All categories

3 years ago

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

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

Hope it helps, if you have any question, comment down below!

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:

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]

Prove that:

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

Proof: 

We know that, by Law of Cosines,

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

Taking LHS

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

Substituting the value of cos A, cos B and cos C,

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

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

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

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

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

On combining the fractions,

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

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

Regrouping the terms,

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

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

$\longmapsto\dfrac{0}{2abc}$

$\longmapsto\bf 0=RHS$

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