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2 years ago

Drag the missing statements and reasons to the correct spot to complete the proof

Mathematics

1 answer:

VARVARA [1.3K]2 years ago

4 0

Answer:

1. Segment MN is congruent to segment OP (Given)

2. Angle MNP is congruent to angle OPN (Given)

3. Segment NP is congruent to segment NP (Reflexive prop.)

4. Triangle MPN is congruent to triangle ONP (SAS)

5. Segment MP is congruent to segment NO (CPCTC)

Hope this helps! :)

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First, find the volume for each bottle. Use 3.14 for pie or the pie button and round to the nearest hundredth. Volume of cylinde

marshall27 [118]

the volume of the cylinder is 572.27 cm³

Explanation:

Given:

radius of the cylinder = 4.5cm

height = 9cm

To get the volume of the cylinder, we will use the formula:

$\begin{gathered} Volume\text{ = \pi r^^b2h} \\ where\text{ \pi= 3.14} \\ r\text{ = radius, h = height} \end{gathered}$ $\begin{gathered} Volume\text{ of the cylinder = 3.14 }\times\text{ 4.5}^2\text{ }\times\text{ 9} \\ Volume\text{ of the cylinder = 3.14 }\times\text{ 182.25} \\ \\ Volume\text{ of the cylinder = 572.265 cm}^3 \end{gathered}$

To the nearest hundredth, the volume of the cylinder is 572.27 cm³

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8 months ago

Which number line shows the correct placement of 2.4 to the second power

MakcuM [25]

Where are the number line choices!?

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2 years ago

Need help very fast pls

Butoxors [25]

Y = 1x -5

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5 0

2 years ago

Read 2 more answers

Which of the following is equal to the rational expression when x≠2 or -6?

Rom4ik [11]

Answer:

B. 3 / (x+6)

Step-by-step explanation:

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5 0

2 years ago

How do I find the integral<br> ∫10(x−1)(x2+9)dxint10/((x-1)(x^2+9))dx ?

defon

$\int\frac{10}{(x-1)(x^2+9)}\ dx=(*)\\\\\frac{10}{(x-1)(x^2+9)}=\frac{A}{x-1}+\frac{Bx+C}{x^2+9}=\frac{A(x^2+9)+(Bx+C)(x-1)}{(x-1)(x^2+9)}\\\\=\frac{Ax^2+9A+Bx^2-Bx+Cx-C}{(x-1)(x^2+9)}=\frac{(A+B)x^2+(-B+C)x+(9A-C)}{(x-1)(x^2+9)}\\\Updownarrow\\10=(A+B)x^2+(-B+C)x+(9A-C)\\\Updownarrow\\A+B=0\ and\ -B+C=0\ and\ 9A-C=10\\A=-B\ and\ C=B\to9(-B)-B=10\to-10B=10\to B=-1\\A=-(-1)=1\ and\ C=-1$

$(*)=\int\left(\frac{1}{x-1}+\frac{-x-1}{x^2+9}\right)\ dx=\int\left(\frac{1}{x-1}-\frac{x+1}{x^2+9}\right)\ dx\\\\=\int\frac{1}{x-1}\ dx-\int\frac{x+1}{x^2+9}\ dx=\int\frac{1}{x-1}-\int\frac{x}{x^2+9}\ dx-\int\frac{1}{x^2+9}\ dx=(**)\\\\\#1\ \int\frac{1}{x-1}\ dx\Rightarrow\left|\begin{array}{ccc}x-1=t\\dx=dt\end{array}\right|\Rightarrow\int\frac{1}{t}\ dt=lnt+C_1=ln(x-1)+C_1$

$\#2\ \int\frac{x}{x^2+9}\ dx\Rightarrow \left|\begin{array}{ccc}x^2+9=u\\2x\ dx=du\\x\ dx=\frac{1}{2}\ du\end{array}\right|\Rightarrow\int\left(\frac{1}{2}\cdot\frac{1}{u}\right)\ du=\frac{1}{2}\int\frac{1}{u}\ du\\\\\\=\frac{1}{2}ln(u)+C_2=\frac{1}{2}ln(x^2+9)+C_2$

$\#3\ \int\frac{1}{x^2+9}\ dx=\int\frac{1}{x^2+3^2}\ dx=\frac{1}{3}tan^{-1}\left(\frac{x}{3}\right)+C_3\\\\therefore:\\\\\#1;\ \#2;\ \#3\Rightarrow(**)=ln(x-1)+C_1-\frac{1}{2}ln(x^2+9)+C_2-\frac{1}{3}tan^{-1}\left(\frac{x}{3}\right)+C_3$

$\boxed{=ln(x-1)-\frac{1}{2}ln(x^2+9)-\frac{1}{3}tan^{-1}\left(\frac{x}{3}\right)+C}$

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3 years ago