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1 year ago

If 5 gallons cost $ 6.60 how much do 29 gallons of gasoline cost

Mathematics

1 answer:

ikadub [295]1 year ago

5 0

Answer:

Should be 38.28

Step-by-step explanation:

Dividing 6.60 by 5 is $1.32, aka the cost of one gallon. multiply by 29, answer is that it costs $38.28 to get 29 gallons of gasoline.

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What is the exact value of tan 30° ? Enter your answer, as a simplified fraction, in the box.

Romashka [77]

Hmmm if you don't have a Unit Circle, this is a good time to get one, many you can find online. Anyhow, check your unit circle for cos(30°) and sin(30°).

$\bf tan(\theta)=\cfrac{sin(\theta)}{cos(\theta)}\qquad \qquad tan(30^o)=\cfrac{sin(30^o)}{cos(30^o)}
\\\\\\
tan(30^o)=\cfrac{\quad \frac{1}{2}\quad }{\frac{\sqrt{3}}{2}}\implies tan(30^o)=\cfrac{1}{2}\cdot \cfrac{2}{\sqrt{3}}\implies tan(30^o)=\cfrac{1}{\sqrt{3}}
\\\\\\
\stackrel{\textit{and now if we rationalize the denominator}}{\cfrac{1}{\sqrt{3}}\cdot \cfrac{\sqrt{3}}{\sqrt{3}}\implies \cfrac{\sqrt{3}}{(\sqrt{3})^2}\implies \cfrac{\sqrt{3}}{3}}$

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2 years ago

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Find the area of the shaded regions below. Give your answer as a completely simplified exact value in terms of π (no approximati

Lera25 [3.4K]

Answer:

(36π -72) cm²

Step-by-step explanation:

The area of a segment that subtends arc α (in radians) is given by ...

A = (1/2)r²·(α - sin(α))

Here, you have r = 12 cm and α = π/2 radians, so the area of the segment is ...

A = (1/2)(12 cm)²·(π/2 -1) = (36π -72) cm²

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2 years ago

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A square has an area of 36 m2. The square is scaled in such a way that its area is increased by 28 m2. What is the approximate f

Roman55 [17]

Answer:

28/36 shd give u the answer

answer in 1:ur answer

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2 years ago

What are the next two numbers 0.06, 0.12, 0.18

Alex Ar [27]

The next two numbers are 0.24 0.36 Your just adding 0.06.
Hope this helped you!

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2 years ago

Use Stokes' Theorem to evaluate C F · dr where C is oriented counterclockwise as viewed from above. F(x, y, z) = xyi + 5zj + 7yk

REY [17]

By Stokes' theorem,

$\displaystyle\int_C\vec F\cdot\mathrm d\vec r=\iint_S(\nabla\times\vec F)\cdot\mathrm d\vec S$

where $S$ is the surface with $C$ as its boundary. The curl is

$\nabla\times\vec F(x,y,z)=2\,\vec\imath-x\,\vec k$

Parameterize $S$ by

$\vec\sigma(u,v)=u\cos v\,\vec\imath+u\sin v\,\vec\jmath+(8-u\cos v)\,\vec k$

with $0\le u\le9$ and $0\le v\le2\pi$ . Then take the normal vector to $S$ to be

$\vec\sigma_u\times\vec\sigma_v=u\,\vec\imath+u\,\vec k$

Then the line integral is equal to the surface integral,

$\displaystyle\iint_S(\nabla\times\vec F)\cdot\mathrm d\vec S=\int_0^{2\pi}\int_0^9(2\,\vec\imath-u\cos v\,\vec k)\cdot(u\,\vec\imath+u\,\vec k)\,\mathrm du\,\mathrm dv$

$\displaystyle=\int_0^{2\pi}\int_0^9(2u-u^2\cos v)\,\mathrm du\,\mathrm dv=\boxed{162\pi}$

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2 years ago

If 5 gallons cost $ 6.60 how much do 29 gallons of gasoline cost​

If 5 gallons cost $ 6.60 how much do 29 gallons of gasoline cost