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2 years ago

11

The cost of renting a car for a day is given by the linear function C(x)=0.07x+39,where C(x) is in dollars and x is the number o

f miles driven.Find the cost of driving the car 210 miles

1 answer:

Bess [88]2 years ago

5 0

Answer:

3ghgfdcvbjjrdcvjjidetyhvcsefgjhvfdtgh

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Which of the following are exterior angles? Check all that apply.

Marina86 [1]

Answer:

A

C

D

E

Step-by-step explanation:

Exterior angles can be described as the angles that are formed between the side of a polygon and the extended adjacent side of the polygon.

Or an exterior angle is the angle that is not inside the triangle formed.

The angles inside the triangle are interior angles.

Exterior angles are :

2

3

4

6

Interior angles are :

1

5

6 0

2 years ago

I need help on this question!!

elena55 [62]

I think the answer is D, 28*11=308 plus the 12*14 equals 476. The answer is D

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3 years ago

Given complex numbers u and v below,what is uv<br><br> u=7i<br> V=6-i

boyakko [2]

Answer:

7 + 42i.

Step-by-step explanation:

u v = 7i(6 - i)

= 42i - 7i^2 But i^2 = -1 so:

= 42i - 7(-1)

= 7 + 42i.

6 0

3 years ago

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.

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2 years ago

Describe a situation in which you could use scaling to solve a real-world problem.

balu736 [363]

Answer:

You want to know how much lager a full sized candy bar is to see if you are getting a good price.

Step-by-step explanation:

6 0

2 years ago