during a bike challenge, riders have to collect various colored ribbons. each 1/2 mile they collect a red ribbon, each 1/8 mile they collect a green ribbon, and each 1/4 mile they collect a blue ribbon. which colors of ribbons will be collected at the 3/4 mile marker?

Answers

Answer 1

Answer: The correct answer is:
green and blue.

Explanation:
We want to see which fractions $(3)/(4)$ is a multiple of. We know that $(3)/(4)$ is a multiple of $(1)/(4)$ , because $(1)/(4)$ *3= $(3)/(4)$ .

We can divide fractions to determine if $(3)/(4)$ is a multiple of $(1)/(2)$ :
$( (3)/(4))/( (1)/(2))$ ;

in order to divide fractions, flip the second one and multiply:
$(3)/(4)$ )* $(2)/(1)$ = $(6)/(4)$ =1 $(1)/(2)$ .
This did not divide evenly, so $(3)/(4)$ is not a multiple of 1/2.

Checking to see if $(3)/(4)$ is a multiple of $(1)/(8)$ , $( (3)/(4))/( (1)/(8))$ ;

flip the second one and multiply:
$(3)/(4)$ * $(8)/(1)$ = $(24)/(4)$ =6.
This divided evenly, so $(3)/(4)$ is a multiple of $(1)/(8)$ .

Answer 2

Answer:

Answer: Hello there!

As we know, each 1/2 mile there is a red ribbon, each 1/8 mile there is a green ribbon, and each 1/4 mile there is a blue ribbon.

We want to know which color of ribbons is in the 3/4 mile marker.

then we need to see if 3/4 is a multiple of some of the previous numbers.

Start with the red ribbons, we need to see the quotient between 3/4 and 1/2, if the result is a whole number, then 3/4 is a multiple of 1/2

this is q = (3/4)/(1/2) = (3/4)*2 = 6/4

here we haven't a whole number, then there is no red ribbon at the 3/4 mile marker.

now with the green ones:

(3/4)/(1/8) = (3/4)*8 = 24/4 = 6

there we have a whole number, this means that at the 3/4 mile marker, we will see the sixth green ribbon, then we have a green ribbon in this marker.

now with the blue:

(3/4)/(1/4) = (3/4)*4 = 3

At the 3/4 mile marker, we will see the third blue ribbon, then there is a blue ribbon in this marker.

Then we know that in the 3/4 mile marker, there are a green and a blue ribbon.

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Prove 4t + t = 5t4t + t

t(4 + 1)

t(5)

5t

What property or properties were used to prove that the expressions are equivalent?
commutative and distributive properties
associative and distributive properties
associative and commutative properties
associative property

Answers

The proof used the distributive property and the associative property of multiplication.

Option B is the correct answer.

What is an expression?

An expression contains one or more terms with addition, subtraction, multiplication, and division.

We always combine the liketerms in an expression when we simplify.

We also keep all the like terms on one side of the expression if we are dealing with two sides of an expression.

Example:

1 + 3x + 4y = 7 is an expression.com

3 + 4 is an expression.

2 x 4 + 6 x 7 – 9 is an expression.

33 + 77 – 88 is an expression.

We have,

To be specific, the distributiveproperty was used to break down the expression 4t + t as:

4t + t = (4 + 1)t ________(1)

And then the associativeproperty was used to rearrange the terms as:

(4 + 1)t = 5t ________(2)

From (1) and (2)

We have shown that 4t + t is equivalent to 5t.

Thus,

The proof used the distributive property and the associative property of multiplication.

Learn more about expressions here:

brainly.com/question/3118662

#SPJ2

Answer:

its A: commutative and distributive properties

Step-by-step explanation:

i got it right

Line segment CD has a length of 3 units. It is translated 2 units to the right on a coordinate plane to obtain line segment C'D'. What is the length of C'D'?

Answers

It's still 3 units long. Being translated changes only the location, not the size

Consider the function represented by the equation 1/2j+1/4K=3 which shows the equation written in function notation with k as the independent variable

Answers

Answer:

The required expression is:

$k=12-2j$

Step-by-step explanation:

We have been given an expression:

$\fraxc{1}{2}j+(1)/(4)k=3$

We have to write the equation in terms when k is independent.

$(1)/(4)k=3-(1)/(2)j$

$\Rightarrow k=4(3-(1)/(2)j)$

$\Rightarrow k=2(6-j)$

$\Rightarrow k=12-2j$

The required expression is:

$k=12-2j$

In a right triangle, find the length of the side not given, if a=8 and b=15

Answers

to find the length of a missing side of a right triangle you have to use the Pythagorean theorem. so 8*8=64 15*15=225 so to find side c you add the two answers 225+64 and the answer is and find the square root which is 17. Hope this helps

   a² + b² = c²
(8)² + (15)² = c²
   64 + 225 = c²
   289 = c²
       17 = c

Which equation has the same soultions as X^2+6x-7=0

Answers

$\n\n\huge\text{The answer is in the attachment !!!}$

$x^(2) +6x-7=0$
Solve by formula : $x_(1.2) = \frac{-b+ \sqrt{ x^(2-4ac) } }{2a}$
$a=1,b=6,c=-7$

$x_(1.2) = \frac{(-6)+ \sqrt{ 6^(2)-4(-7) } }{2}$

$x_(1.2)= \frac{\sqrt{(-6)+ \sqrt{ 6^(2)+28 } } }{2}$

$x_(1.2) = ((-6)+ √(36+28 ) )/(2)$

$x_(1.2) = ((-6)+ √(64) )/(2)$

$x_(1.2) = ((-6)+8)/(2)$

First solution:
$x_(1) = ((-6)+8)/(2)$

$x_(1)= (\not 2)/(\not 2)$

$x_(1) =1$
Second solution:
$x_(2) = (-6-8)/(2)$

$x_(2) =-14:2$

$x_(2) =-7$

The answer is: $x_(1) =1$ , $x_(2) =-7$

Select the expression that is equivalent to the polynomial given below. (4x + 3)(3x - 8)

a.) 12x2 - 23x - 24
b.) 12x2 + 41x + 24
c.) 12x2 - 23x + 24
d.) 12x2 - 41x - 24

which would be the best option?