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Svetach [21]
2 years ago
10

Emily wants to hang a painting in a gallery. The painting and frame must have an area of 31 square feet. The painting is 5 feet

wide by 6 feet long. Which quadratic equation can be used to determine the thickness of the frame, x?
Image of a frame 5 feet wide by 6 feet long, with an of x on the bottom and right sides

4x2 + 22x − 1 = 0
4x2 + 22x + 31 = 0
x2 + 11x − 1 = 0
x2 + 11x + 31 = 0
Mathematics
2 answers:
seropon [69]2 years ago
6 0
Area=legnth times width=31
the frame must go around the frame with equal legnth, x
so
the total area (meaning the frame +picture) is 5+x by 6+x
31=(5+x)(6+x)
expand
31=30+11x+x^2
minus 31 both sides
0=-1+11x+x^2
x^2+11x-1=0

answer is third one
yarga [219]2 years ago
4 0

Answer:   4x^{2} + 22x − 1 = 0


Step-by-step explanation:

Start off with the standard formula to finding area with polynomials and quadratic functions.

(2x + 5)(2x + 6)

Work out like a normal binomial.

4x^{2} + 22x + 30 - 31

(You're subtracting 31 to take out the minimum of what the frame has to be)

4x^{2} + 22x - 1 = 0

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Step-by-step explanation:

Part 1

For this case we know the following info: The length, l cm, of a simple pendulum is directly proportional to the square of its period (time taken to complete one oscillation), T seconds.

L \propto T^2

Using the condition given:

2.205 m = K (3)^2

K = 0.245 \approx \frac{g}{4\pi^2}

So then if we want to create an equation we need to do this:

L = K T^2

With K a constant. For this case the period of a pendulumn is given by this general expression:

T = 2\pi \sqrt{\frac{L}{g}}

Where L is the length in m and g the gravity g = 9.8 \frac{m}{s^2}.

Part 2

For this case using the function in part a we got:

T = 2\pi \sqrt{\frac{L}{g}}

If we square both sides of the equation we got:

T^2 = 4 \pi^2 \frac{L}{g}

And solving for L we got:

L = \frac{g T^2}{4 \pi^2}

Replacing we got:

L =\frac{9.8 \frac{m}{s^2} (5s)^2}{4 \pi^2} = 6.206m

Part 3

For this case using the function in part a we got:

T = 2\pi \sqrt{\frac{L}{g}}

Replacing we got:

T = 2\pi \sqrt{\frac{0.98m}{9.8\frac{m}{s^2}}}= 1.987 s

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