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pickupchik [31]
3 years ago
6

Need some help well give Brainliest if you answer right

Mathematics
1 answer:
xenn [34]3 years ago
3 0

If this is asking 5% of $530 then the answer is 26.5

If this is asking that 530 is 5% then the answer is 10600

You might be interested in
5. Is the sequence arithmetic? If so, identify the common difference.
HACTEHA [7]
5.
aritmetic is same increase per term
14 to 21 is 7
21 to 42 is 21
7≠21
no


6.
decreases by 4 each time
an=a1+d(n-1)
an=10-4(n-1)
n=12 for 12th term
a12=10-4(12-1)
a12=10-4(11)
a12=10-44
a12=-34

it is -34



7.
missing term is y
22+n=y
y+n=34
22+n+n=34
22+2n=34
minus 22 from both sides
2n=12
divide both sides by 2
n=6
22+n=y
22+6=y
28=y

28 is answer



5. no
6. -34
7. 28
6 0
3 years ago
Read 2 more answers
In a survey of 3,500 people who owned a certain type of​ car, 1,225 said they would buy that type of car again. What percent of
Elis [28]
3,500 x .35 = 1,225.
So move the decimal to the right 2 places & that gives you 35% of the people were satisfied with the car.
I would not buy that car!
7 0
3 years ago
Consider the leading term of y = −2x2 − 3x + 3. What is the end behavior of the graph?
Tresset [83]
Y= -4 -3x +3 
y= -3x -1 
3 0
3 years ago
a. Write and simplify the integral that gives the arc length of the following curve on the given integral. b If necessary, use t
WINSTONCH [101]

Answer:

\int\limits^{\pi/2} _0 (1+4cos^{2} (2x)dx

Step-by-step explanation:

Arc length is calculated by dividing the arcs in to small segments ds

By pythagoren theorem

ds^2=dx^2+dy^2

then integrate ds to get arc length.

We are given a function as

y = sin 2x in the interval  [0, pi/2]

To find arc length in the interval

Arc length s =\int\limits^{\pi/2} _0 {1+y'^2} \, dx \\=\int\limits^{\pi/2} _0 (1+4cos^{2} (2x) )dx

Hence arc length would be

B) \int\limits^{\pi/2} _0 (1+4cos^{2} (2x)dx

6 0
3 years ago
Hello please help will give brainliest!
jeka57 [31]
The answer is A. because the first part shows the ratio 3:2 and the other 2 shows equivalent ratios.
8 0
4 years ago
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