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weeeeeb [17]
3 years ago
5

Mr.Johnson leaves on a trip planning to travel 24 mph

Mathematics
2 answers:
dmitriy555 [2]3 years ago
5 0
In conclusion, she is very slow
Artist 52 [7]3 years ago
4 0
Ummm is this a full question?
You might be interested in
The experimental probability that Pedro will hit a hole in one at the golf course is 2/5. If Pedro goes golfing 40 times in a ye
Sati [7]
I'm pretty sure you would need to multiply 2/5 by 40.

So, if you multiply 2/5 by 40, you need to turn 40 into a fraction with 1 being the denominator.

2/5 x 40/1 = 80/5

Since the product is an improper fraction, you would simplify it to a whole number.

80/5 = 16. 

He can expect to hit it 16 times to get a hole in one.

I hope I am right and have a great day.

6 0
3 years ago
Read 2 more answers
<img src="https://tex.z-dn.net/?f=789%20%5Ctimes%20%20%7B8%7D%5E%7B2%7D%20%20%3D%20" id="TexFormula1" title="789 \times {8}^{2}
Nikolay [14]

Answer:

50496

Explanation:

789 × 8 × 8 = 50496

------------------------------------------------------------------------------------

<em>Hope it helps, if you have any question, comment down below!</em>

5 0
2 years ago
Read 2 more answers
If the fifth term in a geometric sequence is 1∕27 and the common ratio is 1∕3, find the explicit formula of the sequence.
ch4aika [34]

Answer:

B

Step-by-step explanation:

8 0
3 years ago
Prove that: (b²-c²/a)CosA+(c²-a²/b)CosB+(a²-b²/c)CosC = 0​
IRISSAK [1]

<u>Prove that:</u>

\:\:\sf\:\:\left(\dfrac{b^2-c^2}{a}\right)\cos A+\left(\dfrac{c^2-a^2}{b}\right)\cos B +\left(\dfrac{a^2-b^2}{c}\right)\cos C=0

<u>Proof: </u>

We know that, by Law of Cosines,

  • \sf \cos A=\dfrac{b^2+c^2-a^2}{2bc}
  • \sf \cos B=\dfrac{c^2+a^2-b^2}{2ca}
  • \sf \cos C=\dfrac{a^2+b^2-c^2}{2ab}

<u>Taking</u><u> </u><u>LHS</u>

\left(\dfrac{b^2-c^2}{a}\right)\cos A+\left(\dfrac{c^2-a^2}{b}\right)\cos B +\left(\dfrac{a^2-b^2}{c}\right)\cos C

<em>Substituting</em> the value of <em>cos A, cos B and cos C,</em>

\longmapsto\left(\dfrac{b^2-c^2}{a}\right)\left(\dfrac{b^2+c^2-a^2}{2bc}\right)+\left(\dfrac{c^2-a^2}{b}\right)\left(\dfrac{c^2+a^2-b^2}{2ca}\right)+\left(\dfrac{a^2-b^2}{c}\right)\left(\dfrac{a^2+b^2-c^2}{2ab}\right)

\longmapsto\left(\dfrac{(b^2-c^2)(b^2+c^2-a^2)}{2abc}\right)+\left(\dfrac{(c^2-a^2)(c^2+a^2-b^2)}{2abc}\right)+\left(\dfrac{(a^2-b^2)(a^2+b^2-c^2)}{2abc}\right)

\longmapsto\left(\dfrac{(b^2-c^2)(b^2+c^2)-(b^2-c^2)(a^2)}{2abc}\right)+\left(\dfrac{(c^2-a^2)(c^2+a^2)-(c^2-a^2)(b^2)}{2abc}\right)+\left(\dfrac{(a^2-b^2)(a^2+b^2)-(a^2-b^2)(c^2)}{2abc}\right)

\longmapsto\left(\dfrac{(b^4-c^4)-(a^2b^2-a^2c^2)}{2abc}\right)+\left(\dfrac{(c^4-a^4)-(b^2c^2-a^2b^2)}{2abc}\right)+\left(\dfrac{(a^4-b^4)-(a^2c^2-b^2c^2)}{2abc}\right)

\longmapsto\dfrac{b^4-c^4-a^2b^2+a^2c^2}{2abc}+\dfrac{c^4-a^4-b^2c^2+a^2b^2}{2abc}+\dfrac{a^4-b^4-a^2c^2+b^2c^2}{2abc}

<em>On combining the fractions,</em>

\longmapsto\dfrac{(b^4-c^4-a^2b^2+a^2c^2)+(c^4-a^4-b^2c^2+a^2b^2)+(a^4-b^4-a^2c^2+b^2c^2)}{2abc}

\longmapsto\dfrac{b^4-c^4-a^2b^2+a^2c^2+c^4-a^4-b^2c^2+a^2b^2+a^4-b^4-a^2c^2+b^2c^2}{2abc}

<em>Regrouping the terms,</em>

\longmapsto\dfrac{(a^4-a^4)+(b^4-b^4)+(c^4-c^4)+(a^2b^2-a^2b^2)+(b^2c^2-b^2c^2)+(a^2c^2-a^2c^2)}{2abc}

\longmapsto\dfrac{(0)+(0)+(0)+(0)+(0)+(0)}{2abc}

\longmapsto\dfrac{0}{2abc}

\longmapsto\bf 0=RHS

LHS = RHS proved.

7 0
2 years ago
Consider the sequence 1, 4, 16, 64, 128, … Which statement describes the sequence? The sequence diverges. The sequence converges
vagabundo [1.1K]

Answer:

The sequence diverges

Step-by-step explanation:

The given sequence is 1, 4, 15, 64, 128, ...

By observation the sequence can be expressed as a sequence of 4 raised to increasing powers of 'x' as follows;

The sequence is 4ˣ, where; x = 0, 1, 2, 3, 4, ..., ∞

Therefore, the values of each subsequent term of the sequent is higher than the term it succeeds, the sequence does not converge to a given limit value and therefore, the sequence diverges.

5 0
2 years ago
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