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2 years ago

Fred buys a video game disk for $8 after a 20% discount. what is the original price?

1 answer:

4 0

I believe this is the answer, im so sorry if im wrong

8 dollars is discount
100-20=80 so
8 dollars =80%
1 dollar=20%
80+20=100
8 + 1 = 9

original price=$9

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goblinko [34]

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3 years ago

Y is directly proportional to x, and y = 45 when x = $600... Find y if x = $1000.

Paul [167]

Answer:

The Answer is: 75.

Step-by-step explanation:

Set up the equation:

45 / 600 = y / 1000

600y = 45 * 1000

600y = 45000

y = 45,000 / 600 = 75.

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3 years ago

Under the Declaration of Independence ordinary citizens have the right to do what and why?

Westkost [7]

Answer:

yes, ordinary netizens have a right to do what they want or what opinion they want to say. Declaration of Indepence is not just for powerful people but also for ordinary netizens in order for them to spek the truth

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2 years ago

(c). It is well known that the rate of flow can be found by measuring the volume of blood that flows past a point in a given tim

aleksklad [387]

(i) Given that

$V(R) = \displaystyle \int_0^R 2\pi K(R^2r-r^3) \, dr$

when R = 0.30 cm and v = (0.30 - 3.33r²) cm/s (which additionally tells us to take K = 1), then

$V(0.30) = \displaystyle \int_0^{0.30} 2\pi \left(0.30-3.33r^2\right)r \, dr \approx \boxed{0.0425}$

and this is a volume so it must be reported with units of cm³.

In Mathematica, you can first define the velocity function with

v[r_] := 0.30 - 3.33r^2

and additionally define the volume function with

V[R_] := Integrate[2 Pi v[r] r, {r, 0, R}]

Then get the desired volume by running V[0.30].

(ii) In full, the volume function is

$\displaystyle \int_0^R 2\pi K(R^2-r^2)r \, dr$

Compute the integral:

$V(R) = \displaystyle \int_0^R 2\pi K(R^2-r^2)r \, dr$

$V(R) = \displaystyle 2\pi K \int_0^R (R^2r-r^3) \, dr$

$V(R) = \displaystyle 2\pi K \left(\frac12 R^2r^2 - \frac14 r^4\right)\bigg_0^R$

$V(R) = \displaystyle 2\pi K \left(\frac{R^4}2- \frac{R^4}4\right)$

$V(R) = \displaystyle \boxed{\frac{\pi KR^4}2}$

In M, redefine the velocity function as

v[r_] := k*(R^2 - r^2)

(you can't use capital K because it's reserved for a built-in function)

Then run

Integrate[2 Pi v[r] r, {r, 0, R}]

This may take a little longer to compute than expected because M tries to generate a result to cover all cases (it doesn't automatically know that R is a real number, for instance). You can make it run faster by including the Assumptions option, as with

Integrate[2 Pi v[r] r, {r, 0, R}, Assumptions -> R > 0]

which ensures that R is positive, and moreover a real number.

5 0

2 years ago

(\tan ^(2)\theta \cos ^(2)\theta -1)/(1+\cos (2\theta ))=

Vitek1552 [10]

(tan²(θ) cos²(θ) - 1) / (1 + cos(2θ))

Recall that

tan(θ) = sin(θ) / cos(θ)

so cos²(θ) cancels with the cos²(θ) in the tan²(θ) term:

(sin²(θ) - 1) / (1 + cos(2θ))

Recall the double angle identity for cosine,

cos(2θ) = 2 cos²(θ) - 1

so the 1 in the denominator also vanishes:

(sin²(θ) - 1) / (2 cos²(θ))

Recall the Pythagorean identity,

cos²(θ) + sin²(θ) = 1

which means

sin²(θ) - 1 = -cos²(θ):

-cos²(θ) / (2 cos²(θ))

Cancel the cos²(θ) terms to end up with

(tan²(θ) cos²(θ) - 1) / (1 + cos(2θ)) = -1/2

7 0

2 years ago