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2 years ago

10

Caroline drew the blueprints of her new kitchen on a coordinate plane. The coordinates of the four corners of her square kitchen

are (2, 4), (-10, 4), (-10, -8), and (2, -8). How many units long is one wall of her kitchen? If needed, create a graph to help answer the question.

1 answer:

olga nikolaevna [1]2 years ago

8 0

Answer:

the wall of her kitchen is 12 units long

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All of the following are equivalent except _____.

Inessa05 [86]

(5_5)x is not equalant

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2 years ago

Read 2 more answers

44. Express each of these system specifications using predicates, quantifiers, and logical connectives. a) Every user has access

DENIUS [597]

Answer:

a. ∀x (User(x) → (∃y (Mailbox(y) ∧ Access(x, y))))

b. FileSystemLocked → ∀x Access(x, SystemMailbox)

c. ∀x ∀y ((Firewall(x) ∧ Diagnostic(x)) → (ProxyServer(y) → Diagnostic(y))

d. ∀x (ThroughputNormal ∧(ProxyServer(x)∧ ¬Diagnostic(x))) → (∃y Router(y)∧Functioning(y))

Step-by-step explanation:

a)

Let the domain be users and mailboxes. Let User(x) be “x is a user”, let Mailbox(y) be “y is a mailbox”, and let Access(x, y) be “x has access to y”.

∀x (User(x) → (∃y (Mailbox(y) ∧ Access(x, y))))

(b)

Let the domain be people in the group. Let Access(x, y) be “x has access to y”. Let FileSystemLocked be the proposition “the file system is locked.” Let System Mailbox be the constant that is the system mailbox.

FileSystemLocked → ∀x Access(x, SystemMailbox)

(c)

Let the domain be all applications. Let Firewall(x) be “x is the firewall”, and let ProxyServer(x) be “x is the proxy server.” Let Diagnostic(x) be “x is in a diagnostic state”.

∀x ∀y ((Firewall(x) ∧ Diagnostic(x)) → (ProxyServer(y) → Diagnostic(y))

(d)

Let the domain be all applications and routers. Let Router(x) be “x is a router”, and let ProxyServer(x) be “x is the proxy server.” Let Diagnostic(x) be “x is in a diagnostic state”. Let ThroughputNormal be “the throughput is between 100kbps and 500 kbps”. Let Functioning(y) be “y is functioning normally”.

∀x (ThroughputNormal ∧(ProxyServer(x)∧ ¬Diagnostic(x))) → (∃y Router(y)∧Functioning(y))

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3 years ago

Solve for x rounded to the nearest hundredth: 10 - 3x = 5

igomit [66]

Answer:

x = 5/3

Step-by-step explanation:

You ALWAYS want your term to be by itself!! So what you need to do is move your constant away from it.

Step 1: Adjust as needed, subtract 10 from both sides of your equation. It should end up like this.

-3x = 5 - 10

Step 2: Solve as needed before continuing equation.

-3x = -5

Step 3: Now you need to get your term by itself. You should divide -3 on both sides of your equation so you should get something like this.

x = -5 / -3

Step 4: You're basically done now! Just simplify! NEGATIVES CROSS EACH OTHER OUT!!!

x = 5/3 OR 1.6

<em>KEY:</em>

<em>Negative and Negative = Positive</em>

<em>Negative and Positive = Negative</em>

<em>Positive and Positive = Positive</em>

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3 years ago

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Answer:

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D shows that 6 is being multiplied by n. Because of the order of operations, we d this first. Then we'd subtract 4.

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2 years ago

You use a line of best fit for a set of data to make a prediction about an unknown value. the correlation coeffecient is -0.833

alina1380 [7]

Answer: The square root of π has attracted attention for almost as long as π itself. When you’re an ancient Greek mathematician studying circles and squares and playing with straightedges and compasses, it’s natural to try to find a circle and a square that have the same area. If you start with the circle and try to find the square, that’s called squaring the circle. If your circle has radius r=1, then its area is πr2 = π, so a square with side-length s has the same area as your circle if s2 = π, that is, if s = sqrt(π). It’s well-known that squaring the circle is impossible in the sense that, if you use the classic Greek tools in the classic Greek manner, you can’t construct a square whose side-length is sqrt(π) (even though you can approximate it as closely as you like); see David Richeson’s new book listed in the References for lots more details about this. But what’s less well-known is that there are (at least!) two other places in mathematics where the square root of π crops up: an infinite product that on its surface makes no sense, and a calculus problem that you can use a surface to solve.

Step-by-step explanation: this is the same paragraph The square root of π has attracted attention for almost as long as π itself. When you’re an ancient Greek mathematician studying circles and squares and playing with straightedges and compasses, it’s natural to try to find a circle and a square that have the same area. If you start with the circle and try to find the square, that’s called squaring the circle. If your circle has radius r=1, then its area is πr2 = π, so a square with side-length s has the same area as your circle if s2 = π, that is, if s = sqrt(π). It’s well-known that squaring the circle is impossible in the sense that, if you use the classic Greek tools in the classic Greek manner, you can’t construct a square whose side-length is sqrt(π) (even though you can approximate it as closely as you like); see David Richeson’s new book listed in the References for lots more details about this. But what’s less well-known is that there are (at least!) two other places in mathematics where the square root of π crops up: an infinite product that on its surface makes no sense, and a calculus problem that you can use a surface to solve.

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2 years ago