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1 year ago

If P(x 3), Q(7, -1) and PQ= 5 units, find the possible value of x.

Mathematics

1 answer:

Neporo4naja [7]1 year ago

5 0

Given :

P(x 3), Q(7, -1) and PQ= 5 .

To Find :

The possible value of x.

Solution :

We know, distance between two points in coordinate plane is given by :

$PQ = \sqrt{(x-7)^2 + ( 3-(-1))^2}\\\\PQ^2 = (x-7)^2 + ( 3-(-1))^2\\\\(x-7)^2 + 4^2 = 5^2\\\\( x- 7)^2 = 3^2 \\\\x - 7 = \pm 3\\\\x = 10 \ and \ x = 4$

Therefore, the possible value of x are 10 and 4.

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Answer:

m∠ABE = 27°

Step-by-step explanation:

* Lets look to the figure to solve the problem

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∴ m∠ABF = m∠CBF = 90°

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∵ m∠ABE = m∠DBE ⇒ have same symbol on the figure

∴ BE is the bisector of angle ABD

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1 year ago

Georgie joins a gym. She pays $25 to sign up and then $15 each monthLabel the y-intercept, the slope, and your axes.

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1 year ago

answer A radio station located 120 miles due east of Collinsville has a listening radius of 100 miles. A straight road joins Col

melamori03 [73]

Answer:

168.7602 miles

Step-by-step explanation:

One way to solve this problem is by using an equation that describes the listening radius of the station, and another for the road, then the points where this two-equation intersect each other will represent when the driver starts and stops listening to the station, and the distance between the points is the miles that the driver will receive the signal.

The equation for the listening radius (the radio station is at (0,0)):

$x^2+y^2=100^2$

The equation for the road that past through the points (-120,0) and (80,100) (Collinsville and Harmony respectively):

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$y-y_1=m(x-x_1)\\y-0=\frac{1}{2}(x-(-120))\\ y=\frac{1}{2}x+60$

Substitutes the value of y in the equation of the circle:

$x^2+(\frac{1}{2}x+60)^2=100^2\\x^2+\frac{1}{4} x^2+60x+3600=10000\\\frac{5}{4} x^2+60x+3600=10000\\\frac{5}{4} x^2+60x+3600-10000=0\\\frac{5}{4} x^2+60x-6400=0\\5 x^2+240x-25600=0\\x^2+48x-5120=0\\$

The formula to solve second-degree equations:

$x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac} }{2a} \\x_{1,2}=\frac{-48\pm\sqrt{48^2-4(1)(-5120)} }{2(1)}\\x_{1,2}=\frac{-48\pm\sqrt{2304+20480} }{2}\\x_{1,2}=\frac{-48\pm\sqrt{22784} }{2}\\x_{1,2}=\frac{-48\pm16\sqrt{89} }{2}\\x_{1,2}=-24\pm8\sqrt{89} \\x_1=-24+8\sqrt{89}\approx51.4718\\x_2=-24-8\sqrt{89}\approx-99.4718\\$

Using the values in x to find the values in y:

$y_1=\frac{1}{2}x_1+60\\y_1=\frac{1}{2}(-24+8\sqrt{89} )+60\\y_1=-12+4\sqrt{89}+60\\ y_1=48+4\sqrt{89}\approx85.7359$

$y_2=\frac{1}{2}x_2+60\\y_2=\frac{1}{2}(-24-8\sqrt{89} )+60\\y_1=-12-4\sqrt{89}+60\\ y_1=48-4\sqrt{89}\approx10.2641$

The distance between the points (51.4718,85.7359) and (-99.4718,10.2641) :

$d=\sqrt{(x_1 -x_2 )^2+(y_1 -y_2)^2} \\d=\sqrt{(-24+8\sqrt{89} -(-24-8\sqrt{89}) )^2+(48+4\sqrt{89} -(48-4\sqrt{89}) )^2}\\d=\sqrt{(-24+8\sqrt{89} +24+8\sqrt{89} )^2+(48+4\sqrt{89} -48+4\sqrt{89} )^2}\\d=\sqrt{(16\sqrt{89} )^2+(8\sqrt{89} )^2}\\d=\sqrt{22784+5696}\\d=\sqrt{28480}\\d=8\sqrt{445}\approx168.7602miles$

4 0

1 year ago

If P(x 3), Q(7, -1) and PQ= 5 units, find the possible value of x.​

If P(x 3), Q(7, -1) and PQ= 5 units, find the possible value of x.