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notsponge [240]
3 years ago
5

Which equation does the graph below represent?

Mathematics
1 answer:
Sergeu [11.5K]3 years ago
8 0

Answer:

y = (½)x

Step-by-step explanation:

(-8,-4) (0,0) and (8,4)

Use any 2 points to find the slope, hence the equation

slope = (y2-y1)/(x2-x1)

= (0-(-4))/(0-(-8))

= 4/8

= ½

y = (½)x + c

When x=0, y=0

0 = (½)(0) + c

c = 0

Equation is:

y = (½)x

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PLEASE HELP ASAP!!
Alekssandra [29.7K]
I think its D. 32.6 s/ft and appoximatley 0.03 ft/s
8 0
3 years ago
I don't understand how to do this. Can someone help?
miskamm [114]
Time used for the call to Zurich: ($5-$1.25)/$0.25=15 minutes
15minutes+3minutes=18minutes
Time used for the call to London: ($7.25-$1.25)/$0.25=24 minutes
24minutes+3minutes=27minutes
Difference between the two calls:27-18=9minutes
3 0
3 years ago
Solve -7x^2+x+9=-6x quadratic formula
myrzilka [38]

Answer:

-7x^2+x+9=-6x

-7x^2+x+9+6x=-6x+6x

-7x^2+7x+9=0

quadratic\: formula

x_{1,\:2}=\frac{-7\pm \sqrt{7^2-4\left(-7\right)\cdot \:9}}{2\left(-7\right)}

\sqrt{-7^{2} -4(-7)\times9} =\sqrt{301}

x_{1,\:2}=\frac{-7\pm \sqrt{301}}{2\left(-7\right)}

\frac{-7+\sqrt{301}}{2\left(-7\right)}

=\frac{-7+\sqrt{301}}{-2\cdot \:7}

=\frac{-7+\sqrt{301}}{-14}

\frac{-7-\sqrt{301}}{2\left(-7\right)}

=\frac{-7-\sqrt{301}}{-2\cdot \:7}

=\frac{-7-\sqrt{301}}{-14}

answer=\frac{7+\sqrt{301}}{14}

OAmalOHopeO

7 0
3 years ago
Rectangle R has varying length l and width w but a constant perimeter of 4 ft. A. Express the area A as a function of l. What do
ivanzaharov [21]
Given:
l = length of the rectangle
w = width of the rectangle
P = 4 ft, constant perimeter

Because the given perimeter is constant,
2(w + l) = 4
w + l = 2
w = 2 - l            (1)

Part A.
The area is
A = w*l 
   = (2 - l)*l
 A  = 2l - l²
This is a quadratic function or a parabola.

Part B.
Write the parabola in standard form.
A = -[l² - 2l]
   = -[ (l -1)² - 1]
   = -(l -1)² + 1
This is a parabola with vertex at (1, 1). Because the leading coefficient is negative the curve is downward, as shown below.

The maximum value occurs at the vertex, so the maximum value of A = 1.
From equation (1), obtain
w = 2 - l = 2 - 1 = 1.
The maximum value of the area occurs when w=1 and l=1 (a square).

Answer:
The area is maximum when l=1 and w=1.
The geometric argument is based on the vertex of the parabola denoting maximum area.

4 0
3 years ago
Can you please help me solve this
Snezhnost [94]

Answer:

yes

Step-by-step explanation:

3 0
3 years ago
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