Answer:
C. RTS
Step-by-step explanation:
When two triangles are congruent, they will have congruent vertices, like each vertex has a partner.
Take a look at the shape of the triangle.
If you look at the triangle on the right, you can see it is almost like a right triangle.
"R" is at the right angle. "S" is on the short side and "T" is on the long side.
In the triangle on the left, "M" is at the right angle. "N" is on the short side and "O" is on the long side.
Thus, the pairs of congruent angles are:
M ≅ R
O ≅ T
N ≅ S
When you write the statement of congruence between triangles, order the letters by their congruent pair.
MON ≅ RTS
y + 11 = –2(x + 1.5)
y=mx+b
multiply the bracket by -2
(-2)(x)=-2x
(-2)(1.5)=-3
y+11=-2x-3
move +11 to the other side
sign changes from +11 to -11
y+11-11=-2x-3-11
y=-2x-14
m is also the slope
b= y-intercept
Answer:
y-intercept=-14
1. divide by the GCF it will give you smaller numbers
2.then factor
3. set each factor to = 0
4. you will have 2 values for x
hope this helps
Try to relax. Your desperation has surely progressed to the point where
you're unable to think clearly, and to agonize over it any further would only
cause you more pain and frustration.
I've never seen this kind of problem before. But I arrived here in a calm state,
having just finished my dinner and spent a few minutes rubbing my dogs, and
I believe I've been able to crack the case.
Consider this: (2)^a negative power = (1/2)^the same power but positive.
So:
Whatever power (2) must be raised to, in order to reach some number 'N',
the same number 'N' can be reached by raising (1/2) to the same power
but negative.
What I just said in that paragraph was: log₂ of(N) = <em>- </em>log(base 1/2) of (N) .
I think that's the big breakthrough here.
The rest is just turning the crank.
Now let's look at the problem:
log₂(x-1) + log(base 1/2) (x-2) = log₂(x)
Subtract log₂(x) from each side:
log₂(x-1) - log₂(x) + log(base 1/2) (x-2) = 0
Subtract log(base 1/2) (x-2) from each side:
log₂(x-1) - log₂(x) = - log(base 1/2) (x-2) Notice the negative on the right.
The left side is the same as log₂[ (x-1)/x ]
==> The right side is the same as +log₂(x-2)
Now you have: log₂[ (x-1)/x ] = +log₂(x-2)
And that ugly [ log to the base of 1/2 ] is gone.
Take the antilog of each side:
(x-1)/x = x-2
Multiply each side by 'x' : x - 1 = x² - 2x
Subtract (x-1) from each side:
x² - 2x - (x-1) = 0
x² - 3x + 1 = 0
Using the quadratic equation, the solutions to that are
x = 2.618
and
x = 0.382 .
I think you have to say that <em>x=2.618</em> is the solution to the original
log problem, and 0.382 has to be discarded, because there's an
(x-2) in the original problem, and (0.382 - 2) is negative, and
there's no such thing as the log of a negative number.
There,now. Doesn't that feel better.
Round off 8.02 to 8 and 5.98 to 6. Then subtract: 8 - 6 = 2 (approximately)
