Answer:
a transformation is a <u>rotation</u><u> </u><u>or</u><u> </u><u>reflection </u> to the <u>posit</u><u>ion</u> or co-ordinates of a figure
Answer:
x = 12
y = 9
12 coins of dimes
9 coins of quartes
Step-by-step explanation:
x ---> coins of dimes
y ---> coins of quartes
x + y = 21
0.1x + 0.25y = 3.45 * 100
x + y = 21 L1
10x + 25y = 345 L2
x = 21 - y
Reemplace in L2
10(21 - y) + 25y = 345
210 - 10y + 25y = 345
15y = 345 - 210
15y = 135
y = 9
x = 21 - y
x = 21 - 9
x = 12
Answer:
a. 2, 3, 6
b. a/b × c = 1
c. a = 2, b = 6, c = 3 or a = 3, b = 6, c = 2
Step-by-step explanation:
Apparently, Erica wants to multiply a fraction by a whole number to get a result of 1. That means the fraction is the multiplicative inverse of the whole number.
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<h3>a.</h3>
There is only one set of three unique numbers among those given such that one of them is a composite number with the other two as factors. Erica must choose 2, 3, and 6.
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<h3>b. </h3>
If the three numbers are represented by a, b, c, Erica wants ...
a/b × c = 1
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<h3>c.</h3>
Multiplying the equation by b gives ...
ac = b
Among the numbers 1–6, only 2×3 = 6 will satisfy this equation. Erica can choose either of ...
a = 2, b = 6, c = 3 ⇒ (2/6)×3 = 1
a = 3, b = 6, c = 2 ⇒ (3/6)×2 = 1
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<h3>d.</h3>
No scoring information is provided.
Multiply -3 to x and get -3x and multiply 4 and -3 and get -12 so then subtract -12 and 15 and get -3 so now it is -3x-3=6-4y so now. You subtract -3 and 6 and get positive 3 so now the final answer is -3x-4y+3
ANSWER
To graph the function

follow the steps below.
1. Find y- intercept by plugging in

.

is on the interval,

, so we substitute in to



Hence the y-intercept is

2. Find x-intercept by setting

This implies that

on

or

on

We now solve for

on each interval,

on

or

on

But observe that

does not belong to

This means it can never be an intercept for this piece-wise function.
Hence our x-intercept is

3. Plotting the boundaries of the interval.
For

on



.
This point

coincides with the x-intercept.


So we have the point

. But note that

does not belong to this interval so we plot this point as a hole.
For

on



So we plot



So we plot

also as a hole.
Plotting all these points we can now graph the function,

See attachment for graph.