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Margaret [11]
3 years ago
8

15

Mathematics
1 answer:
IceJOKER [234]3 years ago
4 0

4x+3x=7x

18 girls-5%

Total no: of girlsx5/100=18

Total-100x18/5-360 girls

4x+3607x

360-7x-4x

360-3x

x=360/3-120

4x=4x120=480

3x=3x120=360

Total no: of students=480+360-840

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Follow the steps above and find c, the total of the payments, and the monthly payment. Choose the right answers.
saveliy_v [14]

To the nearest penny, c = $6,252.50

Total of payments = amount financed + c = $26,752.50

Total of payments ÷ number of payments = monthly payment = $445.88

5 0
2 years ago
A group of people signed up for a bus tour that costs a total of $1120. If they can increase the size of their group by 7 people
mr_godi [17]

Answer: 160

Step-by-step explanation:

7 0
3 years ago
The zoom feature on a camera lens allows you dilate what appears on the display when you change from 100% to 100%, the new image
mel-nik [20]

Answer:

When you change from 100% to 200%, the new image on your screen is an enlargement of the previous image with a scale factor of 2.

6 0
3 years ago
Help calculus module 8 DBQ<br><br> please show work
igor_vitrenko [27]

1. The four subintervals are [0, 2], [2, 3], [3, 7], and [7, 8]. We construct trapezoids with "heights" equal to the lengths of each subinterval - 2, 1, 4, and 1, respectively - and the average of the corresponding "bases" equal to the average of the values of R(t) at the endpoints of each subinterval. The sum is then

\dfrac{R(0)+R(2)}2(2-0)+\dfrac{R(2)+R(3)}2(3-2)+\dfrac{R(3)+R(7)}2(7-3)+\dfrac{R(7)+R(8)}2(7-8)=\boxed{24.83}

which is measured in units of gallons, hence representing the amount of water that flows into the tank.

2. Since R is differentiable, the mean value theorem holds on any subinterval of its domain. Then for any interval [a,b], it guarantees the existence of some c\in(a,b) such that

\dfrac{R(b)-R(a)}{b-a)=R'(c)

Computing the difference quotient over each subinterval above gives values of 0.275, 0.3, 0.3, and 0.26. But just because these values are non-zero doesn't guarantee that there is definitely no such c for which R'(c)=0. I would chalk this up to not having enough information.

3. R(t) gives the rate of water flow, and R(t)\approx W(t), so that the average rate of water flow over [0, 8] is the average value of W(t), given by the integral

R_{\rm avg}=\displaystyle\frac1{8-0}\int_0^8\ln(t^2+7)\,\mathrm dt

If doing this by hand, you can integrate by parts, setting

u=\ln(t^2+7)\implies\mathrm du=\dfrac{2t}{t^2+7}\,\mathrm dt

\mathrm dv=\mathrm dt\implies v=t

R_{\rm avg}=\displaystyle\frac18\left(t\ln(t^2+7)\bigg|_{t=0}^{t=8}-\int_0^8\frac{2t^2}{t^2+7}\,\mathrm dt\right)

For the remaining integral, consider the trigonometric substitution t=\sqrt 7\tan s, so that \mathrm dt=\sqrt 7\sec^2s\,\mathrm ds. Then

R_{\rm avg}=\displaystyle\ln71-\frac{\sqrt7}4\int_0^{\tan^{-1}(8/\sqrt7)}\frac{7\tan^2s}{7\tan^2s+7}\sec^2s\,\mathrm ds

R_{\rm avg}=\displaystyle\ln71-\frac{\sqrt7}4\int_0^{\tan^{-1}(8/\sqrt7)}\tan^2s\,\mathrm ds

R_{\rm avg}=\displaystyle\ln71-\frac{\sqrt7}4\int_0^{\tan^{-1}(8/\sqrt7)}(\sec^2s-1)\,\mathrm ds

R_{\rm avg}=\displaystyle\ln71-\frac{\sqrt7}4\left(\tan s-s\right)\bigg|_{s=0}^{s=\tan^{-1}(8/\sqrt7)}

R_{\rm avg}=\displaystyle\ln71-\frac{\sqrt7}4\left(\tan\left(\tan^{-1}\frac8{\sqrt7}\right)-\tan^{-1}\frac8{\sqrt7}\right)

\boxed{R_{\rm avg}=\displaystyle\ln71-2+\frac{\sqrt7}4\tan^{-1}\frac8{\sqrt7}}

or approximately 3.0904, measured in gallons per hour (because this is the average value of R).

4. By the fundamental theorem of calculus,

g'(x)=f(x)

and g(x) is increasing whenever g'(x)=f(x)>0. This happens over the interval (-2, 3), since f(x)=3 on [-2, 0), and -x+3>0 on [0, 3).

5. First, by additivity of the definite integral,

\displaystyle\int_{-2}^xf(t)\,\mathrm dt=\int_{-2}^0f(t)\,\mathrm dt+\int_0^xf(t)\,\mathrm dt

Over the interval [-2, 0), we have f(x)=3, and over the interval [0, 6], f(x)=-x+3. So the integral above is

\displaystyle\int_{-2}^03\,\mathrm dt+\int_0^x(-t+3)\,\mathrm dt=3t\bigg|_{t=-2}^{t=0}+\left(-\dfrac{t^2}2+3t\right)\bigg|_{t=0}^{t=x}=\boxed{6+3x-\dfrac{x^2}2}

6 0
3 years ago
A sports team is selling circular patches for their fans. Each patch has a
atroni [7]

Answer: 63.59cm²

Step-by-step explanation:

The area covered by the pitch will be gotten by finding the value of the area of the circle and this will be:

Area of q circle = πr²

where,

π = 3.14

r = radius = diameter/2 = 9cm/2 = 4.5cm

Area = πr² = 3.14 × 4.5²

= 3.14 × 20.25

= 63.59cm²

Therefore, the area covered by the pitch is 63.59cm²

4 0
3 years ago
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