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3 years ago

Whats is 0.000001 written as a power of ten

1 answer:

8 0

You need to move the decimal point to be at 1.

Since this is a decimal, the power will be negative.

I had to move the decimal 6 places.

So,

1^-6

I hope this helps!
~kaikers

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5(x+3)−2x Which expression correctly combines the like terms? 3x+15 3x+3 3x2+15 3x2+3

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3x+15 , this expression correctly combines the like terms

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3 years ago

(m^(6)n^(3))/(m^(2)n^(6))

mario62 [17]

Answer:

$(m^4)/(n^3)$

Step-by-step explanation:

See attachment

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2 years ago

Mikheal wanted to rewrite the conversion factor "1 yard = 0.914 meters" to create a conversion factor to convert meters to yards

Levart [38]

Divide both 1 yard and 0.914 meters by 1. 1x = 0.914y
Now, since you want 1 meter = "variable"
1 meter times 1 yard divided by 0.914 meters.
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3 years ago

Can someone please answer this question please answer it correctly please show work please help me I need it

Lena [83]

Answer:

B) Step 2

Step-by-step explanation:

Raul thought that 5 - 7 was 2, but it is actually -2.

If this answer is correct, please make me Brainliest!

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3 years ago

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:

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))

4 0

3 years ago