When writing an equation, your “y-intercept” is whatever the value of y is when x =

Could you please help me with this problem?

NeTakaya

First Question:

Number of trapezoids - 6

Perimeter - 72

3 + 3 + 3 + 6 = 15

3 + 3 + 3 + 3 + 6 + 6 = 24

3 + 3 + 6 + 3 + 3 + 6 + 3 + 6 = 33

15 + 24 + 33 = 72

Second Question:

Number of cubes - 6

Surface area - 24 (probably 480 since it said area)

first cube - 6 ones

second cube - 8 ones

Third cube - 10 ones

you could do 6 + 8 + 10 = 24 or add up all the ones (1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1) and you will still get 24

(It said area, and area means to multiply so maybe its 6 x 8 x 10 = 480 and not 6 + 8 + 10)

Anyways have a good day/night :) I might be wrong, but I hope I helped at least a little

3 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

3 years ago

(x-6)² + (x + 5)² = 16

Phoenix [80]

Answer:

answer A : (2 -5)

Step-by-step explanation:

the midipoint P,Q is the center A of this circle (because PQ is a diameter)

use : (x-6)²+(x+5)² = 16 you have : A(6 , -5) now let Q(x,y) :

(x+10)/2 = 6

(y-5)/2 = - 5

x+10 = 12

y-5 = -10

so : x = 2 and y = - 5....answer A : (2 -5)

6 0

3 years ago

PLSS HELP ILL GIVE BRAINLIEST IF YOU HELP ME AND EXPLAIN UR ANSWER

kherson [118]

Answer:

It is the second choice.

Step-by-step explanation:

Divide both side by pi r squared. Boom.

3 0

2 years ago

Read 2 more answers

Dan draws a circle with a radius of 8 inches. Question 1 Part A: What is the circumference of Dan's circle, inches? If needed, r

geniusboy [140]

Answer:

The circumference of Dan's circle is approximately 50.27 inches.

Step-by-step explanation:

The length of the circumference ( $s$ ), measured in inches, is determined by the formula:

$s = 2\pi\cdot r$ (1)

Where $r$ is the radius, measured in inches.

If we know that $r = 8\,in$ , then the length of the circumference is:

$s = 2\pi\cdot (8\,in)$

$s \approx 50.27\,in$

The circumference of Dan's circle is approximately 50.27 inches.

8 0

3 years ago