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Bad White [126]

3 years ago

13

Rodney is helping to make hamburgers for the football game. He's made 10 hamburgers so far. His coach asked him to make at least

30 hamburgers but no more than 80. Solve the inequality and interpret the solution.

1 answer:

anastassius [24]3 years ago

3 0

Answer:

40 I had this question before

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Find the area of a square if the perimeter is equal to 38 in.

Misha Larkins [42]

Hello! I would love to help!

Alright! We know that perimeter is basically adding all the sides of a shape together. We also know that a square has 4 sides, so 38 is the result of 4 sides being added. Now, we also know that all sides of a square are even. So, to find one side of a square, we just need to do 38/4

38/4=9.5

Alright. So now we know one side of a square is 9.5 in. We know that in order to find area, we have to multiply length and width. We know that since all sides of a square are even, 9.5 is both our length and width. So, in order to find the area, we just have to multiply 9.5 by 9.5

9.5*9.5=90.25

Alright! So we have determined your answer!!!

90.25 or in fraction form 90 1/4 inches squared!

Hope this helped! Comment if you have questions!!

8 0

3 years ago

The volume of the cube below is 1/8 cm2.

Katen [24]

9514 1404 393

Answer:

C) x = 1/2 cm

Step-by-step explanation:

Apparently, we're to assume the units of x are centimeters. The volume of the cube is the product of its dimensions:

V = x³ = 1/8

Then the value of x is ...

x = ∛(1/8) = 1/∛8

x = 1/2

_____

<em>Additional comment</em>

It can be useful to memorize the squares and cubes of small integers. The cubes you see most often are ...

1³ = 1; 2³ = 8; 3³ = 27; 4³ = 64; 5³ = 125; 6³ = 216; 7³ = 343

7 0

3 years ago

Can someone help me out on this math problem?

denis-greek [22]

1 ,3, 1

The first one is 1 the second is 3 and the last one is 1

8 0

2 years ago

A tank contains 180 gallons of water and 15 oz of salt. water containing a salt concentration of 17(1+15sint) oz/gal flows into

Stels [109]

Let A(t) denote the amount of salt (in ounces, oz) in the tank at time t (in minutes, min).

Salt flows in at a rate of

$\dfrac{dA}{dt}_{\rm in} = \left(17 (1 + 15 \sin(t)) \dfrac{\rm oz}{\rm gal}\right) \left(8\dfrac{\rm gal}{\rm min}\right) = 136 (1 + 15 \sin(t)) \dfrac{\rm oz}{\min}$

and flows out at a rate of

$\dfrac{dA}{dt}_{\rm out} = \left(\dfrac{A(t) \, \mathrm{oz}}{180 \,\mathrm{gal} + \left(8\frac{\rm gal}{\rm min} - 8\frac{\rm gal}{\rm min}\right) (t \, \mathrm{min})}\right) \left(8 \dfrac{\rm gal}{\rm min}\right) = \dfrac{A(t)}{180} \dfrac{\rm oz}{\rm min}$

so that the net rate of change in the amount of salt in the tank is given by the linear differential equation

$\dfrac{dA}{dt} = \dfrac{dA}{dt}_{\rm in} - \dfrac{dA}{dt}_{\rm out} \iff \dfrac{dA}{dt} + \dfrac{A(t)}{180} = 136 (1 + 15 \sin(t))$

Multiply both sides by the integrating factor, $e^{t/180}$ , and rewrite the left side as the derivative of a product.

$e^{t/180} \dfrac{dA}{dt} + e^{t/180} \dfrac{A(t)}{180} = 136 e^{t/180} (1 + 15 \sin(t))$

$\dfrac d{dt}\left[e^{t/180} A(t)\right] = 136 e^{t/180} (1 + 15 \sin(t))$

Integrate both sides with respect to t (integrate the right side by parts):

$\displaystyle \int \frac d{dt}\left[e^{t/180} A(t)\right] \, dt = 136 \int e^{t/180} (1 + 15 \sin(t)) \, dt$

$\displaystyle e^{t/180} A(t) = \left(24,480 - \frac{66,096,000}{32,401} \cos(t) + \frac{367,200}{32,401} \sin(t)\right) e^{t/180} + C$

Solve for A(t) :

$\displaystyle A(t) = 24,480 - \frac{66,096,000}{32,401} \cos(t) + \frac{367,200}{32,401} \sin(t) + C e^{-t/180}$

The tank starts with A(0) = 15 oz of salt; use this to solve for the constant C.

$\displaystyle 15 = 24,480 - \frac{66,096,000}{32,401} + C \implies C = -\dfrac{726,594,465}{32,401}$

So,

$\displaystyle A(t) = 24,480 - \frac{66,096,000}{32,401} \cos(t) + \frac{367,200}{32,401} \sin(t) - \frac{726,594,465}{32,401} e^{-t/180}$

Recall the angle-sum identity for cosine:

$R \cos(x-\theta) = R \cos(\theta) \cos(x) + R \sin(\theta) \sin(x)$

so that we can condense the trigonometric terms in A(t). Solve for R and θ :

$R \cos(\theta) = -\dfrac{66,096,000}{32,401}$

$R \sin(\theta) = \dfrac{367,200}{32,401}$

Recall the Pythagorean identity and definition of tangent,

$\cos^2(x) + \sin^2(x) = 1$

$\tan(x) = \dfrac{\sin(x)}{\cos(x)}$

Then

$R^2 \cos^2(\theta) + R^2 \sin^2(\theta) = R^2 = \dfrac{134,835,840,000}{32,401} \implies R = \dfrac{367,200}{\sqrt{32,401}}$

and

$\dfrac{R \sin(\theta)}{R \cos(\theta)} = \tan(\theta) = -\dfrac{367,200}{66,096,000} = -\dfrac1{180} \\\\ \implies \theta = -\tan^{-1}\left(\dfrac1{180}\right) = -\cot^{-1}(180)$

so we can rewrite A(t) as

$\displaystyle A(t) = 24,480 + \frac{367,200}{\sqrt{32,401}} \cos\left(t + \cot^{-1}(180)\right) - \frac{726,594,465}{32,401} e^{-t/180}$

As t goes to infinity, the exponential term will converge to zero. Meanwhile the cosine term will oscillate between -1 and 1, so that A(t) will oscillate about the constant level of 24,480 oz between the extreme values of

$24,480 - \dfrac{267,200}{\sqrt{32,401}} \approx 22,995.6 \,\mathrm{oz}$

and

$24,480 + \dfrac{267,200}{\sqrt{32,401}} \approx 25,964.4 \,\mathrm{oz}$

which is to say, with amplitude

$2 \times \dfrac{267,200}{\sqrt{32,401}} \approx \mathbf{2,968.84 \,oz}$

6 0

2 years ago

BC is perpendicular to AC

FromTheMoon [43]

Answer:

In ∆ABC, AD is perpendicular while BC and BE are perpendicular to AC.

Step-by-step explanation:

5 0

3 years ago