Find the x-intercept and y-intercept of the line. 7x+5y=-15

Mathematics

1 answer:

riadik2000 [5.3K]3 years ago

5 0

Answer:

The x-intercept is (-2 1/7, 0).

The y-intercept is (0, -3).

Step-by-step explanation:

To find the y-intercept, substitute x with zero.

To find the x-intercept, substitute y with zero.

Finding the y-intercept by substituting x with zero:

7*0 + 5y = -15

5y = -15

y = -3

The y-intercept is (0, -3).

Finding the x-intercept by substituting y with zero:

7x + 5*0 = -15

7x = -15

x = -15/7

x = -2 1/7

The x-intercept is (-2 1/7, 0).

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20.2057 Units.

Step-by-step explanation:

First, we determine the length of the cable.

Distance between Centerville (8,0) and point (x,0) is given as:

$\sqrt{(8-x)}^2=8-x$

Distance between point (x,0) and Springfield(0,7) is:

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Distance between point (x,0) and Shelvyfield(0,-7) is:

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Therefore the Length of the Cable L(x)

$L(x)=(8-x)+2\sqrt{(7)^2+(x)^{2}}$

To find the critical point, we set the derivative of L(x)=0

$L^{'}(x)=-1+\frac{2x}{\sqrt{\left( 49 - x^{2}\right) }}$

$\frac{-\sqrt{\left( 49 - x^{2}\right)}+2x}{\sqrt{\left( 49 - x^{2}\right)}}=0\\-\sqrt{\left( 49 - x^{2}\right)}+2x=0\\\sqrt{\left( 49 - x^{2}\right)}=2x\\(\sqrt{\left( 49 - x^{2}\right)})^2=(2x)^2\\49 - x^{2}=4x^2\\49=5x^2\\x^2=\frac{49}{5}\\x= 3.1305$

To verify that L(x) has a minimum at this critical number we compute the second derivative L″(x) and find its value at the critical number.

$L^{''}(x)=\frac{\left( 98\right) \,\sqrt{\left( 49 - x^{2}\right) }}{{\left( 7 - x\right) }^{2}\,{\left( 7+x\right) }^{2}}\\At \:x=3.1305, L^{''}=0.3993$