9514 1404 393
Answer:
Step-by-step explanation:
Let d represent the number of daffodil bulbs you buy. Then 60-d is the number of crocus bulbs. The total cost will be ...
0.45(60 -d) +0.65(d) = 35.00
27 +0.20d = 35
0.20d = 8 . . . . . . . . . subtract 27
d = 40 . . . . . . . . . . multiply by 5
60 -40 = 20 . . . . . crocus bulbs
You can buy 20 crocus bulbs and 40 daffodil bulbs for $35.
Problem 7)
The answer is choice B. Only graph 2 contains an Euler circuit.
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To have a Euler circuit, each vertex must have an even number of paths connecting to it. This does not happen with graph 1 since vertex A and vertex D have an odd number of vertices (3 each). The odd vertex count makes it impossible to travel back to the starting point, while making sure to only use each edge one time only.
With graph 2, each vertex has exactly two edges attached to it. So an Euler circuit is possible here.
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Problem 8)
The answer is choice B) 5
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Work Shown:
abc base 2 = (a*2^2 + b*2^1 + c*2^0) base 10
101 base 2 = (1*2^2 + 0*2^1 + 1*2^0) base 10
101 base 2 = (1*4 + 0*2 + 1*1) base 10
101 base 2 = (4 + 0 + 1) base 10
101 base 2 = 5 base 10
Answer:
store A- 4
Store B-1
Store C-3
Store D- 2
Step-by-step explanation:
gotta help a brotha out if i was correct Drop a sub at iToxicMakersYT
Answer: Choice C
x intercept is (4,0)
y intercept is (0,2)
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Work Shown:
To find the x intercept, we plug in y = 0 and solve for x
2x+4y = 8
2x+4(0) = 8
2x+0 = 8
2x = 8
x = 8/2
x = 4
We have x = 4 pair up with y = 0. So we have (x,y) = (4,0) as the x intercept.
This is the location where the graph crosses the x axis.
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The y intercept is a similar story, but we use x = 0 to find y.
2x+4y = 8
2(0)+4y = 8
0+4y = 8
4y = 8
y = 8/4
y = 2
The value x = 0 leads to y = 2. This gets us (x,y) = (0,2) as the y intercept.
This is the location where the graph crosses the y axis.
Answer:
4x / (2x^2+1)
Step-by-step explanation:
g(x)=ln(2x^2+1)
Using u substitution
u= 2x^2 +1
du = 4x dx
g'(x) = d/du ( g(u) du)
We know that the derivative of ln(u) = 1/u since this is always positive
= 1/ u* du
Substituting x back into the equation
g'(x) = 1/(2x^2+1) * 4x
= 4x / (2x^2+1)
