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AfilCa [17]
3 years ago
7

May someone help me ASAP?! 24x^4+92x=11x

Mathematics
1 answer:
Lapatulllka [165]3 years ago
5 0

Answer:

x=0 or x= -3/2

Step-by-step explanation:

Step 1: Subtract 11x from both sides.

24x4+92x−11x=11x−11x

24x4+81x=0

Step 2: Factor left side of equation.

3x(2x+3)(4x2−6x+9)=0

Step 3: Set factors equal to 0.

3x=0 or 2x+3=0 or 4x2−6x+9=0

x=0 or x= -3/2

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I have been stuck on this forever plz help <br> What is the surface area of this cylinder
Andrei [34K]

Answer:

1865.16 m^2

Step-by-step explanation:

first let 's find radius of a circle

radius=diameter/2

=18/2

=9 m

area of a cylinder=2πrh+2πr2

=2*3.14*9*24 + 2*3.14*(9)^2

=1356.48 + 2*3.14*81

=1356.48 + 508.68

=1865.16 m^2

6 0
3 years ago
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What are the coordinates of the point 3/4 of the way from A to B?
ZanzabumX [31]

Answer:

(-3.5,1)

Step-by-step explanation:

There are 7 boxes between A and B. So I figure if you take 3/4 of 7 is about 5.

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3 years ago
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Derivative of y=(3x+5)^3​
UkoKoshka [18]

\\ \sf\longmapsto \dfrac{dy}{dt}

\\ \sf\longmapsto \dfrac{d}{dt}(3x+5)^3

\\ \sf\longmapsto \{3(3x+5)\}^2

\\ \sf\longmapsto (9x+15)^2

\\ \sf\longmapsto 9x^2+2(9x)(15)+(15)^2

\\ \sf\longmapsto 81x^2+270x+225

6 0
3 years ago
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Help me please !<br> 4,5 and 6
andreev551 [17]

9514 1404 393

Answer:

  4a. ∠V≅∠Y

  4b. TU ≅ WX

  5. No; no applicable postulate

  6. see below

Step-by-step explanation:

<h3>4.</h3>

a. When you use the ASA postulate, you are claiming you have shown two angles and the side between them to be congruent. Here, you're given side TV and angle T are congruent to their counterparts, sides WY and angle W. The angle at the other end of segment TV is angle V. Its counterpart is the other end of segment WY from angle W. In order to use ASA, we must show ...

  ∠V≅∠Y

__

b. When you use the SAS postulate, you are claiming you have shown two sides and the angle between them are congruent. The angle T is between sides TV and TU. The angle congruent to that, ∠W, is between sides WY and WX. Then the missing congruence that must be shown is ...

  TU ≅ WX

__

<h3>5.</h3>

The marked congruences are for two sides and a non-included angle. There is no SSA postulate for proving congruence. (In fact, there are two different possible triangles that have the given dimensions. This can be seen in the fact that the given angle is opposite the shortest of the given sides.)

  "No, we cannot prove they are congruent because none of the five postulates or theorems can be used."

__

<h3>6.</h3>

The first statement/reason is always the list of "given" statements.

1. ∠A≅∠D, AC≅DC . . . . given

2. . . . . vertical angles are congruent

3. . . . . ASA postulate

4. . . . . CPCTC

8 0
3 years ago
1. Solve for w in the equation W/5 = 25<br> A.W = 5<br> B.W = 125<br> C. W = 30<br> D.W = 20
vekshin1

Answer: <em>w = 125</em>

Explanation: Since <em>w</em> is being divided by 5, to solve for <em>w</em>,

multiply both sides of the equation by 5.

On the left side, the 5's will cancel

and on the right side, 25(5) is 125.

So <em>w = 125</em>.

<u />

<u>Please do not do this problem in your head.</u>

Show the work that it takes to get <em>w</em> by itself.

3 0
3 years ago
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