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2 years ago

6

please help now it kinda important i get it done soon or i cant go to a dance for my school tomorrow. im begging u ill give u 5

stars and a thanks and maby even a showt out thats how much this means to me.

1 answer:

Tasya [4]2 years ago

8 0

Four apples.

Fifty percent - or half - of eight is four

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Rhonda had a piece of ribbon that was 5 feet 3 inches long. She removed a 5-inch piece to use in her art project. What is the le

nikitadnepr [17]

Answer:

4 feet 10 inches

5 0

4 years ago

Read 2 more answers

What’s the missing side round to the nearest tenth?

Step2247 [10]

Sine of angle = opposite ÷ hypotenuse

sin 20 = $\frac{14}{x}$

x = $\frac{14}{sin 20}$

x = 40.9 (rounded up to nearest tenth)

4 0

3 years ago

What is the length of side AB? Round your answer to the nearest tenth.

ELEN [110]

Answer:

AB ≈ 2.8 units

Step-by-step explanation:

Calculate the length using the distance formula

d = √ (x₂ - x₁ )² + (y₂ - y₁ )²

with (x₁, y₁ ) = A(1, 1) and (x₂, y₂ ) = B(3, 3)

AB = $\sqrt{(3-1)^2+(3-1)^2}$

= $\sqrt{2^2+2^2}$

= $\sqrt{4+4}$

= $\sqrt{8}$ ≈ 2.8 units ( to the nearest tenth )

3 0

3 years ago

How many x-intercepts would the function shown below have?<br> f(x) = -5x (2x + 1)? (x² – 6)(x² + 9)

BabaBlast [244]

Step-by-step explanation:

Here we have

f

(

x

)

=

2

x

2

(

x

2

−

9

)

, which can be factorized as

f

(

x

)

=

2

x

2

(

x

+

3

)

(

x

−

3

)

As there is no common factor between numerator and denominator, there s no hole.

Further vertical asymptotes are

x

=

−

3

and

x

=

3

and as

f

(

x

)

=

2

x

2

(

x

2

−

9

)

=

2

1

−

9

x

2

, as

x

→

∞

,

f

(

x

)

→

2

, hence horizontal asymptote is

y

=

2

.

Observe that

f

(

−

x

)

=

f

(

x

)

and hence graph is symmetric w.r.t.

y

-axis. Further as

x

=

0

,

f

(

x

)

=

0

. Using calculas we can find that at

(

0

,

0

)

there is a local maxima as

d

y

d

x

=

−

36

x

(

x

2

−

9

)

2

and at

x

=

0

it is

0

. Further while for

x

<

−

3

and

x

>

3

, function is positive, for

−

3

<

x

<

3

function is negative.

Now take a few values of

x

say

{

−

10

,

−

7

,

−

4

,

−

2

,

−

1

,

1

,

2

,

4

,

7

,

10

}

and corresponding values of

f

(

x

)

are

{

2

18

91

,

2

9

20

,

4

4

7

,

−

1

3

5

,

−

1

4

,

−

1

4

,

−

1

3

5

,

4

4

7

,

2

9

20

,

2

18

91

}

5 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