The answer is C.
<span>C. A markup has no limit, but a markdown is limited to 100% or less. </span>
Answer:
23. x = 4; DE = 44
24. x = 25; DS = 28
Step-by-step explanation:
23. Point S is the midpoint of DE, so ...
DS = SE
3x +10 = 6x -2
12 = 3x . . . . . . . . . add 2-3x
4 = x . . . . . . . . . . . divide by 3
Then DS has length ...
DS = 3x +10 = 12 +10 = 22
and DE is twice that length, so ...
DE = 44
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24. DS is half the length of DE, so is ...
DS = DE/2 = 56/2
DS = 28
Then x can be found from ...
DS = x +3
28 -3 = x = 25 . . . . . substitute value for DS
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<em>Comment on problem 24</em>
Sometimes it is easier to work parts of a problem out of sequence. Here, finding DS first makes finding x easier.
Answer:
A) 5x-6y=3
B) 7y=2x+8
B) -2x + 7y = 8 then multiplying "B" by 2.5
B) -5x +17.5y = 20 then adding this to A
A) 5x -6y = 3
11.5y = 23
y = 2
x = 3
Step-by-step explanation:
Answer: Total cost is $1.75
Explanation:
Since we have given that
Number of postcards bought by Doug's family= 7
Cost of each postcard including tax = $0.25
Total cost is given by
Now, we want to show this in number line :
On the number line put all the number with a difference of 0.25 and run the number line seven times and we get the answer.
Each step is of length 0.25 and there are 7 jumps to reach required answer.
So, after 7 jumps we reach at 1.75 which is the required answer.
Hence total cost is $1.75
First we need to find the gradient of K
which is y1-y2/x1-x2
(-1,3) and (5,-2)
so it becomes 3-(-2)/-1-5
m=-5/6
when two lines are perpendicular their gradients multiply to make -1
that means the gradient of L has to be 6/5
we can substitute the point on L (5,-2) and the gradient of 6/5 into y=mx+c
-2 = (6/5) x 5 + c
c = -8
the equation of line L is y= 6x/5 -8
