Answer:
Possible locker numbers for Selma are: 3, 5 and 29
Step-by-step explanation:
First of all, let us have a look at the definition of a prime number.
A <em>prime number</em> is a number which is divisible either by 1 or the number itself.
No other number divides the prime number other than 1 and the number itself.
Now, let us factorize the given number 435 and let us observe the what all possibilities are there.
Factorizing 145 further:
No further factors are possible.
Therefore 3, 5 and 29 are the prime factors of 435.
So, the answer is:
Possible locker numbers for Selma are: 3, 5 and 29
They intersect at a right angle
Just by comparing the plots of f(x) and g(x), it's clear that g(x) is just some positive scalar multiple of f(x), so that for some constant k, we have
g(x) = k • f(x) = kx² = (√k x)²
The plot of the transformed function g(x) = (√k x)² passes through the point (1, 4), which means
g(1) = (√k • 1)² = 4
and it follows that k = 4. So g(x) = 4x² = (2x)² and B is the correct choice.
Answer:
the graph is on the photo
Step-by-step explanation:
Answer: x = 0 y = –3
Step-by-step explanation:
We can solve this system by substitution.
use the second equation as a value for x.
Then substitute that value in place of x in the first equation and solve for y.
6x - 5y = 15 becomes
6(y + 3) - 5y = 15 Distribute 6 × parentheses
6y + 18 - 5y = 15 combine like terms
6y -5y + 18 = 15
y + 18 = 15 Subtract 18 from both sides.
y =15 -18
y = –3
Use this value for y in either equation to solve for x
6x - 5(-3) = 15
6x + 15 = 15 Subtract 15 from both sides, divide by 6 (seems silly!)
x = 0
OR in he second equation,
x = -3 + 3
Again, x = 0
