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zlopas [31]
2 years ago
7

Determine whether each point lies on the graph of the equation.

Mathematics
1 answer:
sladkih [1.3K]2 years ago
3 0
Points 2,2
2=square root of 6-2
2=square root of 4
2=2
Yes point lies on graph

Points 6,0
0=square root of 6-6
0=0
Yes point lies on graph
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100 more than the difference between 88 and 66
asambeis [7]
Wouldn't the answer be 122
5 0
3 years ago
Read 2 more answers
I need all 5 of them please
aliina [53]
1.) 8.88= 4.44 (x -7)
8.88 = 4.44x - 31.08
+31.08 +31.08
39.96 = 4.44x
39.96 / 4.44x
9 = x
4 0
2 years ago
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A public beach in Florida charges non residents $8 per day for a fishing license and $2.50 per day for live bait. Florida reside
melamori03 [73]

For a non-resident:

C_{nr}=8d+2.5d=10.5d

For a resident:

C_{r}=114+1d

We want to know the number of days d such that the two costs are equal ie \{d|C_{nr}(d)=C_{r}(d)\}

To do this set the two costs equal to each other:

C_{nr}=C_{r}

10.5d=114+1d

And solve for d:

10.5d-1d=114+1d-1d

9.5d=114

d=\frac{114}{9.5}

d=12

The costs will be equal when a resident and a non-resident each use the beach for 12 days.



3 0
3 years ago
I need help with this one please
Semmy [17]

Answer:

A

Step-by-step explanation:

a geometric sequence is where we multiply a factor from element to element.

a1 = $900

a2 = 981 = a1 × f = 900 ×

a3 = 1069.29 = a2 × f = a1 × f × f = s1 × f²

an = 900 \times  {f}^{n - 1}

so, now let's try and get f.

remember, 981 = 900 × f

f = 981/900 = 109/100 = 1.09

just to control, we check for s3 :

900 × (1.09)² = 900 × 1.1881 = 1069.29

correct.

so,

a13 = 900 × (1.09)¹² = 2,531.398304

s13 is then the sum of all a1, ..., a13

there is a nice formula for sums of finite sequences

s13 = 900 × (1-f¹³) / (1-f) = 900×(1-(1.09)¹³) / (1-1.09) =

= 900×(1-3.065804612) / (-0.09) =

= 900×(-2.065804612) / (-0.09) = 20,658.04612

.

7 0
3 years ago
Please help!!!! thank you!!!!!
iris [78.8K]
A parallel set of lines are two lines that do not touch one another, but are on the same plane and are equidistant forever. Think about two highways that run next to each other but never touch or cross. CF, therefore, could be parallel to BE, AH, and DG.
8 0
3 years ago
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