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3 years ago

Round to the nearest whole number. 6.4

Mathematics

1 answer:

lara31 [8.8K]3 years ago

4 0

Answer:

Step-by-step explanation:

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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

MARK YOU BRAINLIST It takes Pervis 20 minutes to ride his bicycle 1.5 miles to school each day. How far can Pervis ride in one h

exis [7]

Answer:

4.5 miles

Step-by-step explanation:

so basically 1 hour equals 60 minutes so we divide 60 by 20 because he can ride 1.5 miles in 20 minutes. so 60/20 is 3 and now all we do is multiply 1.5 by 3 which gives us 4.5.

4 0

3 years ago

A suit case measures 24 inches long and the diagonal is 26 inches long. How much material is needed to cover one side of the sui

ASHA 777 [7]

It helps to draw diagrams when working through problems like this.

Given a rectangular prism (the suitcase), drawing a diagonal through its side will give you two right triangles. You can use the Pythagorean Theorem (a² + b² = c², where c is the hypotenuse of a right triangle) to find the length of the unknown side. The hypotenuse of the triangle is the diagonal since it is the longest side, and it is opposite the right angle.

a² + b² = c²

a² + 24² = 26²

a² + 576 = 676

a² = 100

a = 10 in

Remember that this is only the side length of the triangle. To find the amount of material needed to cover one side of the suitcase, we need its area. We can find the area using the formula for area of a rectangle: A = bh.

A = bh

A = 10*24

A = 240 in²

<h3>Answer:</h3>

240 in²

8 0

3 years ago

NEED HELP!!!<br> see picture*****

Lana71 [14]

Answer:

1. -10

2. $x=1\pm\sqrt{2}i$

3. -1+2i

4. -3-7i

5. 13

6. rectangular coordinates are (-4.3,-2.5)

7. rectangular coordinates are (-2.5,4.3)

8. x^2 + y^2 = 8y

9. Polar coordinates of point (-3,0) are (3,180°)

10. Polar coordinates of point (1,1) are (√2,45°)

Step-by-step explanation:

1) Simplify (2+3i)^2 + (2-3i)^2

Using formula (a+b)^2 = a^2+2ab+b^2

=((2)^2+2(2)(3i)+(3i)^2)+((2)^2-2(2)(3i)+(3i)^2)

=(4+12i+9i^2)+(4-12i+9i^2)

We know that i^2=-1

=(4+12i+9(-1))+(4-12i+9(-1))

=(4+12i-9)+(4-12i-9)

=(-5+12i)+(-5-12i)

=5+12i-5-12i

=-10

2. Solve x^2-2x+3 = 0

Using quadratic formula to find value of x

a=1, b=-2 and c=3

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\x=\frac{-(-2)\pm\sqrt{(-2)^2-4(1)(3)}}{2(1)}\\x=\frac{2\pm\sqrt{4-12}}{2}\\x=\frac{2\pm\sqrt{-8}}{2}\\x=\frac{2\pm2\sqrt{-2}}{2}\\x=\frac{2\pm2\sqrt{2}i}{2}\\x=2(\frac{1\pm\sqrt{2}i}{2})\\x=1\pm\sqrt{2}i$

3. If u =1+3i and v =-2-i what is u+v

u+v = (1+3i)+(-2-i)

u+v = 1+3i-2-i

u+v = 1-2+3i-i

u+v = -1+2i

4. if u = 3-4i and v = 3i+6 what is u-v

u-v = (3-4i)-(3i+6)

u-v = 3-4i-3i-6

u-v = 3-6-4i-3i

u-v = -3-7i

5. if u=(3+2i) and v=(3-2i) what is uv?

uv = (3+2i)(3-2i)

uv = 3(3-2i)+2i(3-2i)

uv = 9-6i+6i-4i^2

uv = 9-4i^2

i^2=-1

uv = 9-4(-1)

uv = 9+4

uv = 13

6. Convert (5, 7π/6) to rectangular form

To convert polar coordinate into rectangular coordinate we use formula:

x = r cos Ф

y = r sin Ф

r = 5, Ф= 7π/6

x = r cos Ф

x = 5 cos (7π/6)

x = -4.3

y = r sin Ф

y = 5 sin (7π/6)

y = -2.5

So rectangular coordinates are (-4.3,-2.5)

7. Convert (5, 2π/3) to rectangular form

To convert polar coordinate into rectangular coordinate we use formula:

x = r cos Ф

y = r sin Ф

r = 5, Ф= 2π/3

x = r cos Ф

x = 5 cos (2π/3)

x = -2.5

y = r sin Ф

y = 5 sin (2π/3)

y = 4.33

So rectangular coordinates are (-2.5,4.33)

8. Convert r=8cosФ to rectangular form

r.r = (8 cos Ф)r

r^2 = 8 (cosФ)(r)

Let (cosФ)(r) = y and we know that r^2 = x^2+y^2

x^2 + y^2 = 8y

9. Convert(-3,0) to polar form

We need to find (r,Ф)

r = √x^2+y^2

r = √(-3)^2+(0)^2

r =√9

r = 3

and tan Ф = y/x

tan Ф = 0/-3

tan Ф = 0

Ф = tan^-1(0)

Ф = 0°

As Coordinates are in 2nd quadrant, so add 180° in the given angle

0+180 = 180°

So,Polar coordinates of point (-3,0) are (3,180°)

10) Convert (1,1) to polar form

We need to find (r,Ф)

r = √x^2+y^2

r = √(1)^2+(1)^2

r =√2

and tan Ф = y/x

tan Ф = 1/1

tan Ф = 1

Ф = tan^-1(1)

Ф = 45°

As Coordinates are in 1st quadrant, so Ф will be as found

So,Polar coordinates of point (1,1) are (√2,45°)

5 0

3 years ago

Margie calculated that she would spend $175 on school supplies this year. She actually spent $97.50 on school supplies. What is

Mandarinka [93]

0.679 because go on a calculator and put money to percent

4 0

4 years ago

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