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3 years ago

WHAT IS THE ANSWER ?? PLEASE HELP !!

Mathematics

2 answers:

Jlenok [28]3 years ago

8 0

Answer:

B. 43°

Step-by-step explanation:

Internal angles of the triangle are:

180°-90°=90° and

180°-133°= 47°

So It is a right triangle and the missing angle is:

p= 90°- 47°= 43°

Neko [114]3 years ago

3 0

Answer:

The answer is B) 43°

Step-by-step explanation:

180° - 133° = 47° (that's one of the angles inside the triangle)

180° - 90° = 90° (the other angle)

Every angle inside a triangle added equals 180°, therefor:

180° - (47° + 90°) = 43°

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Is the volume of a triangular prism one-third the volume of a rectangular prism with the same base area and height?

Vlad [161]

Nope

Step-by-step explanation:

Volume of rectangular prism :

= Base Area × h = l × w × h

Volume of triangular prism :

= Base Area × h

Cuz the base area of triangular prism, the volume is :

(½ × l × w) × h

= ½ lwh

So, the volume of a triangular prism is half of a rectangular prism

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3 years ago

The drawing is composed of a rectangle and a semicircle find the area of the figure to the nearest unit.

zimovet [89]

The area equation for a rectangle is length times width. And the area equation for a half (or semi) circle is (1/2) times pi (or 3.14) times the radius squared.

So, you have to add the area of the rectangle and semi-circle together.

Rectangle: A = 18*10 = 180 cm^2
Semi-circle: A = (1/2)*3.14*(18/2)^2 = 127.17 cm^2

Add those together to get 307.17cm^2 or choice C.

6 0

3 years ago

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Find the surface area of the cylinder and round to the nearest tenth. (It is recommended that you use the π button on your calcu

elena-s [515]

The surface area of the cylinder is 18.84 square feet

<h3>How to determine the surface area?</h3>

The given parameters are

Height, h = 2 ft

Diameter, d = 2 ft

The radius (r) is half of the diameter (d)

This is calculated as:

Radius = Diameter/2

So, we have:

r = d/2

Substitute 2 for d

r = 2/2

Evaluate the quotient i.e. divide 2 by 1

r = 1

The surface area is then calculated using the following formula

A = 2πr² + 2πrh

Substitute the given values in the above equation

So, we have:

A = 2 * 3.14 * 1^2 + 2 * 3.14 * 1 * 2

Evaluate the exponents

A = 2 * 3.14 * 1 + 2 * 3.14 * 1 * 2

Evaluate the products

A = 6.28 + 12.56

Evaluate the sum

A = 18.84

Hence, the surface area of the cylinder with the given height and radius is 18.84 square feet

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2 years ago

Write the equation in slope intercept form<br> 2x+y=6

Rina8888 [55]

Answer: the slope intercept form of 2x+y=6 is y=-2+6

Step-by-step explanation:

2x+y=6 (subtract 2x from both sides)

-2x -2x (the -2x and the 2x cancel leaving...)

<u><em>y=-2x+6</em></u> →( y equals negative two x plus six)

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3 years ago

Exercise 3.9.101: Find a particular solution to x 0 = 5x + 4y+ t, y 0 = x + 8y−t, a) using integrating factor method, b) using e

enot [183]

In matrix form, the ODE is given by

$\underbrace{\begin{bmatrix}x'\\y'\end{bmatrix}}_{\vec x'}=\underbrace{\begin{bmatrix}5&4\\1&8\end{bmatrix}}_A\underbrace{\begin{bmatrix}x\\y\end{bmatrix}}_{\vec x}+t\underbrace{\begin{bmatrix}1\\-1\end{bmatrix}}_{\vec f}$

a. Move $A\vec x$ to the left side and multiply both sides by the integrating factor, the matrix exponential of $-A$ , $e^{-At}$ :

$e^{-At}\vec x'-Ae^{-At}\vec x=te^{-At}\vec f$

Condense the left side as the derivative of a product:

$\left(e^{-At}\vec x\right)=te^{-At}\vec f$

Integrate both sides and multipy by $e^{At}$ to solve for $\vec x$ :

$e^{-At}\vec x=\displaystyle\left(\int te^{-At}\,\mathrm dt\right)\vec f\implies\vec x=\displaystyle e^{At}\left(\int te^{-At}\,\mathrm dt\right)\vec f$

Finding $e^{\pm At}$ requires that we diagonalize $A$ .

$A$ has eigenvalues 4 and 9, with corresponding eigenvectors $\begin{bmatrix}-4&1\end{bmatrix}^\top$ and $\begin{bmatrix}1&1\end{bmatrix}^\top$ (explanation for this in part (b)), so we have

$A=\begin{bmatrix}-4&1\\1&1\end{bmatrix}\begin{bmatrix}4&0\\0&9\end{bmatrix}\begin{bmatrix}-4&1\\1&1\end{bmatrix}^{-1}$

$\implies A^n=\begin{bmatrix}-4&1\\1&1\end{bmatrix}\begin{bmatrix}4^n&0\\0&9^n\end{bmatrix}\begin{bmatrix}-4&1\\1&1\end{bmatrix}^{-1}$

$\implies A^n=\dfrac15\begin{bmatrix}4^{n+1}+9^n&4\cdot9^n-4^{n+1}\\9^n-4^n&4^n+4\cdot9^n\end{bmatrix}$

$\implies e^{\pm At}=\dfrac15\begin{bmatrix}4e^{\pm4t}+e^{\pm9t}&4e^{\pm9t}-4e^{\pm4t}\\e^{\pm9t}-e^{\pm4t}&e^{\pm4t}+4e^{\pm9t}\end{bmatrix}$

$\implies\vec x=\dfrac15e^{At}\begin{bmatrix}C_1\\C_2\end{bmatrix}-\dfrac1{216}\begin{bmatrix}72t+20\\-36t-7\end{bmatrix}$

b. Find the eigenvalues of $A$ :

$\det(A-\lambda I_2)=\begin{vmatrix}5-\lambda&4\\1&8-\lambda\end{vmatrix}=\lambda^2-13\lambda+36=0$

$\implies(\lambda-4)(\lambda-9)=0\implies\lambda_1=4,\lambda_2=9$

Let $\vec\eta=\begin{bmatrix}\eta_1&\eta_2\end{bmatrix}^\top$ and $\vec\theta=\begin{bmatrix}\theta_1&\theta_2\end{bmatrix}^\top$ be the corresponding eigenvectors.

For $\lambda_1=4$ , we have

$\begin{bmatrix}1&4\\1&4\end{bmatrix}\begin{bmatrix}\eta_1\\\eta_2\end{bmatrix}=\begin{bmatrix}0\\0\end{bmatrix}$

which means we can pick $\eta_1=-4$ and $\eta_2=1$ .

For $\lambda_2=9$ , we have

$\begin{bmatrix}-4&4\\1&-1\end{bmatrix}\begin{bmatrix}\theta_1\\\theta_2\end{bmatrix}=\begin{bmatrix}0\\0\end{bmatrix}$

so we pick $\theta_1=\theta_2=1$ .

Then the characteristic solution to the system is

$\vec x_c=C_1e^{\lambda_1t}\vec\eta+C_2e^{\lambda_2t}\vec\theta$

$\vec x_c=C_1e^{4t}\begin{bmatrix}-4\\1\end{bmatrix}+C_2e^{9t}\begin{bmatrix}1\\1\end{bmatrix}$

c. Now we find the particular solution with undetermined coefficients.

The nonhomogeneous part of the ODE is a linear function, so we can start with assuming a particular solution of the form

$\vec x_p=\vec at+\vec b\implies\vec x_p'=\vec a$

Substituting these into the system gives

$\begin{bmatrix}a_1\\a_2\end{bmatrix}=\begin{bmatrix}5&4\\1&8\end{bmatrix}\left(\begin{bmatrix}a_1\\a_2\end{bmatrix}t+\begin{bmatrix}b_1\\b_2\end{bmatrix}\right)+\begin{bmatrix}1\\-1\end{bmatrix}t$

$\begin{bmatrix}a_1\\a_2\end{bmatrix}=\begin{bmatrix}5&4\\1&8\end{bmatrix}\begin{bmatrix}a_1t+b_1\\a_2t+b_2\end{bmatrix}+\begin{bmatrix}t\\-t\end{bmatrix}$

$\begin{bmatrix}a_1\\a_2\end{bmatrix}=\begin{bmatrix}(5a_1+4a_2+1)t+(5b_1+4b_2)\\(a_1+8a_2-1)t+(b_1+8b_2)\end{bmatrix}$

$\implies\begin{cases}5a_1+4a_2=-1\\5b_1+4b_2=a_1\\a_1+8a_2=1\\b_1+8b_2=a_2\end{cases}\implies a_1=-\dfrac13,a_2=\dfrac16,b_1=-\dfrac5{54},b_2=\dfrac7{216}$

Put everything together to get a solution

$\vec x=\vec x_c+\vec x_p$

that should match the solution in part (a).

8 0

4 years ago