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3 years ago

Which group of numbers even or odd include more prime numbers? Why?

Mathematics

2 answers:

serg [7]3 years ago

7 0

The First group i.e 1-10 include more prime numbers .(2,3,5,7,1)

Because it has the basic inputs of every number
Like 15 consists of 3×5

NikAS [45]3 years ago

6 0

Answer:

Odd numbers.

Step-by-step explanation:

There is one even prime number - 2.

All the rest primes are odd numbers.

Therefore odd numbers include more primes.

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Find missing angle.

Vsevolod [243]

I don’t really know exactly to what degree you need to put in your solution. But I rounded to the nearest tenth degree.
A = 112 degrees
B= 28 degrees (180-112-40 bc all sides of a triangle must equal 180)
C= 40
a= 27.6 (this is the side opposite of angle A)
b= 14 (side opposite b)
c= 19.2 (side opposite c)

HOW TO SOLVE:
c= law of sines, so c/sin(40) = 14/sin(28), multiply both sides by sin(40) so c can be isolated and solved for. c = 14sin(40)/sin(28). Plug into calculator then get answer. c is approximately 19.2.
a = law of sines again, so a/sin(112) = 14/sin(28). Multiply both sides again by sin(112) then solve. a = 14sin(112)/sin(28). Calculator again. a is around 27.6

7 0

3 years ago

This question has several parts that must be completed sequentially. If you skip a part of the question, you will not receive an

sleet_krkn [62]

Answer:

-6re−r [sin(6θ) - cos(6θ)]

Step-by-step explanation:

the Jacobian is ∂(x, y) /∂(r, θ) = δx/δθ × δy/δr - δx/δr × δy/δθ

x = e−r sin(6θ), y = er cos(6θ)

δx/δθ = -6rcos(6θ)e−r sin(6θ), δx/δr = -sin(6θ)e−r sin(6θ)

δy/δθ = -6rsin(6θ)er cos(6θ), δy/δr = cos(6θ)er cos(6θ)

∂(x, y) /∂(r, θ) = δx/δθ × δy/δr - δx/δr × δy/δθ

= -6rcos(6θ)e−r sin(6θ) × cos(6θ)er cos(6θ) - [-sin(6θ)e−r sin(6θ) × -6rsin(6θ)er cos(6θ)]

= -6rcos²(6θ)e−r (sin(6θ) - cos(6θ)) - 6rsin²(6θ)e−r (sin(6θ) - cos(6θ))

= -6re−r (sin(6θ) - cos(6θ)) [cos²(6θ) + sin²(6θ)]

= -6re−r [sin(6θ) - cos(6θ)] since [cos²(6θ) + sin²(6θ)] = 1

6 0

3 years ago

85.87 J or heat energy is added to a 34.8 g mass of substance the temperature rises from 21.76°C

Andru [333]

Answer:

The required specific heat is 196.94 joule per kg per °C

Step-by-step explanation:

Given as :

The heat generated = Q = 85.87 J

Mass of substance (m)= 34.8 gram = 0.0348 kg

Change in temperature = T2 - T1 = 34.29°C - 21.76°C = 12.53°C

Let the specific heat = S

Now we know that

Heat = Mass × specific heat × change in temperature

Or, Q = msΔt

Or, 85.87 = (0.0348 kg ) × S × 12.53°C

Or , 85.87 = 0.4360 × S

Or, S = $\frac{85.87}{0.4360}$

∴ S = 196.94 joule per kg per °C

Hence the required specific heat is 196.94 joule per kg per °C Answer

7 0

2 years ago

A crowd of people stood on both sides of the street to view a 4th of July Parade.

Dmitriy789 [7]

Given :

Length of parade , $L = 3\ miles=3\times 5280\ miles = 15840\ miles$ .

Breadth of parade , $B=2\times 15=30\ ft$ .

5 people can fit into a $25\ ft^2$ area.

1 mile = 5280 ft .

To Find :

The size of the crowd.

Solution :

Area of parade ,

$A=15840\times 30\ ft^2\\\\A=475200\ ft^2$

Also , 5 people can fit in $25\ ft^2$ area .

So , number of people fit in A is :

$n = \dfrac{475200\times 5}{25}\\\\n=95040$

Therefore , 95040 can fit in the parade .

Hence , this the required solution .

6 0

3 years ago

How do you simplify this math problem 5(1 + 3x) ?

lubasha [3.4K]

you need to distribute the 5. multiply each term in the parentheses by 5:

5(1) + 5(3x)

5 + 15x

5 0

3 years ago