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Arisa [49]
3 years ago
5

The results of a survey are shown below. In the survey, 12 students said that they would like to learn French. How many of the s

tudents surveyed would like to learn Spanish?

Mathematics
2 answers:
Ivenika [448]3 years ago
7 0

Answer:

18

Step-by-step explanation:

24 divided by 12 is 2. 36 divided by 2 is 18.

Hope this is right.

myrzilka [38]3 years ago
5 0

Answer:

18 students would like to learn Spanish

Step-by-step explanation:

first find the whole

24%=0.24

12/0.24=50

next find how many students would like to learn spanish

36%=0.36

50*0.36=18

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What is 1+1? Need help with my son's online school I never went to school so no one has ever taught me math​
goldfiish [28.3K]

Answer:

2

Step-by-step explanation:

.3.

4 0
3 years ago
Use Order of Operations to simplify. 42 – (28 ÷ 7) + 111
Mariulka [41]
do parentheses first which gives you 42-4+111 then aubtract to get 38+111 which equals 149
8 0
3 years ago
Read 2 more answers
Please helpp <br> find the missing length indicated
Murrr4er [49]

Answer:

i think its C: none of the answers are correct

7 0
2 years ago
Read 2 more answers
3. If 3, x, y, 18, are in arithmetic progression, (A, P) find the value of x and y
Luba_88 [7]

Answer:

Below in bold

Step-by-step explanation:

The sequence is:

3, x, y, 18

If this is an A P then

x - 3 = y - x  

2x - y = 3      (A)  and

y - x = 18 - y

2y - x = 18      (B)

Multiply  (A) by 2:

4x - 2y = 6    (C)

Adding B and C:

3x = 24

x = 8.

and

2y - 8 = 18

2y = 26

y = 13.

So x = 8 and y = 13.

b)   ar + ar^2 = 6ar^3  where a = first term and r = common ratio

Divide by a:

r + r^2 = 6r^3

6r^3 - r^2 - r = 0

r(6r^2 - r - 1) = 0

r(3r  + 1)(2r  - 1) = 0

So the 2 possible values of r

= -1/3 and 1/2.

i) The common ratio is positive so it must be 1/2.

Second term ar = 8

1/2 a = 8

a = 16.

So the first 6 terms are:

16, 8, 4, 2, 1, 1/2.

5 0
2 years ago
I really really really need help!!!!
Yakvenalex [24]
For f to be continuous at x=1, you need to have the limit from either side as x\to1 to be the same.

\displaystyle\lim_{x\to1^-}f(x)=\lim_{x\to1^-}(|x-1|+2)=2
\displaystyle\lim_{x\to1^+}f(x)=\lim_{x\to1^+}(ax^2+bx)=a+b

If a=2 and b=3, then the limit from the right would be 2+3=5\neq2, so the answer to part (1) is no, the function would not be continuous under those conditions.

This basically answers part (2). For the function to be continuous, you need to satisfy the relation a+b=2.

Part (c) is done similarly to part (1). This time, you need to limits from either side as x\to2 to match. You have

\displaystyle\lim_{x\to2^-}f(x)=\lim_{x\to2^-}(ax^2+bx)=4a+2b
\displaystyle\lim_{x\to2^+}f(x)=\lim_{x\to2^+}(5x-10)=0

So, a and b have to satisfy the relation 4a+2b=0, or 2a+b=0.

Part (4) is done by solving the system of equations above for a and b. I'll leave that to you, as well as part (5) since that's just drawing your findings.
8 0
2 years ago
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