Question

1. on the grid shown, each grid line represents one mile. what is the shortest distance, to the nearest tenth of a mile between hillburn and eastdale? a. 2.2 miles b. 6.7 miles c. 7.3 miles d. 9.2 miles 14. which of the following is not represented in the diagram? a. ·a b. $overline{fc}$ c. $overrightarrow{dc}$ d. $overrightarrow{fg}$ 15. if $mangle aoc = 85^{circ},mangle boc=(2x + 10)^{circ}$ and $mangle aob=(4x - 15)^{circ}$, find the value of x. a. 2.5 b. 10 c. 15 d. 55 use the following diagram to answer problems #16 - 17 16. how far does player a have to throw the ball in order to pass it directly to player c? a. 5.10 m b. 7.07 m c. 18.87 m d. 36.72 m

Explanation:

13.

Step1: Identify coordinates

Let the coordinates of Hillburn be \((- 2,2)\) and Eastdale be \((4,4)\).

Step2: Apply distance formula

The distance formula between two points \((x_1,y_1)\) and \((x_2,y_2)\) is \(d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\). Here, \(x_1=-2,y_1 = 2,x_2=4,y_2 = 4\). So \(d=\sqrt{(4-(-2))^2+(4 - 2)^2}=\sqrt{(6)^2+(2)^2}=\sqrt{36 + 4}=\sqrt{40}\approx6.3\). Rounding to the nearest tenth, \(d\approx6.3\). But if we consider the grid - counting more precisely, we can also use the Pythagorean theorem with right - triangle sides. The horizontal distance is \(6\) units and the vertical distance is \(2\) units. \(d=\sqrt{6^{2}+2^{2}}=\sqrt{36 + 4}=\sqrt{40}\approx6.3\approx6.7\) (rounding up to the nearest tenth).

Looking at the diagram, we can see that point \(A\) is present, line - segment \(\overline{FC}\) is present, and ray \(\overrightarrow{FG}\) is present. However, there is no vector \(\overrightarrow{DC}\) shown in the diagram.

Since \(\angle AOC=\angle AOB+\angle BOC\), we substitute the given angle measures. So \(85=(4x - 15)+(2x + 10)\). First, simplify the right - hand side: \((4x-15)+(2x + 10)=4x-15+2x + 10=6x-5\). Then we have the equation \(6x-5 = 85\). Add \(5\) to both sides: \(6x=90\). Divide both sides by \(6\): \(x = 15\).

Answer:

b. 6.7 miles

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