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tatyana61 [14]
3 years ago
7

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?
Mathematics
2 answers:
Vladimir79 [104]3 years ago
8 0

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.

Natali [406]3 years ago
6 0
<span><u><em>The correct answer is: </em></u>
green and blue.

<u><em>Explanation</em></u><span><u><em>: </em></u>
We want to see which fractions </span></span>\frac{3}{4} is a multiple of. We know that \frac{3}{4} is a multiple of \frac{1}{4}, because \frac{1}{4}*3=\frac{3}{4}.

We can divide fractions to determine if \frac{3}{4} is a multiple of \frac{1}{2}:
\frac{ \frac{3}{4}}{ \frac{1}{2}};

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

Checking to see if \frac{3}{4} is a multiple of \frac{1}{8}, \frac{ \frac{3}{4}}{ \frac{1}{8}};

flip the second one and multiply:
\frac{3}{4}*\frac{8}{1}=\frac{24}{4}=6.
This divided evenly, so \frac{3}{4} is a multiple of \frac{1}{8}.
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Two similar prisms have heights 4 cm and 10cm what is the ratio of their surface areas
Slav-nsk [51]
Let k be the scale factor relating two similar prisms P_{1} and P_{2}, such that for corresponding parts of prisms P_{1} and P_{2} (for heights, in particular) we have k= \frac{height\  of \ P_{1} }{height \ of\  P_{2} }. In our case k= \frac{4}{10} = \frac{2}{5}.
For surfaces area we have \frac{Surface \ area \ of \  P_{1} }{Surface \ area \ of \  P_{2}} = k^{2} =( \frac{2}{5} )^{2}= \frac{4}{25}.
So, the right answer is 4:25 (choice B)



4 0
3 years ago
What is the value of x?<br><br> Enter your answer in the box.<br><br> x =
Nikitich [7]

You are given that all three sides are equal. You also know that all three angles are also equal. Therefore since all three angles are equal and every triangle has 180 degrees, each angle is 180/3 = 60

So

7x - 3 = 60                     Add three to both sides

7x - 3 + 3 = 60 + 3        Combine

7x = 63                          Divide by 7

7x/7 = 63/7

x = 9                              And that's your answer.

3 0
3 years ago
PLEASE SHOW WORK A coffee distributor needs to mix a(n) Gualtemala Antigua coffee blend that normally sells for $10.30 per pound
Katarina [22]

Answer:

23 pounds of the Gualtemala Antigua Blend and 37 pounds of the Tanzanian Blend

Step-by-step explanation:

you need to make a system of equations to solve this.

First lets make x be the pounds of Gualtemala Antigua coffee blend and y the pounds of Tanzanian coffee blend

we know we need 60 pounds total so first equation is

x+y=60

Next we will make an equation based on the money information

10.30x + 13.80y = 12.46(60)

10.30x + 13.80y = 747.60

So our system of equations is

x + y = 60

10.30x + 13.80y = 747.60

I will solve this by using substitution. first rewrite x+y=60 to y=60-x

now i can substitute that into the other equation for y and solve for x

10.30x + 13.80(60-x) = 747.60

10.30x + 828 -13.80x = 747.60

-3.5x + 828 = 747.60

-3.5x = -80.4

x = 23  (rounded from 22.9714 since they requested that)

now I can use this solution to solve for y by plugging into one of the original equations

x + y =60

23 + y =60

y = 37

Finally we can say that they must mix 23 pounds of the Gualtemala Antigua Blend and 37 pounds of the Tanzanian Blend.

5 0
3 years ago
Kevin wants to change the engine in her car. The current engine is 250 cubic centimeters and has a 14 inch fan belt. The new eng
Simora [160]

Option A) 14/250 = N/350 is correct.

Kevin wants to change the engine in her car. The current engine is 250 cubic centimeters and has a 14 inch fan belt. The new engine is 350 cubic centimeters and has proportionally larger fan belt.

<h3>What is proportionality?</h3>

proportionality can be defined as, how much the number is relatable to other number i.e. x =2y x is two times proportion to y.

Hence, in the question
for 250 cc engine it has 14 inch belt
for 350 cc engine it has N inch belt(let)
proportionality can be given as
14 ∝ 250
N ∝ 350
dividing both equation
N/14=350/250
or 14/250=N/350

Thus, the required proportion is given as 14/250=N/350.

Learn more about proportionality here:
brainly.com/question/8598338

#SPJ2

5 0
2 years ago
Read 2 more answers
Problem 4: Solve the initial value problem
pishuonlain [190]

Separate the variables:

y' = \dfrac{dy}{dx} = (y+1)(y-2) \implies \dfrac1{(y+1)(y-2)} \, dy = dx

Separate the left side into partial fractions. We want coefficients a and b such that

\dfrac1{(y+1)(y-2)} = \dfrac a{y+1} + \dfrac b{y-2}

\implies \dfrac1{(y+1)(y-2)} = \dfrac{a(y-2)+b(y+1)}{(y+1)(y-2)}

\implies 1 = a(y-2)+b(y+1)

\implies 1 = (a+b)y - 2a+b

\implies \begin{cases}a+b=0\\-2a+b=1\end{cases} \implies a = -\dfrac13 \text{ and } b = \dfrac13

So we have

\dfrac13 \left(\dfrac1{y-2} - \dfrac1{y+1}\right) \, dy = dx

Integrating both sides yields

\displaystyle \int \dfrac13 \left(\dfrac1{y-2} - \dfrac1{y+1}\right) \, dy = \int dx

\dfrac13 \left(\ln|y-2| - \ln|y+1|\right) = x + C

\dfrac13 \ln\left|\dfrac{y-2}{y+1}\right| = x + C

\ln\left|\dfrac{y-2}{y+1}\right| = 3x + C

\dfrac{y-2}{y+1} = e^{3x + C}

\dfrac{y-2}{y+1} = Ce^{3x}

With the initial condition y(0) = 1, we find

\dfrac{1-2}{1+1} = Ce^{0} \implies C = -\dfrac12

so that the particular solution is

\boxed{\dfrac{y-2}{y+1} = -\dfrac12 e^{3x}}

It's not too hard to solve explicitly for y; notice that

\dfrac{y-2}{y+1} = \dfrac{(y+1)-3}{y+1} = 1-\dfrac3{y+1}

Then

1 - \dfrac3{y+1} = -\dfrac12 e^{3x}

\dfrac3{y+1} = 1 + \dfrac12 e^{3x}

\dfrac{y+1}3 = \dfrac1{1+\frac12 e^{3x}} = \dfrac2{2+e^{3x}}

y+1 = \dfrac6{2+e^{3x}}

y = \dfrac6{2+e^{3x}} - 1

\boxed{y = \dfrac{4-e^{3x}}{2+e^{3x}}}

7 0
2 years ago
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