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3 years ago

8

What is the volume of this triangular prism

2 answers:

OleMash [197]3 years ago

6 0

Answer:

18.1 x 22.4 x 28/2 =5676.16

Sav [38]3 years ago

3 0

The answer to this question is C)

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Marsha used 6 cups of flour to make 4 trays of

PolarNik [594]

Answer:

9

Step-by-step explanation:

4/6 = 1.5, 1.5(6)= 9

6 0

2 years ago

Solve the given initial-value problem. the de is of the form dy dx = f(ax + by + c), which is given in (5) of section 2.5. dy dx

shutvik [7]

$\dfrac{\mathrm dy}{\mathrm dx}=\cos(x+y)$

Let $v=x+y$ , so that $\dfrac{\mathrm dv}{\mathrm dx}-1=\dfrac{\mathrm dy}{\mathrm dx}$ :

$\dfrac{\mathrm dv}{\mathrm dx}=\cos v+1$

Now the ODE is separable, and we have

$\dfrac{\mathrm dv}{1+\cos v}=\mathrm dx$

Integrating both sides gives

$\displaystyle\int\frac{\mathrm dv}{1+\cos v}=\int\mathrm dx$

For the integral on the left, rewrite the integrand as

$\dfrac1{1+\cos v}\cdot\dfrac{1-\cos v}{1-\cos v}=\dfrac{1-\cos v}{1-\cos^2v}=\csc^2v-\csc v\cot v$

Then

$\displaystyle\int\frac{\mathrm dv}{1+\cos v}=-\cot v+\csc v+C$

and so

$\csc v-\cot v=x+C$

$\csc(x+y)-\cot(x+y)=x+C$

Given that $y(0)=\dfrac\pi2$ , we find

$\csc\left(0+\dfrac\pi2\right)-\cot\left(0+\dfrac\pi2\right)=0+C\implies C=1$

so that the particular solution to this IVP is

$\csc(x+y)-\cot(x+y)=x+1$

5 0

2 years ago

Do the following angles form a triangle

Lemur [1.5K]

Answer:

We need a picture or link

Step-by-step explanation:

3 0

2 years ago

What is a factor of 15 but not a multiple of 3

11111nata11111 [884]

45
15 is a multiple of 3
30 is a multiple of 3
45 is NOT a multiple of 3
60 is a multiple of 3
Ect.

7 0

3 years ago

PLEASE SHOW WORK

CaHeK987 [17]

(C)

Step-by-step explanation:

The volume of the conical pile is given by

$V = \dfrac{\pi}{3}r^2h$

Taking the derivative of V with respect to time, we get

$\dfrac{dV}{dt} = \dfrac{\pi}{3}\dfrac{d}{dt}(r^2h)$

$\:\:\:\:\:\:\:= \dfrac{\pi}{3}\left(2rh\dfrac{dr}{dt} + r^2\dfrac{dh}{dt}\right)$

Since r is always equal to h, we can set

$\dfrac{dr}{dt} = \dfrac{dh}{dt}$

so that our expression for dV/dt becomes

$\dfrac{dV}{dt} = \dfrac{\pi}{3}\left(3r^2\dfrac{dh}{dt}\right)$

$\:\:\:\:\:\:\:= \pi r^2\dfrac{dh}{dt}$

Solving for dh/dt, we get

$\dfrac{dh}{dt} = \dfrac{1}{\pi r^2}\dfrac{dV}{dt}$

$\:\:\:\:\:\:\:= \dfrac{1}{9\pi\:\text{m}^2}(36\:\text{m}^3\text{/s})$

$\:\:\:\:\:\:\:= \dfrac{4}{\pi}\:\text{m/s}$

7 0

3 years ago