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3 years ago

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Item 8 The cost (in dollars) of making b bracelets is represented by 4 5b. The cost (in dollars) of making b necklaces is repres

ented by 8b 6. Write a polynomial in standard form that represents how much more it costs to make b necklaces than b bracelets.? The polynomial is .

1 answer:

grin007 [14]3 years ago

4 0

Answer:

3b+2

Explanation:

First your question has so many mistakes

What I understand

bracelets cost 4+5b

necklaces cost 8b+6

First we have to make an equaion

=(8b+6)-(4+5b)

simplify it

=8b+6-4-5b

=3b+2 Answer

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How many molecules are in 3 moles of CO2

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Answer:

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Two point charges of +2.0 μC and -6.0 μC are located on the x-axis at x = -1.0 cm and x = +2.0 cm respectively. Where should a t

lyudmila [28]

Answer:

x = -3 cm

Explanation:

The electrical potential is the sum of the potentials of each charge

V = k ∑ $q_{i} / r_{i}$

let's apply this to our case where the potential is V = 0 for x = 0

0 = k (q₁ / (x₁-0) + q₂ / (x₂-0) + q₃ / (x₃-0))

in our case

q₁ = + 2.0 10⁻⁶ C

q₂ = - 6.0 10⁻⁶ C

q₃ = + 3.0 10⁻⁶ C

x₁ = -1.0 cm = 1.0 10⁻² m

x₂ = +2.0 cm = 2.0 10⁻² m

we substitute in the equation

0 = k (2 10⁻⁶ / 1 10⁻² - 6 10⁻⁶ / 2 10⁻² + 3 10⁻⁶ / x)

3 10⁻⁶ / x = 2 10⁻⁴ - 3 10⁻⁴

3 10⁻⁶ / x = -1 10⁻⁴

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3 years ago

Car enthusiasts often lower their cars closer to the ground as a matter of style. James wants to lower his car by replacing all

Alexandra [31]

Answer:

$\Delta h=0.0364\ m=3.64\ cm$

Explanation:

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change in stiffness constant of the spring on replacing the original springs, $\delta k=5355\ N.m^{-1}$

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initial length of the original car-spring before compression, $l_i=12\ cm=0.12\ m$

final length of the original car-spring after compression, $l_f=8.55\ cm=0.0855\ m$

So, weight of the car:

$w=m.g$

$w=1455\times 9.81$

$w=14273.55\ N$

<u>Now the spring constant of original spring:</u>

$w=4k_o.(l_i-l_f)$ (since 4 springs are in parallel)

$14273.55=4k_o\times (0.12-0.0855)$

$k_o=103431.522\ N.m^{-1}$

<u>So the stiffness constant of the new springs:</u>

$k_n=k_o-\delta k$

$k_n=103431.522-5355$

$k_n=98076.522\ N.m^{-1}$

<u>Now the height lowered:</u>

$w=k_n.4\Delta h$ (since 4 springs are in parallel)

$14273.55=4\times98076.522\times \Delta h$

$\Delta h=0.0364\ m=3.64\ cm$

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3 years ago

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anygoal [31]

Answer:

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Studentka2010 [4]

(a) 18.9 m/s

The motion of the stone consists of two independent motions:

- A horizontal motion at constant speed

- A vertical motion with constant acceleration ( $g=9.8 m/s^2$ ) downward

We can calculate the components of the initial velocity of the stone as it is launched from the ground:

$u_x = v_0 cos \theta = (25.0)(cos 41.0^{\circ})=18.9 m/s\\u_y = v_0 sin \theta = (25.0)(sin 41.0^{\circ})=16.4 m/s$

The horizontal velocity remains constant, while the vertical velocity changes due to the acceleration along the vertical direction.

When the stone reaches the top of its parabolic path, the vertical velocity has became zero (because it is changing direction): so the speed of the stone is simply equal to the horizontal velocity, therefore

$v=18.9 m/s$

(b) 22.2 m/s

We can solve this part by analyzing the vertical motion only first. In fact, the vertical velocity at any height h during the motion is given by

$v_y^2 - u_y^2 = 2ah$ (1)

where

$u_y = 16.4 m/s$ is the initial vertical velocity

$v_y$ is the vertical velocity at height h

$a=g=-9.8 m/s^2$ is the acceleration due to gravity (negative because it is downward)

At the top of the parabolic path, $v_y = 0$ , so we can use the equation to find the maximum height

$h_{max} = \frac{-u_y^2}{2a}=\frac{-(16.4)^2}{2(-9.8)}=13.7 m$

So, at half of the maximum height,

$h = \frac{13.7}{2}=6.9 m$

And so we can use again eq(1) to find the vertical velocity at h = 6.9 m:

$v_y = \sqrt{u_y^2 + 2ah}=\sqrt{(16.4)^2+2(-9.8)(6.9)}=11.6 m/s$

And so, the speed of the stone at half of the maximum height is

$v=\sqrt{v_x^2+v_y^2}=\sqrt{18.9^2+11.6^2}=22.2 m/s$

(c) 17.4% faster

We said that the speed at the top of the trajectory (part a) is

$v_1 = 18.9 m/s$

while the speed at half of the maximum height (part b) is

$v_2 = 22.2 m/s$

So the difference is

$\Delta v = v_2 - v_2 = 22.2 - 18.9 = 3.3 m/s$

And so, in percentage,

$\frac{\Delta v}{v_1} \cdot 100 = \frac{3.3}{18.9}\cdot 100=17.4\%$

So, the stone in part (b) is moving 17.4% faster than in part (a).

4 0

4 years ago