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scZoUnD [109]
3 years ago
14

Solve for x, in terms of y: 2/3x + 1/5y = 2

Mathematics
1 answer:
lakkis [162]3 years ago
4 0

Answer:

x = 3 - 3/10y

Step-by-step explanation:

2/3x + 1/5y = 2

Subtract 1/5y from each side

2/3x + 1/5y -1/5 y= 2-1/5 y

2/3x = 2-1/5y

Multiply by 3/2 to isolate x

3/2 * 2/3x = 3/2 (2-1/5y)

x = 3/2 (2-1/5y)

Distribute the 3/2

x = 3 - 3/10y

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find a number such that if 8 is subtracted from 9 times the number , the result is 6 more than twice the number​
zheka24 [161]

Answer:

x=2

Step-by-step explanation:

let the number be x

9x-8=6+2x

9x-2x=6+8

7x=14

x=14/7

x=2

8 0
3 years ago
Given that angle a = 54° and angle b = 48°, work out x.
Arisa [49]
If it’s a triangle then it has to equal 180
so 48+54+x=180
48+54=102
180-102=x

x=78
3 0
2 years ago
HELPPP RIGHT NOW PLEASEEEEE
Nadusha1986 [10]

The value of the variable x will be 37°. Then the measure of the angle ∠QNP will be 105°.

<h3>What is an angle?</h3>

The angle is the distance between the intersecting lines or surfaces. The angle is also expressed in degrees. The angle is 360 degrees for one complete spin.

Supplementary angle - Two angles are said to be supplementary angles if their sum is 180 degrees.

The measure of angles ∠MNQ = (2x + 1)° and ∠QNP = (3x - 6)°.

Then the value of x will be

We know that the angles ∠MNQ and ∠QNP are supplementary angles. Then the value of the variable x will be

∠MNQ + ∠QNP = 180°

(2x + 1)° + (3x - 6)° = 180°

5x - 5° = 180°

5x = 185°

x = 37°

Then the measure of the angle ∠QNP will be

∠QNP = [3(37) - 6]°

∠QNP = 111° - 6°

∠QNP = 105°

The value of the variable x will be 37°. Then the measure of the angle ∠QNP will be 105°.

More about the angled link is given below.

brainly.com/question/15767203

#SPJ1

4 0
1 year ago
What operation would you use to solve x/4=20?
umka21 [38]

Answer:

Multiplication

Step-by-step explanation:

x=20•4

x=80

7 0
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))

4 0
3 years ago
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