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Fittoniya [83]
3 years ago
10

Solve the equation E = IR for I

Mathematics
1 answer:
alexgriva [62]3 years ago
3 0

Answer:

I=\frac{E}{R}; R\neq 0

Step-by-step explanation:

E=IR\\\\IR=E\\\\\frac{IR=E}{R}\\\\ \boxed{I=\frac{E}{R}; R\neq 0}

Hope this helps.

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How do I simply this, thanks!
stepan [7]
(-a^3b^2*-a^-2b^-3)^-2/2a^2b^-3 = a^4b^9/2a^8b^4 =b^5/2a^4 so your answer is b^5/2a^4
7 0
4 years ago
The equation 13x + 1 = 5 – x can be solved by graphing y = 13x + 1 and y = 5 – x.
shusha [124]

Answer:

The graph shows that when x =3 both

\frac{1}{3}x + 1

and 5 – x equal 2

Step-by-step explanation:

The equation 13x + 1 = 5 – x can be solved by graphing y = 13x + 1 and y = 5 – x.

From the graph, the two lines representing the two equations intersect at (3,2).

This implies that the solution to the system

y =  \frac{1}{3} x + 1

and

y=5-x

is x=3, y=2

Hence the value of the two functions equals 2 at x=3.

7 0
3 years ago
B) Prove or disprove by example or by exhaustive checking that every integer greater than 1 is either prime or sum of two primes
lisabon 2012 [21]

Answer:

Every integer greater than 1 is either prime or composite.

Step-by-step explanation:

Let us negate this statement.

Assumption - any number greater than 1, is neither prime nor composite.

A prime number has only 2 divisors, and a composite number has at least 3.

Let a number be X such that X >1.

Now X is divisible by 1 (Any integer is divisible by 1)

And X is divisible by X (X/X = 1, if X is not zero)

So by this, we see X has at least 2 divisors, X is either prime, or composite if it has more divisors!

So our assumption is wrong, and every integer greater than 1 is either prime or composite.

6 0
3 years ago
hey can someone please confirm my answer? i got D but it was a guess because i’m not confident in my graphing skills yet
NikAS [45]
Correct answer would be c, since the question is only asking for the solution and not the x valued (plotted left and right) it would be where both segments meet
8 0
3 years ago
Read 2 more answers
2. There are 1800 students enrolled at Springfield College, and 65% of them are female.
Kamila [148]

Answer:

a) 1170

b) 35%

c) 630

Step-by-step explanation:

6 0
3 years ago
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