What the answer rate you 5 star

WHAT NUMBER OCCURS FREQUENTLY<br><br><br><br><br>ILL GIVE BRAINLIEST<br>

xxMikexx [17]

Answer:

I think it's 2 because it has a larger number of dots than other numbers.

Step-by-step explanation:

Hope this helps!!!

5 0

2 years ago

Read 2 more answers

Tired of WRESTLING, need you to pend this problem to the canvas... -1/2~3/4= ? "Divide"

Alexxx [7]

-2/3 is the simplified form of the expression -1/2 ÷ 3/4.

<h3>What is the simplified form of the given expression?</h3>

Given the expression in the question;

-1/2 ÷ 3/4

To divide by a fraction, multiply by its reciprocal.

-1/2 ÷ 3/4

Reciprocal of 3/4 is 4/3

Hence

-1/2 × 4/3

Next, we cancel out the common factor of 2

Factor out 2 from 4

-1/2 × 2(2)/3

-1/1 × 2/3

( -1 × 2 ) / ( 1 × 3 )

(-2) / (3)

-2/3

Therefore, the simplified form is -2/3

Learn how to solve more fraction problems here: brainly.com/question/1627825

#SPJ1

8 0

1 year ago

Which does NOT have a value of 4 when

12345 [234]

Answer:

3 beause when you dividend it -5.6+4(3) .

7 0

3 years ago

Tell how you know when you have found the prime factorization of a number

sweet-ann [11.9K]

When you end up with prime numbers for example
If you factored out 12 you could say 3x4. Now 3 is a prime number so it cant be broken down any more. You can break down 4 into 2x2. 2 is a prime number so cant be broken down any further. When 12 is factored out it is 2x2x3

7 0

3 years ago

Read 2 more answers

A wire b units long is cut into two pieces. One piece is bent into an equilateral triangle and the other is bent into a circle.

mezya [45]

1. Divide wire b in parts x and b-x.

2. Bend the b-x piece to form a triangle with side (b-x)/3

There are many ways to find the area of the equilateral triangle. One is by the formula A= $\frac{1}{2}sin60^{o}side*side= \frac{1}{2} \frac{ \sqrt{3} }{2} (\frac{b-x}{3}) ^{2}= \frac{ \sqrt{3} }{36}(b-x)^{2}$
A= $\frac{ \sqrt{3} }{36}(b-x)^{2}=\frac{ \sqrt{3} }{36}( b^{2}-2bx+ x^{2} )=\frac{ \sqrt{3} }{36}b^{2}-\frac{ \sqrt{3} }{18}bx+ \frac{ \sqrt{3} }{36}x^{2}$

Another way is apply the formula A=1/2*base*altitude,
where the altitude can be found by applying the pythagorean theorem on the triangle with hypothenuse (b-x)/3 and side (b-x)/6

3. Let x be the circumference of the circle.

$2 \pi r=x$

so $r= \frac{x}{2 \pi }$

Area of circle = $\pi r^{2}= \pi ( \frac{x}{2 \pi } )^{2} = \frac{ \pi }{ 4 \pi ^{2} }* x^{2} = \frac{1}{4 \pi } x^{2}$

4. Let f(x)= $\frac{ \sqrt{3} }{36}b^{2}-\frac{ \sqrt{3} }{18}bx+ \frac{ \sqrt{3} }{36}x^{2}+\frac{1}{4 \pi } x^{2}$

be the function of the sum of the areas of the triangle and circle.

5. f(x) is a minimum means f'(x)=0

f'(x)= $\frac{ -\sqrt{3} }{18}b+ \frac{ \sqrt{3} }{18}x+\frac{1}{2 \pi } x$ =0

$\frac{ -\sqrt{3} }{18}b+ \frac{ \sqrt{3} }{18}x+\frac{1}{2 \pi } x=0$

$(\frac{ \sqrt{3} }{18}+\frac{1}{2 \pi }) x=\frac{ \sqrt{3} }{18}b$

$x= \frac{\frac{ \sqrt{3} }{18}b}{(\frac{ \sqrt{3} }{18}+\frac{1}{2 \pi }) }$

6. So one part is $\frac{\frac{ \sqrt{3} }{18}b}{(\frac{ \sqrt{3} }{18}+\frac{1}{2 \pi }) }$ and the other part is b- $\frac{\frac{ \sqrt{3} }{18}b}{(\frac{ \sqrt{3} }{18}+\frac{1}{2 \pi }) }$

4 0

3 years ago

Read 2 more answers