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ruslelena [56]
2 years ago
9

Solve for v: 56= 7v + v simplify your answer as much as possible​

Mathematics
2 answers:
IRISSAK [1]2 years ago
5 0

Given equation :

{\qquad   \sf    \dashrightarrow{ 56= 7v + v }}

We can rewrite the equation as :

{\qquad   \sf    \dashrightarrow{  7v + v = 56 }}

Adding like terms :

{\qquad   \sf    \dashrightarrow{ 8v = 56 }}

Dividing both sides by 8 :

{\qquad   \sf    \dashrightarrow{  \dfrac{8v}{8}  =  \dfrac{56}{8}  }}

{\qquad   \bf    \dashrightarrow{ v = 7 }}

Henceforth,

  • The value of v is 7.
bulgar [2K]2 years ago
4 0

Answer:

v=7

Step-by-step explanation:

i just used a math website for my answers and told it to simplify. try using that website. i always get an 90-100's on my tests from using that

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3 years ago
What is the domain and range of y=x^2?
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Hi there! the best way of solving this is picturing out what the graph might look like. Let's assume you had the graph of a parabola y=x^2. You know that for every x you substitute, there'd always be a value for y. Thus, the domain is ALL REAL NUMBERS or from -INFINITY to + INFINITY. The range on the other hand is different. We know that any number raised to the second power will always yield a positive integer or 0. Thus, y=x^2 won't have any negative y-values as the graph opens upward. Therefore, the range is: ALL REAL NUMBERS GREATER THAN OR EQUAL TO 0. or simply: 0 to +INFINITY. 

<span>On the other hand, a cubic function y=x^3 is quite different from the parabola. For any x that we plug in to the function, we'd always get a value for y, thus there are no restrictions. And the domain is ALL REAL NUMBERS or from -INFINITY to + INFINITY. For the y-values, the case would be quite similar but different to that of the y=x^2. Since a negative number raised to the third power gives us negative values, then the graph would cover positive and negative values for y. Thus, the range is ALL REAL NUMBERS or from -INFINITY to + INFINITY. Good luck!!!:D</span>
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3 years ago
Read 2 more answers
How can 12 x 50 help find 12 x 49
Dafna1 [17]
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2 years ago
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