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3 years ago

Troy drives 65 miles west and then drives 72 miles north. How far is he from his starting point?

Mathematics

1 answer:

iragen [17]3 years ago

4 0

Answer:

132 miles away from the start

Step-by-step explanation:

First Troy went to the left 65 miles so he starts off 65 miles from the start then he goes 72 miles right so then he is 132 miles away

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What is the distance between the points (5, 1) and (-3,-5)?

Reil [10]

Answer: THIRD OPTION.

Step-by-step explanation:

For this exercise you need to use the formula for calculate the distance between two points. This is:

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Given the following points:

$(5, 1)$ and $(-3,-5)$

You can identify that:

$x_2=-3\\x_1=5\\\\y_2=-5\\y_1=1$

Knowing these values, the next step is to substitute them into the equation for calculate the distance between two points and then evaluate.

Therefore, the distance between $(5, 1)$ and $(-3,-5)$ is:

$d=\sqrt{(-3-5)^2+(-5-1)^2}\\\\d=10$

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3 years ago

Find the measure of one interior angle of a regular 14-gon.

lozanna [386]

$\bf \textit{sum of all interior angles in a polygon}\\\\ n\theta =180(n-2)~~ \begin{cases} n=\stackrel{polygon}{sides}\\ \theta = angle~in\\ \qquad degrees\\[-0.5em] \hrulefill\\ n = 14\\ \end{cases}\implies 14\theta =180(14-2) \\\\\\ 14\theta =180(12)\implies 14\theta =2160\implies \theta =\cfrac{2160}{14}\implies \theta \approx 154.3^o$

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3 years ago

WILL MAKE BRAINLIEST!!Determine if the two triangles are congruent if they are state the postulate or theorem that provides they

Dmitry [639]

Answer:

<h2>Yes they are congruent by SSS criterion </h2>

Step-by-step explanation:

AC=DC. (Given in figure)

AB=BD. (Given in figure

BC=BC. (Common)

that means triangle BAC is congruent to triangle BDC by SSS criterion.

5 0

1 year ago

Simplify this expression. 8+3(60/5)

ddd [48]

Answer:44

Step-by-step explanation: 60/5 evaluates to 12

Multiply 3 and 12

3*(60/5) evaluates to 36

8+3*(60/5) evaluates to 44

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3 years ago

Read 2 more answers

Prove A-(BnC) = (A-B)U(A-C), explain with an example

NikAS [45]

Answer:

Prove set equality by showing that for any element $x$ , $x \in (A \backslash (B \cap C))$ if and only if $x \in ((A \backslash B) \cup (A \backslash C))$ .

Example:

$A = \lbrace 0,\, 1,\, 2,\, 3 \rbrace$ .

$B = \lbrace0,\, 1 \rbrace$ .

$C = \lbrace0,\, 2 \rbrace$ .

$\begin{aligned} & A \backslash (B \cap C) \\ =\; & \lbrace 0,\, 1,\, 2,\, 3 \rbrace \backslash \lbrace 0 \rbrace \\ =\; & \lbrace 1,\, 2,\, 3 \rbrace \end{aligned}$ .

$\begin{aligned}& (A \backslash B) \cup (A \backslash C) \\ =\; & \lbrace 2,\, 3\rbrace \cup \lbrace 1,\, 3 \rbrace \\ =\; & \lbrace 1,\, 2,\, 3 \rbrace\end{aligned}$ .

Step-by-step explanation:

Proof for $[x \in (A \backslash (B \cap C))] \implies [x \in ((A \backslash B) \cup (A \backslash C))]$ for any element $x$ :

Assume that $x \in (A \backslash (B \cap C))$ . Thus, $x \in A$ and $x \not \in (B \cap C)$ .

Since $x \not \in (B \cap C)$ , either $x \not \in B$ or $x \not \in C$ (or both.)

If $x \not \in B$ , then combined with $x \in A$ , $x \in (A \backslash B)$ .
Similarly, if $x \not \in C$ , then combined with $x \in A$ , $x \in (A \backslash C)$ .

Thus, either $x \in (A \backslash B)$ or $x \in (A \backslash C)$ (or both.)

Therefore, $x \in ((A \backslash B) \cup (A \backslash C))$ as required.

Proof for $[x \in ((A \backslash B) \cup (A \backslash C))] \implies [x \in (A \backslash (B \cap C))]$ :

Assume that $x \in ((A \backslash B) \cup (A \backslash C))$ . Thus, either $x \in (A \backslash B)$ or $x \in (A \backslash C)$ (or both.)

If $x \in (A \backslash B)$ , then $x \in A$ and $x \not \in B$ . Notice that $(x \not \in B) \implies (x \not \in (B \cap C))$ since the contrapositive of that statement, $(x \in (B \cap C)) \implies (x \in B)$ , is true. Therefore, $x \not \in (B \cap C)$ and thus $x \in A \backslash (B \cap C)$ .
Otherwise, if $x \in A \backslash C$ , then $x \in A$ and $x \not \in C$ . Similarly, $x \not \in C \!$ implies $x \not \in (B \cap C)$ . Therefore, $x \in A \backslash (B \cap C)$ .

Either way, $x \in A \backslash (B \cap C)$ .

Therefore, $x \in ((A \backslash B) \cup (A \backslash C))$ implies $x \in A \backslash (B \cap C)$ , as required.

8 0

2 years ago