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UNO [17]
3 years ago
5

ANSWER THIS QUESTION FOR 10 POINTS AND BRAINLEST

Mathematics
2 answers:
JulijaS [17]3 years ago
5 0

Answer:

ABC

Step-by-step explanation:

Elanso [62]3 years ago
4 0
The answer is going to be ABC
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Semmy [17]
The correct answer is x = 3
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2 years ago
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Solve for x<br><br> 15x^2+16x= -4
Klio2033 [76]
15x²+16x+4 =0      (ax² +bx +c=0)

Δ = b²-4ac =256 - 4×15×4 =16

x1 = (-b+√Δ) / 2a = (-16+√16) / 30 =( -16+4) / 30 = -12/30 = - 2/5
x2 = (-b -√Δ) / 2a = (-16 -√16) / 30 = (-16 -4) /30 = -20/30 = -2/3
4 0
2 years ago
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WORTH 10 POINTS! 1)susan can type 4 pages of text in 10 minutes.Assuming she types at a constant rate,write the linear equation
Minchanka [31]
1) pages per minute is the rate. the rate is the slope pages/minutes = .4
y=.4x or y=2/5(x)
2)birdhouses per day is the slope. = .6
y=.6x or y=3/5(x)
3) miles per hour is the slope. 500/8=62.5
y=62.5x
4) square feet per minute is the slope. 40/9 =4.444444444
y=4.4444444444x or y=40/9(x)
5)miles per minutes is the rate. 5/41 is the slope. 
y=5/41(x)
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8 0
2 years ago
Find the dimensions of the rectangle
hoa [83]
(a) The sum of length and width is half the perimeter, so the breadth in terms of length (x) in cm will be
.. breadth = 34 -x


(b) Then the expression for the area (in cm^2) is
.. area = length*breadth
.. 253 = (x)(34 -x)
.. x^2 -34x +253 = 0 . . . . . subtract the right side, eliminate parentheses
.. (x -11)(x -23) = 0 . . . . . . . factor
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The dimensions of the rectangle are 11 cm by 23 cm.
7 0
3 years ago
the volume v of a right circular cylinder of radius r and heigh h is V = pi r^2 h 1. how is dV/dt related to dr/dt if h is const
laiz [17]
In general, the volume

V=\pi r^2h

has total derivative

\dfrac{\mathrm dV}{\mathrm dt}=\pi\left(2rh\dfrac{\mathrm dr}{\mathrm dt}+r^2\dfrac{\mathrm dh}{\mathrm dt}\right)

If the cylinder's height is kept constant, then \dfrac{\mathrm dh}{\mathrm dt}=0 and we have

\dfrac{\mathrm dV}{\mathrm dt}=2\pi rh\dfrac{\mathrm dt}{\mathrm dt}

which is to say, \dfrac{\mathrm dV}{\mathrm dt} and \dfrac{\mathrm dr}{\mathrm dt} are directly proportional by a factor equivalent to the lateral surface area of the cylinder (2\pi r h).

Meanwhile, if the cylinder's radius is kept fixed, then

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

since \dfrac{\mathrm dr}{\mathrm dt}=0. In other words, \dfrac{\mathrm dV}{\mathrm dt} and \dfrac{\mathrm dh}{\mathrm dt} are directly proportional by a factor of the surface area of the cylinder's circular face (\pi r^2).

Finally, the general case (r and h not constant), you can see from the total derivative that \dfrac{\mathrm dV}{\mathrm dt} is affected by both \dfrac{\mathrm dh}{\mathrm dt} and \dfrac{\mathrm dr}{\mathrm dt} in combination.
8 0
2 years ago
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