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Blizzard [7]
3 years ago
11

Find x in this porportion 5/2x =25/4

Mathematics
2 answers:
Andrei [34K]3 years ago
6 0

Answer:

x = \frac{2}{5}

Step-by-step explanation:

Given

\frac{5}{2x} = \frac{25}{4} ( cross- multiply )

25 × 2x = 5 × 4

50x = 20 ( divide both sides by 50 )

x = \frac{20}{50} = \frac{2}{5}

lions [1.4K]3 years ago
3 0

Answer:

x = 2/5

Step-by-step explanation:

5/2x} = 25/4 (cross- multiply)

25 × 2x = 5 × 4

50x = 20 (divide both sides by 50)

x = 20/50

x = 2/5

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Two cars leave Phoenix and travel along roads 90 degrees apart. If Car 1 leaves 30 minutes earlier than Car 2 and averages 42 mp
romanna [79]

Answer:

C. 210 miles

Step-by-step explanation:

We have been given that two cars leave Phoenix and travel along roads 90 degrees apart. Car 1 leaves 30 minutes earlier than Car 2 and averages 42 mph.

We will use distance formula and Pythagoras theorem to solve our given problem.

\text{Distance}=\text{Speed}\times \text{Time}

\text{Distance covered by Car 1 in 3.5 hours}=\frac{42\text{ Miles}}{\text{Hour}}\times \text{3.5 hour}

\text{Distance covered by Car 1 in 3.5 hours}=147\text{ Miles}.

Since car 1 leaves 30 minutes before car 2, so car 2 will travel for only 3 hours when car 1 will travel for 3.5 hours.

\text{Distance covered by Car 2 in 3 hours}=\frac{50\text{ Miles}}{\text{Hour}}\times \text{3 hour}

\text{Distance covered by Car 2 in 3 hours}=150\text{ Miles}

Since both car travel along roads 90 degree apart, therefore, the distance between both cars after Car 1 has traveled 3.5 hours would be hypotenuse with legs 147 and 150.

\text{Distance between both cars}=\sqrt{147^2+150^2}

\text{Distance between both cars}=\sqrt{21609+22500}

\text{Distance between both cars}=\sqrt{44109}

\text{Distance between both cars}=210.021427\approx 210

Therefore, the both cars will be 210 miles apart and option C is the correct choice.

3 0
3 years ago
24 rainy days out of 36 days
GarryVolchara [31]
⅔ of the days are rainy days
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3 years ago
John read the first 114 pages of a novel, which was 3 pages less than start fraction, 1/3 1, divided by, 3, end fraction of the
anyanavicka [17]
HERE IS YOUR ANSWER


<span>114 = (1/3)p - 3  
HOPED I HELPED AND SMASH THAT THANKS BUTTON AND ADD ME AS A FRIEND</span>
3 0
3 years ago
The minimum number of triangles needed to construct the polygon Sophia made is 3. Which could show Sophia’s polygon?
schepotkina [342]

I think the answer is 6 or 9  Because the minimum is 3 I hope this helps.

6 0
3 years ago
Read 2 more answers
Looking at the two quadratic functions below (1 &amp; 2), answer the following questions.
algol [13]
Part A:

Given parabola (1) to be f(x) =-(x+12)^2 -6, and parabola (2) to be f(x)=13(x-4)^2+1

Notice that parabola (2) is stretched horizontally by a factor of 13 which is greater than 1. This means that parabola (2) is further away from the x-axis than parabola (1). (i.e. parabola (2) is more 'vertical' than parabola (1).

Therefore, parabola (1) is wider than parabola (2).



Part B:

A parabola open up when the coefficient of the quadratic term (the squared term) is positive and opens down when the coefficient of the quadratic term is negative.

Given parabola (1) to be f(x) =-(x+12)^2 -6, and parabola (2) to be f(x)=13(x-4)^2+1

Notice that the coefficient of the quadratic term is positive for parabola (2) and negative for parabola (1), therefore, the parabola that will open down is parabola (1).



Part C:

For any function, f(x), the graph of the function is moved p places to the left when p is added to x (i.e. f(x + p)) and moves p places to the right when p is subtracted from x (i.e. f(x - p)).

Given parabola (1) to be f(x) =-(x+12)^2 -6, and parabola (2) to be f(x)=13(x-4)^2+1

Notice that in parabola (1), 12 is added to x, which means that the graph of the parent function is shifted 12 places to the left while in parabola (2), 4 is subtracted from x, which means that the graph of the parent function is shifted 4 places to the right.

Therefore, the parabola that would be furthest left on the x-axis is parabola (1).



Part D:

For any function, f(x), the graph of the function is moved q places up when q is added to the function (i.e. f(x) + q) and moves q places down when q is subtracted from the function (i.e. f(x) - q).

Given parabola (1) to be f(x) =-(x+12)^2 -6, and parabola (2) to be f(x)=13(x-4)^2+1

Notice that in parabola (2), 1 is added to the function, which means that the graph of the parent function is shifted 1 place up while in parabola (1), 6 is subtracted from the function, which means that the graph of the parent function is shifted 6 places down.

Therefore, the parabola that would be highest on the y-axis is parabola (2).
7 0
4 years ago
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