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2 years ago

6

Hey can you please help me out posted picture

2 answers:

Harman [31]2 years ago

4 0

For each language we have:
Number of people who speak Spanish:
Spanish = 8 + 4 = 12
Number of people who speak Chinese:
Chinese = 5 + 6 = 11
Number of people who speak both languages:
Both = 3 + 2 = 5
Adding we have:
12 + 11 + 5 = 28
Answer:
28
option A

Anna007 [38]2 years ago

3 0

A because you have to do 3+2+6+5+4+8

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What is the rate of change?<br> the second 1 and second 4?

Mumz [18]

Idek but I do know that 12x8= 72

5 0

2 years ago

All of the middle school mathletes solve math problems at the same pace. Working together, six mathletes complete all of the sum

guapka [62]

Answer:

5 1/2 hours

Step-by-step explanation:

by a factor of 5 means times 4

6*4=24

22/4=5.5 hours

8 0

3 years ago

Look at the picture for the question please show work.

lukranit [14]

$A=\dfrac{1}{2}ah\\
a=25 \text{ mm}\\
h=14 \text{ mm}\\\\
A=\dfrac{1}{2}\cdot25\cdot14\\
A=175 \text{ mm}^2 \Rightarrow C$

6 0

3 years ago

Which ordered pair would form a proportional relationship with the point graphed below?

vampirchik [111]

Hey there! I'm happy o help!

To get from 60 to -20, you multiply by -1/3. This is our constant of proportionality. Let's see which points below follow this.

-10(-1/3)=3 1/3≠30

30(-1/3)=-10≠-15

-30(-1/3)=10

80(-1/3)=-26 2/3≠-30

Therefore, the correct answer is (-30,10).

Have a wonderful day! :D

6 0

3 years ago

Prove the following by induction. In each case, n is apositive integer.<br> 2^n ≤ 2^n+1 - 2^n-1 -1.

frutty [35]

<h2>Answer with explanation:</h2>

We are asked to prove by the method of mathematical induction that:

$2^n\leq 2^{n+1}-2^{n-1}-1$

where n is a positive integer.

Let us take n=1

then we have:

$2^1\leq 2^{1+1}-2^{1-1}-1\\\\i.e.\\\\2\leq 2^2-2^{0}-1\\\\i.e.\\2\leq 4-1-1\\\\i.e.\\\\2\leq 4-2\\\\i.e.\\\\2\leq 2$

Hence, the result is true for n=1.

Let us assume that the result is true for n=k

i.e.

$2^k\leq 2^{k+1}-2^{k-1}-1$

Now, we have to prove the result for n=k+1

i.e.

<u>To prove:</u> $2^{k+1}\leq 2^{(k+1)+1}-2^{(k+1)-1}-1$

Let us take n=k+1

Hence, we have:

$2^{k+1}=2^k\cdot 2\\\\i.e.\\\\2^{k+1}\leq 2\cdot (2^{k+1}-2^{k-1}-1)$

( Since, the result was true for n=k )

Hence, we have:

$2^{k+1}\leq 2^{k+1}\cdot 2-2^{k-1}\cdot 2-2\cdot 1\\\\i.e.\\\\2^{k+1}\leq 2^{(k+1)+1}-2^{k-1+1}-2\\\\i.e.\\\\2^{k+1}\leq 2^{(k+1)+1}-2^{(k+1)-1}-2$

Also, we know that:

$-2$

(

Since, for n=k+1 being a positive integer we have:

$2^{(k+1)+1}-2^{(k+1)-1}>0$ )

Hence, we have finally,

$2^{k+1}\leq 2^{(k+1)+1}-2^{(k+1)-1}-1$

Hence, the result holds true for n=k+1

Hence, we may infer that the result is true for all n belonging to positive integer.

i.e.

$2^n\leq 2^{n+1}-2^{n-1}-1$ where n is a positive integer.

6 0

3 years ago