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2 years ago

6

twelve and three tenths more than five and thirteen thousandths of a number d is equal to fifteen and three hundred two thousand

ths. what is the value of d

1 answer:

AVprozaik [17]2 years ago

6 0

5.013d + 12.3 = 15.302
5.013d = 15.302 - 12.3
5.013d = 3.002
d = 3.002 / 5.013
d = 0.599 (thats rounded) <==

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camilla paid $1.15 per pound for x pounds of tomatoes. She paid for her tomatoes with $10 and received $4.48 in change. How many

Alika [10]

Answer:

$x=4.8\ pounds$

Step-by-step explanation:

Let

x-----> the number of pounds of tomatoes Camilla bought

we know that

The equation that represent this situation is equal to

$1.15x=10-4.48$

Solve for x

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3 years ago

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hichkok12 [17]

Answer:

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Step-by-step explanation:

0=(x+2)^2-25

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1 year ago

Read 2 more answers

Family has an annual income of $23 comma 400. Of this, one fourth

denis23 [38]

Amount spent for food = $5850, housing = $4680, clothing = $2340, savings = $2600, taxes = $5850, expenses = $2080.

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

Step 1:

Given details are Total Income = $23,400

Step 2:

Amount spent for food, F= 1/4th of 23,400 = $1/4 \times 23400$

= $5850

Step 3:

Amount spent for housing, H = 1/5th of 23,400 = $1/5 \times23400$

= $4680

Step 4:

Amount spent for clothing, C = 1/10th of 23,400 = $1/10 \times 23400$ = $2340

Step 5:

Amount spent for savings, S = 1/9th of 13,400 = $1/9\times 23400$ = $2600

Step 6:

Amount spent for taxes, T = 1/4th of 23,400 = $1/4 \times 23400$ =$5850

Step 7:

Find amount spent on other expenses by totaling all expenditures and subtracting from the annual income.

Annual Income = F + H + C + S + T + E

Amount spent on other expenses, E = 23400 - (5850 + 4680 + 2340 + 2600 + 5850)

E = 23400 - 21320

E = $2080

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2 years ago

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natima [27]

Answer:

b hopefully this helps you with work

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2 years ago

The graphs of the polar curves r = 4 and r = 3 + 2cosθ are shown in the figure above. The curves intersect at θ = π/3 and θ = 5π

Gennadij [26K]

(a)

$\displaystyle \frac{1}{2} \cdot \int_{\frac{\pi}{3}}^{\frac{5\pi}{3}} \left(4^2 - (3 + 2\cos\theta)^2 \right) \, d\theta$

or, via symmetry

$\displaystyle\frac{1}{2} \cdot 2 \int_{\frac{\pi}{3}}^{\pi} \left(4^2 - (3 + 2\cos\theta)^2 \right) \, d\theta$

____________

(b)

By the chain rule:

$\displaystyle \frac{dy}{dx} = \frac{ dy/ d\theta}{ dx/ d\theta}$

For polar coordinates, x = rcosθ and y = rsinθ. Since
<span>r = 3 + 2cosθ, it follows that

$x = (3 + 2\cos\theta) \cos \theta \\ 
y = (3 + 2\cos\theta) \sin \theta$

Differentiating with respect to theta

$\begin{aligned}
\displaystyle \frac{dy}{dx} &= \frac{ dy/ d\theta}{ dx/ d\theta} \\
&= \frac{(3 + 2\cos\theta)(\cos\theta) + (-2\sin\theta)(\sin\theta)}{(3 + 2\cos\theta)(-\sin\theta) + (-2\sin\theta)(\cos\theta)} \\ \\
\left.\frac{dy}{dx}\right_{\theta = \frac{\pi}{2}}
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\end{aligned}$

2/3 is the slope

____________

(c)

"</span><span>distance between the particle and the origin increases at a constant rate of 3 units per second" implies dr/dt = 3

A</span>ngle θ and r are related via <span>r = 3 + 2cosθ, so implicitly differentiating with respect to time

</span><span /> $\displaystyle\frac{dr}{dt} = -2\sin\theta \frac{d\theta}{dt} \quad \stackrel{\theta = \pi/3}{\implies} \quad 3 = -2\left( \frac{\sqrt{3}}{2}}\right) \frac{d\theta}{dt} \implies \\ \\ \frac{d\theta}{dt} = -\sqrt{3} \text{ radians per second}$

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3 years ago