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TiliK225 [7]
3 years ago
14

Linear equations for Pre-IB Algebra 2. Solve for s, R(s + t) = st.

Mathematics
1 answer:
Anastaziya [24]3 years ago
3 0

R(s + t) = st

Rs + Rt = st

Rs - st = -Rt

s ( R - t) = -Rt

*s = - Rt / (R - t) Multiply through by -1 on the right.

*s = Rt / (t - R) Put in the brackets in your answer. What you have been writing means

s = Rt/r - R you need the brackets. The stars mean that either could be an answer. I think your problem could be the brackets. Now I'm reasonably certain this is correct.

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A weight lifter can bench press 189 kg. How many milligrams (mg) is this?
Inessa [10]

A weight lifter can bench press 189 kg. How many milligrams (mg) is this?

Answer: We know that:

1kg=1,000,000 mg

We are given, that a weight lifter can bench press 189 kg

Therefore, a weight lifter can bench press:

189 \times 1,000,000 mg

189,000,000mg

18.9 \times 10^{7} mg

Therefore, 18.9 \times 10^{7} mg is equal to the 189 kg

3 0
3 years ago
Prove or disprove (from i=0 to n) sum([2i]^4) <= (4n)^4. If true use induction, else give the smallest value of n that it doe
ddd [48]

Answer:

The statement is true for every n between 0 and 77 and it is false for n\geq 78

Step-by-step explanation:

First, observe that, for n=0 and n=1 the statement is true:

For n=0: \sum^{n}_{i=0} (2i)^4=0 \leq 0=(4n)^4

For n=1: \sum^{n}_{i=0} (2i)^4=16 \leq 256=(4n)^4

From this point we will assume that n\geq 2

As we can see, \sum^{n}_{i=0} (2i)^4=\sum^{n}_{i=0} 16i^4=16\sum^{n}_{i=0} i^4 and (4n)^4=256n^4. Then,

\sum^{n}_{i=0} (2i)^4 \leq(4n)^4 \iff \sum^{n}_{i=0} i^4 \leq 16n^4

Now, we will use the formula for the sum of the first 4th powers:

\sum^{n}_{i=0} i^4=\frac{n^5}{5} +\frac{n^4}{2} +\frac{n^3}{3}-\frac{n}{30}=\frac{6n^5+15n^4+10n^3-n}{30}

Therefore:

\sum^{n}_{i=0} i^4 \leq 16n^4 \iff \frac{6n^5+15n^4+10n^3-n}{30} \leq 16n^4 \\\\ \iff 6n^5+10n^3-n \leq 465n^4 \iff 465n^4-6n^5-10n^3+n\geq 0

and, because n \geq 0,

465n^4-6n^5-10n^3+n\geq 0 \iff n(465n^3-6n^4-10n^2+1)\geq 0 \\\iff 465n^3-6n^4-10n^2+1\geq 0 \iff 465n^3-6n^4-10n^2\geq -1\\\iff n^2(465n-6n^2-10)\geq -1

Observe that, because n \geq 2 and is an integer,

n^2(465n-6n^2-10)\geq -1 \iff 465n-6n^2-10 \geq 0 \iff n(465-6n) \geq 10\\\iff 465-6n \geq 0 \iff n \leq \frac{465}{6}=\frac{155}{2}=77.5

In concusion, the statement is true if and only if n is a non negative integer such that n\leq 77

So, 78 is the smallest value of n that does not satisfy the inequality.

Note: If you compute  (4n)^4- \sum^{n}_{i=0} (2i)^4 for 77 and 78 you will obtain:

(4n)^4- \sum^{n}_{i=0} (2i)^4=53810064

(4n)^4- \sum^{n}_{i=0} (2i)^4=-61754992

7 0
3 years ago
Will someone help me with this please
ANEK [815]

the first answer choice, associative property of addition

hope this helps. gl!

8 0
3 years ago
Read 2 more answers
Can anyone help me do this ​
SVETLANKA909090 [29]

Answer:

b) 42,610

Step-by-step explanation:

46089 - 3479 = 42610

4 0
2 years ago
Read 2 more answers
Which equation is correctly setup to solve for x?
AURORKA [14]

The formula for the hypotenuse of a triangle is:

[a squared] plus [b squared] = c squared

a and b represent the two sides of the triangle.

You would have to square 8 and 4, add them, then take the square root of it to find the length of the hypotenuse.

<h2>The first equation is correct.</h2>
5 0
2 years ago
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