of line y-2 = 5(x+4)

therefore y = 2+ 5x+20

y = 5x + 22is the eqaution of required line.

when two lines are perpendiculer then

$m_{1} m_{2}=-1$

where $m_{1} and m_{2}$ are slope of the lines therefore

m×5=-1

therefore m= $\frac{-1}{5}$

therefore eqaution of line passing through (-4,2) and with slope m= $\frac{-1}{5}$ is given by $y - 2= \frac{-1}{5} (x+4)$

$y= \frac{-1}{5} x+\frac{6}{5}$

3 0

3 years ago

Please help thank you.

alekssr [168]

Hello!

First you have to find BO

BC = 27
AO = CO
CO = 24

BO = 27 + 24

BO = 51

Now we use the Pythagorean Theorem

24^2 + b^2 = 51^2

Square the numbers

576 + b^2 = 2601

Subtract 576 from both sides

b^2 = 2025

Take the square root

b = 45

The answer is A) 45

Hope this helps!

6 0

3 years ago

Identify the standard form of the equation by completing the square.

OLEGan [10]

Answer:

$\dfrac{(x-1)^2}{9}-\dfrac{(y-2)^2}{4}=1$

Step-by-step explanation:

Given equation:

$4x^2-9y^2-8x+36y-68=0$

This is an equation for a horizontal hyperbola.

To complete the square for a hyperbola

Arrange the equation so all the terms with variables are on the left side and the constant is on the right side.

$\implies 4x^2-8x-9y^2+36y=68$

Factor out the coefficient of the x² term and the y² term.

$\implies 4(x^2-2x)-9(y^2-4y)=68$

Add the square of half the coefficient of x and y inside the parentheses of the left side, and add the distributed values to the right side:

$\implies 4\left(x^2-2x+\left(\dfrac{-2}{2}\right)^2\right)-9\left(y^2-4y+\left(\dfrac{-4}{2}\right)^2\right)=68+4\left(\dfrac{-2}{2}\right)^2-9\left(\dfrac{-4}{2}\right)^2$

$\implies 4\left(x^2-2x+1\right)-9\left(y^2-4y+4\right)=36$

Factor the two perfect trinomials on the left side:

$\implies 4(x-1)^2-9(y-2)^2=36$

Divide both sides by the number of the right side so the right side equals 1:

$\implies \dfrac{4(x-1)^2}{36}-\dfrac{9(y-2)^2}{36}=\dfrac{36}{36}$

Simplify:

$\implies \dfrac{(x-1)^2}{9}-\dfrac{(y-2)^2}{4}=1$

Therefore, this is the standard equation for a horizontal hyperbola with:

center = (1, 2)
vertices = (-2, 2) and (4, 2)
co-vertices = (1, 0) and (1, 4)
$\textsf{Asymptotes}: \quad y = -\dfrac{2}{3}x+\dfrac{8}{3} \textsf{ and }y=\dfrac{2}{3}x+\dfrac{4}{3}$
$\textsf{Foci}: \quad (1-\sqrt{13}, 2) \textsf{ and }(1+\sqrt{13}, 2)$

4 0

1 year ago