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ElenaW [278]
3 years ago
13

Please help!!! I really don’t get this only I get where the 3 are but please help me for the 4/5 and -0.4!

Mathematics
1 answer:
Naily [24]3 years ago
8 0

Answer:

-0.4 should be in between 0 and -0.5, however be as close to -0.5 as possible while not being ON -0.5.

3 4/5 should be in between 3 and 4, so you can split 4/5 into 8/10 which are equal to each other. Then you can pretend to split the 3 and 4 gap into 10 equal pieces. After you split the gap, count 8 parts to the left of 3.

Step-by-step explanation:

I hope you understand what I mean.

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Please include an explanation !!
Nonamiya [84]

Answer:

D

Step-by-step explanation:

Volume for a sphere: 4/3(pi)r^3

4/3(pi)(1x1x1)= 4/3pi

hope this helped :)

3 0
3 years ago
If 19.3% of children and adolescents are obese (14.4M), what is the total number of children and adolescents overall? Please i n
Anna35 [415]

Answer:

74,611,399

Step-by-step explanation:

Percentage of obese children and adolescent = 19.3%

19.3% = 14.4 million

The overall number of children and adolescent equals a total percentage of 100%

Hence,

If 19.3 % = 14.4 million

100% = x

Cross multiply :

19.3% x = 100% * 14.4 million

0.193x = 14400000

x = 14400000 / 0.193

x = 74611398.964

Total number of children and adolescent = 74,611,399

4 0
3 years ago
If p varies inversely as q and p =8 when q=10,find p when q=4
Akimi4 [234]

Answer:

20

Step-by-step explanation:

When two numbers vary inversely, they always multiply to the same product, in this case, it is 80, because it is 8 x 10(product of two known values of p and q). This means, for q = 4, we can write:

p * 4 = 80

This means that p is equal to 20.

7 0
3 years ago
Which trig ratio uses hypotenuse and adjacent side
e-lub [12.9K]

Answer:

It's cosine.

Step-by-step explanation:

Cosine is Adjacent / Hypotenuse.

7 0
4 years ago
Read 2 more answers
If S_1=1,S_2=8 and S_n=S_n-1+2S_n-2 whenever n≥2. Show that S_n=3⋅2n−1+2(−1)n for all n≥1.
Snezhnost [94]

You can try to show this by induction:

• According to the given closed form, we have S_1=3\times2^{1-1}+2(-1)^1=3-2=1, which agrees with the initial value <em>S</em>₁ = 1.

• Assume the closed form is correct for all <em>n</em> up to <em>n</em> = <em>k</em>. In particular, we assume

S_{k-1}=3\times2^{(k-1)-1}+2(-1)^{k-1}=3\times2^{k-2}+2(-1)^{k-1}

and

S_k=3\times2^{k-1}+2(-1)^k

We want to then use this assumption to show the closed form is correct for <em>n</em> = <em>k</em> + 1, or

S_{k+1}=3\times2^{(k+1)-1}+2(-1)^{k+1}=3\times2^k+2(-1)^{k+1}

From the given recurrence, we know

S_{k+1}=S_k+2S_{k-1}

so that

S_{k+1}=3\times2^{k-1}+2(-1)^k + 2\left(3\times2^{k-2}+2(-1)^{k-1}\right)

S_{k+1}=3\times2^{k-1}+2(-1)^k + 3\times2^{k-1}+4(-1)^{k-1}

S_{k+1}=2\times3\times2^{k-1}+(-1)^k\left(2+4(-1)^{-1}\right)

S_{k+1}=3\times2^k-2(-1)^k

S_{k+1}=3\times2^k+2(-1)(-1)^k

\boxed{S_{k+1}=3\times2^k+2(-1)^{k+1}}

which is what we needed. QED

6 0
3 years ago
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