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Natalija [7]
3 years ago
8

A line passes throughout (-3,-2) and is perpendicular to 3x-2y=7. What is the equation of the line in slope-intercept form? A. Y

=3/2X B. Y=3/2x+4 C. y=-2/3x-4 D. y=/2/3x
Mathematics
1 answer:
marissa [1.9K]3 years ago
6 0
3x - 2y = 7
-2y = -3x + 7
y = 3/2x - 7/2....slope of this line is 3/2. A perpendicular line will have a negative reciprocal slope. All that means is " flip " the slope and change the sign. So the slope we need is -2/3 (see how I flipped the slope and changed the sign)

y = mx + b
slope(m) = -2/3
(-3,-2).....x = -3 and y = -2
now we sub and find b, the y int
-2 = -2/3(-3) = b
-2 = 2 + b
-2 - 2 = b
-4 = b

so ur perpendicular equation is : y = -2/3x - 4 <==
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Answer:

C, Length = 39.4 cm; width = 7.9 cm

Step-by-step explanation:

The permimeter of a rectangle is 2 times the width + 2 times the length or 2(l + w)

The length is 4 times w plus 7.8.

So:  2w+2(4w+7.8)=94.6

=  2w+8w+15.6=94.6  (Factor out the brackets)

=  10w+15.6=94.6  (Combine like terms)

=  10w=79  (Subtract 15.6 from both sides)

=  w=7.9  (Divide both by 10).

So, the width is 7.9 cm and the length is 39.4 cm (4 * 7.9 + 7.8)

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3 years ago
George wants to put carpeting in a rectangular living room and a square bedroom.The length and width of the living room is 12 fe
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Since George is going to put carpeting in two different rooms, one rectangular and one square, you need to find the area of each room, add the sum and then multiply the total area needed by the cost per square foot of carpet.  The area of a rectangle is: a = lw, where l=length and w = width.  The area of a square is: a = s², where s = side of the square.  Take the sum of these two areas would be (12 x 18) + 13² or 213 + 169 = 382.  Since the cost of the carpet is the same per square foot, you can multiply the entire sum by the cost of $3.50 or:  3.5((12 x 18) + 13²) or 3.5(382) = $1,337.

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A giant tank in a shape of an inverted cone is filled with oil. the height of the tank is 1.5 metre and its radius is 1 metre. t
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The given height of the cylinder of 1.5 m, and radius of 1 m, and the rate

of dripping of 110 cm³/s gives the following values.

1) The rate of change of the oil's radius when the radius is 0.5 m is r' ≈ <u>9.34 × 10⁻⁵ m/s</u>

2) The rate of change of the oil's height when the height is 20 cm is h' ≈ <u>1.97 × 10⁻³ m/s</u>

3) The rate the oil radius is changing when the radius is 10 cm is approximately <u>0.175 m/s</u>

<h3>How can the rate of change of the radius & height be found?</h3>

The given parameters are;

Height of the tank, h = 1.5 m

Radius of the tank, r = 1 m

Rate at which the oil is dripping from the tank = 110 cm³/s = 0.00011 m³/s

1) \hspace{0.15 cm}V = \frac{1}{3} \cdot \pi \cdot r^2 \cdot h

From the shape of the tank, we have;

\dfrac{h}{r} = \dfrac{1.5}{1}

Which gives;

h = 1.5·r

V = \mathbf{\frac{1}{3} \cdot \pi \cdot r^2 \cdot (1.5 \cdot r)}

\dfrac{d}{dr} V =\dfrac{d}{dr}  \left( \dfrac{1}{3} \cdot \pi \cdot r^2 \cdot (1.5 \cdot r)\right) = \dfrac{3}{2} \cdot \pi  \cdot r^2

\dfrac{dV}{dt} = \dfrac{dV}{dr} \times \dfrac{dr}{dt}

\dfrac{dr}{dt} = \mathbf{\dfrac{\dfrac{dV}{dt} }{\dfrac{dV}{dr} }}

\dfrac{dV}{dt} = 0.00011

Which gives;

\dfrac{dr}{dt} = \mathbf{ \dfrac{0.00011 }{\dfrac{3}{2} \cdot \pi  \cdot r^2}}

When r = 0.5 m, we have;

\dfrac{dr}{dt} = \dfrac{0.00011 }{\dfrac{3}{2} \times\pi  \times 0.5^2} \approx  9.34 \times 10^{-5}

The rate of change of the oil's radius when the radius is 0.5 m is r' ≈ <u>9.34 × 10⁻⁵ m/s</u>

2) When the height is 20 cm, we have;

h = 1.5·r

r = \dfrac{h}{1.5}

V = \mathbf{\frac{1}{3} \cdot \pi \cdot \left(\dfrac{h}{1.5} \right) ^2 \cdot h}

r = 20 cm ÷ 1.5 = 13.\overline3 cm = 0.1\overline3 m

Which gives;

\dfrac{dr}{dt} = \dfrac{0.00011 }{\dfrac{3}{2} \times\pi  \times 0.1 \overline{3}^2} \approx  \mathbf{1.313 \times 10^{-3}}

\dfrac{d}{dh} V = \dfrac{d}{dh}  \left(\dfrac{4}{27} \cdot \pi  \cdot h^3 \right) = \dfrac{4 \cdot \pi  \cdot h^2}{9}

\dfrac{dV}{dt} = \dfrac{dV}{dh} \times \dfrac{dh}{dt}

\dfrac{dh}{dt} = \dfrac{\dfrac{dV}{dt} }{\dfrac{dV}{dh} }<em />

\dfrac{dh}{dt} = \mathbf{\dfrac{0.00011}{\dfrac{4 \cdot \pi  \cdot h^2}{9}}}

When the height is 20 cm = 0.2 m, we have;

\dfrac{dh}{dt} = \dfrac{0.00011}{\dfrac{4 \times \pi  \times 0.2^2}{9}} \approx \mathbf{1.97 \times 10^{-3}}

The rate of change of the oil's height when the height is 20 cm is h' ≈ <u>1.97 × 10⁻³ m/s</u>

3) The volume of the slick, V = π·r²·h

Where;

h = The height of the slick = 0.1 cm = 0.001 m

Therefore;

V = 0.001·π·r²

\dfrac{dV}{dr} = \mathbf{ 0.002 \cdot \pi \cdot r}

\dfrac{dr}{dt} = \mathbf{\dfrac{0.00011 }{0.002 \cdot \pi  \cdot r}}

When the radius is 10 cm = 0.1 m, we have;

\dfrac{dr}{dt} = \dfrac{0.00011 }{0.002 \times \pi  \times 0.1} \approx \mathbf{0.175}

The rate the oil radius is changing when the radius is 10 cm is approximately <u>0.175 m</u>

Learn more about the rules of differentiation here:

brainly.com/question/20433457

brainly.com/question/13502804

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