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3 years ago

Simplify this expression 6w+4+8w-1

Mathematics

2 answers:

FrozenT [24]3 years ago

6 0

Answer:

14w + 3

Step-by-step explanation:

6w + 4 + 8w -1 (Given)

14w + 3 (combine like terms)

"The Kid Laroi" - Rapper-Songwriter.

den301095 [7]3 years ago

5 0

Answer:

14w+3

Step-by-step explanation:

6w+4+8w-1

(6w+8w)+(4-1)

14w+3

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Which expressions below are equivalent to

Yuri [45]

Answer:

Step-by-step explanation:

2(2x + 1) = (2)(2x) + (2)(1) = 4x + 2 → A

used distributive property

2(2x + 1) = 2(1 + 2x) → B

used commutative property

2(2x + 1) = (2x + 1) + (2x + 1) = 2x + 1 + 2x + 1 → D

used 2a = a + a

2(2x + 1) = 4x + 2 = x + x + x + x + 1 + 1 → E

used 4x = x + x + x + x and 2 = 1 + 1

6 0

3 years ago

Find the percent of change 1970's: $0.79 2010's: $3.72

tatyana61 [14]

Answer:

471%

Step-by-step explanation:

If you divide 3.72 by .79 you get 4.7088...

Multiply by 100 to get your percentage is 470.88...

Rounded to nearest percentage is 471

hope this helps

7 0

2 years ago

Supposed original quantity is 179 five and the new quantity is 126. Write an expression that represents the percent change. Find

vlabodo [156]

Answer:

This is a -29.6% change

Step-by-step explanation:

To find the percent change, use the percent change equation.

(New - Old)/(Old) * 100 = Percent Change

(126 - 179)/(179) * 100 = Percent Change

-53/179 * 100 = Percent Change

-.296 * 100 = Percent Change

-29.6% = Percent Change

4 0

3 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

I really need help thx

scZoUnD [109]

5) 2*5
   over which equals 10/40 or 1/4
5*8

6)8*3=24
   15*10=150 answer is 24/150

7)2*12=24
9*13=117 answer 8/39

8)7*12=84
   8*1=8 answer 8 and 1/2 or 84/8

9)4*5=20
   12*1=12 answer 1and 2/3 or 20/12

10) 4*4=16
9*9=81 answer 16/81

11)3*7=21
14*9=126 answer 1/6

12)8*24=192
   9*11=99 answer 1 and 31/33 or 192/99

8 0

3 years ago