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3 years ago

What is the value of 5 in 43.245

1 answer:

6 0

4➡ tens
3 ➡ ones
----------------
2➡tenths
4➡hundredths
5➡ thousandths

Anything behind the decimal point ends with a "ths" ("th" )

So, the value of "5" might be 'thousandths'

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Suppose that you have a goal to drive two miles at an average rate of 60 miles per hour (mph). during the first mile you travel

sveta [45]

Your goal requires you complete the distance in 2 minutes. If you travel at half the target speed, you will have used up the allowed time traveling the first half of the distance.

You need to travel the second mile at infinite speed to average 60 mph.

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3 years ago

Identity the value for a, b, and c. <br>2x^2=-7x-9

castortr0y [4]

Is there more information to this? Or do you just want x solved

6 0

3 years ago

PLEASE HELP QUICK A prism has 2 congruent hexagonal bases like the one shown. Each hexagon is made from 2 congruent isosceles tr

Mademuasel [1]

<h3>Answer: C) 6 units</h3>

==========================================================

Explanation:

Let's find the area of the hexagon. It's composed of two identical (aka congruent) trapezoids.

Each trapezoid has two parallel bases of 4+4 = 8 and 5 units. The height is 3. The area of one trapezoid is

area = height(base1+base2)/2

area = 3*(8+5)/2

area = 19.5

which doubles to 2*19.5 = 39 to represent the area of the entire hexagon

--------------------------------

The volume of any prism is found through this formula

volume = (area of base)*(height of prism)

We just found the area of the base to be 39. The height is unknown, so we'll call it h. The volume is given to be 234.

We end up with this equation

234 = 39h

which solves to h = 6 after dividing both sides by 39. This prism has a height of 6 units.

5 0

3 years ago

Read 2 more answers

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

The pitchers that Isaac bought for the party each hold 2 quarts of liquid. If he makes 15 pints of smoothie and has three pitche

saveliy_v [14]

1 pint = 0.5 quart
15 pints = 15 x 0.5 = 7.5 quarts

3 pitchers can hold 2 x 3 = 6 quarts.

Amount remaining = 7.5 - 6 = 1.5 quarts = 1.5/0.5 = 3 pints of smoothie

7 0

3 years ago

Read 3 more answers