What is the rule for the reflection?

aleksley [76]

You take the amount paid and divide it by the amount of cheese. So 10.50 divided by 2.5 equals 4.2 and 12.60 divided by 3 equals 4.2 meaning that one pound of cheese costs $4.20

6 0

4 years ago

When asked to write the domain for the equation y= |x - 6| + 3, Mary wrote the Domain is y < 3. What mistake(s) did Mary make

castortr0y [4]

Answer:

Ok, the domain is the set of values that we can input in a function.

In this case, we have:

y = Ix - 6I + 3.

Notice that there is no restriction here, x can actually take any value, then the domain will be the set of all real numbers.

The correct domain is x, x ∈ R

Now, if we had (for example) something like:

y = Ix - 6I < 3

Now we have a restriction in the domain because we can not have y equal or larger than 3.

To find the domain, we can break the absolute value:

Ix - 6I < 3

is equivalent to:

-3 < x - 6 < 3

now let's add 6 in each side.

-3 + 6 < x - 6 + 6 < 3 + 6

3 < x < 9

That will be the domain in that case.

6 0

3 years ago

Given that point A is the midpoint of line segment ED, where E( -1,9) and A(4,5), what is the location of point D?

yuradex [85]

Answer:

D(9, 1)

Step-by-step explanation:

A being the midpoint means its coordinates satisfy ...

A = (E + D)/2

Solving for D, we find ...

D = 2A -E

= 2(4, 5) -(-1, 9) = (2·4 +1, 2·5 -9)

D = (9, 1)

8 0

4 years ago

A motorboat takes 5 hours to travel 100mi going upstream. The return trip takes 2 hours going downstream. What is the rate of th

Temka [501]

The boat went 5hours upstream, let's say it has a "still water" speed rate of "b", it went 100miles... however, going upstream is going against the current, let's say the current has a speed rate of "c"

so, when the boat was going up, it wasn't really going "b" fast, it was going " b - c " fast, because the current was eroding speed from it

now, when coming down, the return trip, well the length is the same, so the distance is also 100miles, it only took 2hrs though, because, the boat wasn't coming down "b" fast, it was coming down " b + c " fast, because the current was adding speed to it, so it came down quicker

now, recall your d = rt, distance = rate * time

$\bf \begin{array}{lccclll}
&distance&rate&time\\
&-----&-----&-----\\
upstream&100&b-c&5\\
downstream&100&b+c&2
\end{array}
\\\\\\
\begin{cases}
100=(b-c)5\\
\qquad \frac{100}{5}=b-c\\
\qquad 20=b-c\\
\qquad 20+c=\boxed{b}\\
100=(b+c)2\\
\qquad 50=b+c\\
\qquad 50=\boxed{20+c}+c
\end{cases}$

solve for "c", to see what's the current's speed

what's "b"? well 20+c = b

4 0

3 years ago

PLEASE HELP! I have a picture of the question below :)

Charra [1.4K]

3. ∠1+∠2=180° and ∠2+∠3=180°

4. they are supplementary

5. ∠1+∠2=180°=∠2+∠3

6. ∠1+∠2=180°=∠2+∠3

∠2 are same in both side so ∠1=180°-∠2 =∠3=180°-∠2

7.

6 0

3 years ago