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avanturin [10]
3 years ago
11

2x + 5y = 17 what is x what is y

Mathematics
1 answer:
andrew-mc [135]3 years ago
8 0

Answer:

x = (17 - 5y)/2

y = (17 - 2x)/5

Step-by-step explanation:

nothing else can be done here.

if you are looking for actual numbers as solution, then you have forgotten to put something important here.

to solve a system with 2 variables we need 2 equations. it cannot be solved with only 1 equation.

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What is -3 the sum of 5
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The solution is 2. -3+5=2
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Sherry has taken 10 quizzes. Her combined total for the 10 quizzes is 83 points. Nine of her quiz scores are shown on the plot.
castortr0y [4]

Answer:

9

Step-by-step explanation:

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4 years ago
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Write an inequality for the given graph.
Aliun [14]
Y=MX+B

M = slope

B = y intercept (number that intersects with the line in the y axis)

y=-3/4x+2

It starts on the y intercept (2) then we travel to the next point (4 on the x axis)

So we travel 3 spaces down (-3) landing on the origin (0,0) then four spaces to the right (4). Getting the slope of -3/4 (rise over run)

Then just plug in these into the equation

y = -3/4 + 2
5 0
3 years ago
A skirt is on sale for 15% the sale price is $48
Ierofanga [76]

Answer:

15% x 48=7.2 so 48-7.2=40.8

Step-by-step explanation:

4 0
4 years ago
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Recall, we have five connectives in propositional logic ¬, ∧, ∨, →, ↔ (negation, conjunction, disjunction, conditional and bicon
Free_Kalibri [48]

Answer:

(a) ¬(p→¬q)

(b) ¬p→q

(c) ¬((p→q)→¬(q→p))

Step-by-step explanation

taking into account the truth table for the conditional connective:

<u>p | q | p→q </u>

T | T |   T    

T | F |   F    

F | T |   T    

F | F |   T    

(a) and (b) can be seen from truth tables:

for (a) <u>p∧q</u>:

<u>p | q | ¬q | p→¬q | ¬(p→¬q) | p∧q</u>

T | T |  F  |   F     |    T       |  T

T | F |  T  |  T      |    F       |  F

F | T |  F  |  T      |    F       |  F

F | F |  T  |  T      |    F       |  F

As they have the same truth table, they are equivalent.

In a similar manner, for (b) p∨q:

<u>p | q | ¬p | ¬p→q | p∨q</u>

T | T |  F  |   T     |    T    

T | F |  F  |   T     |    T    

F | T |  T  |   T     |    T    

F | F |  T  |   F     |    F    

again, the truth tables are the same.

For (c)p↔q, we have to remember that p ↔ q can be written as (p→q)∧(q→p). By replacing p with (p→q) and q with (q→p) in the answer for part (a) we can change the ∧ connector to an equivalent using ¬ and →. Doing this we get ¬((p→q)→¬(q→p))

4 0
3 years ago
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