Which is a graph of g(c) = (0.5)x+3^ -4

THERE IS 2 QUESTIONS PLEASE HELP ME!

lukranit [14]

Answer:

Number 2 is 3:4.

Number 1 is d.

Step-by-step explanation:

Boxers and collies added together gives you 12:16. Divide them both by 4 and you get 3:4.

<

The number of student in class 4 (6) divided by the total number of kids in the entire class (18) gives you 0.33333333333. Multiply that by 100 and it gives you 33.3333333333. As a percentage it is 33 1/2 %

6 0

2 years ago

State the slope and the y-intercept for the graph of the equation. y= -2

kotegsom [21]

Answer: 0; -2

Step-by-step explanation:

Data: y-intercept=x and slope=x

Equation=y=-2

Only step, Point out the y-intercept and slope

-2=y-intercept

0=slope

Reason: Since there is no visible slope, the placeholder would be 0 therefore the slope would be 0, -2 is the only number left so that means the only number that can be y-intercept is -2.

That is why the slope is 0 and the y-intercept is -2

I hope this helps!

5 0

2 years ago

An airplane descends 1.5 miles to an elevation of 5.25 miles. Write and solve an equation to find the elevation of the plane bef

DerKrebs [107]

2.25 is the elevation of the plane before its decent

7 0

2 years ago

A string passing over a smooth pulley carries a stone at one end. While its other end is attached to a vibrating tuning fork and

nasty-shy [4]

Answer:

correct option is C) 2.8

Step-by-step explanation:

given data

string vibrates form = 8 loops

in water loop formed = 10 loops

solution

we consider mass of stone = m

string length = l

frequency of tuning = f

volume = v

density of stone = $\rho$

case (1)

when 8 loop form with 2 adjacent node is $\frac{\lambda }{2}$

so here

$l = \frac{8 \lambda _1}{2}$ ..............1

$l = 4 \lambda_1\\\\\lambda_1 = \frac{l}{4}$

and we know velocity is express as

velocity = frequency × wavelength .....................2

$\sqrt{\frac{Tension}{mass\ per\ unit \length }}$ = f × $\lambda_1$

here tension = mg

so

$\sqrt{\frac{mg}{\mu}}$ = f × $\lambda_1$ ..........................3

and

case (2)

when 8 loop form with 2 adjacent node is $\frac{\lambda }{2}$

$l = \frac{10 \lambda _1}{2}$ ..............4

$l = 5 \lambda_1\\\\\lambda_1 = \frac{l}{5}$

when block is immersed

equilibrium eq will be

Tenion + force of buoyancy = mg

T + v × $\rho$ × g = mg

and

T = v × $\rho$ - v × $\rho$ × g

from equation 2

f × $\lambda_2$ = f × $\frac{1}{5}$

$\sqrt{\frac{v\rho _{stone} g - v\rho _{water} g}{\mu}} = f \times \frac{1}{5}$ .......................5

now we divide eq 5 by the eq 3

$\sqrt{\frac{vg (\rho _{stone} - \rho _{water})}{\mu vg \times \rho _{stone}}} = \frac{fl}{5} \times \frac{4}{fl}$

solve irt we get

$1 - \frac{\rho _{stone}}{\rho _{water}} = \frac{16}{25}$

so

relative density $\frac{\rho _{stone}}{\rho _{water}} = \frac{25}{9}$

relative density = 2.78 ≈ 2.8

so correct option is C) 2.8

3 0

2 years ago

Write the equation of the line that is the perpendicular bisector of the segment with endpoints (4, 1) and (2, -5)

Sedbober [7]

Answer: $\bold{y=-\dfrac{1}{3}x-1}$

<u>Step-by-step explanation:</u>

(4, 1) & (2, -5)

First, find the slope (m) and then the perpendicular (opposite reciprocal) slope:

$m=\dfrac{y_2-y_1}{x_2-x_1}\\\\\\m=\dfrac{-5-1}{2-4} = \dfrac{-6}{-2}=3\quad \rightarrow \quad m_{\perp}=-\dfrac{1}{3}\\$

Next, find the midpoint of (4, 1) and (2, -5):

$Midpoint=\bigg(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\bigg)\\\\\\.\qquad \qquad=\bigg(\dfrac{4+2}{2},\dfrac{1-5}{2}\bigg)\\\\\\.\qquad \qquad=\bigg(\dfrac{6}{2},\dfrac{-4}{2}\bigg)\\\\\\.\qquad \qquad=(3, -2)$

Lastly, input the perpendicular slope and the midpoint into the Point-Slope formula to find the equation of the line:

$y - y_1 = m_{\perp}(x - x_1)\\\\y - (-2) = -\dfrac{1}{3}(x - 3)\\\\y + 2=-\dfrac{1}{3}x +1\\\\y =-\dfrac{1}{3}x - 1\\$

6 0

2 years ago