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2 years ago

-1 + 2x - x = x - 8 + (-x)

Mathematics

1 answer:

CaHeK987 [17]2 years ago

6 0

Answer:

x = -7

Step-by-step explanation:

-1 + x = -8

x = -7

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Lauren keeps records of the distances she travels in a taxi and what she

Scrat [10]

Answer:

m = 2.25

Step-by-step explanation:

In a standard equation of y = mx + b, m is the slope. In this situation, the slope is the price paid per mile.

To find the slope of the line, we use the slope formula: (y₂ - y₁) / (x₂ - x₁). I will use the points (3, 8.25) & (11, 26.25)

Plug in these values:

(26.25 - 8.25) / (11 - 3)

Simplify the parentheses.

= (18) / (8)

Simplify the fraction.

18/8

= 9/4

This is your slope.

In a dollar value, it is 9/4 of a dollar, or $2.25 per mile.

3 0

3 years ago

Section 5.2 Problem 6: Find the general solution <img src="https://tex.z-dn.net/?f=y%27%27%20%2B%206y%27%20%2B%2010y%2

mihalych1998 [28]

Answer:

$y=e^{-3t}(A\: cos\: t+B\:sin\:t)$

Step-by-step explanation:

Given Second-Order Homogenous Differential Equation

$y''+6y'+10y=0$

Use Auxiliary Equation

$m^2+6m+10=0\\\\(m+3)^2+1=0\\\\(m+3)^2=-1\\\\m+3=\pm i\\\\m=-3\pm i$

General Solution

$y=e^{pt}(A\: cos\: qt+B\:sin\:qt)\\\\y=e^{-3t}(A\: cos\: t+B\:sin\:t)$

Note that the DE has two distinct complex solutions $p\pm qi$ where $A$ and $B$ are arbitrary constants.

4 0

2 years ago

HELP ME PLEASEEEEEEEEEEEE

laila [671]

Answer:

1080 in^2

Step-by-step explanation:

8 0

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