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dmitriy555 [2]
3 years ago
8

This year ⅗ of the students in Fionna’s class identify as girls. Fionna notices that ⅔ of these girls wear braces. Only ½ of the

boys wear braces. What fraction of all the students in the class wear braces?
Mathematics
1 answer:
Tju [1.3M]3 years ago
5 0

Answer:

³/5

Step-by-step explanation:

Total = 5 units

Girls = 3 units

~if ⅔ of the girls wear braces

= 2units of girs wear braces

Boys = 2 units

~if ½ of the boys wear braces

= 1unit of boys wear braces

Hence, 2units + 1unit =3 units of the class wears braces.

Total pupils = 5units

total that wear braces out of the whole class = 3/5

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5 0
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Dan says that 3/5 is the same as 3.5. Is he correct
adoni [48]
No he is not correct because if you divide 5 into 3 you would get 0.6
3/5 is a fraction while 3.5 is a decimal
5 0
2 years ago
$200 at 7% per year for 9 months
Ganezh [65]
the answer is 126
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2 x 7= 14
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4 0
3 years ago
Can someone pleas explain what is means
kipiarov [429]

Answer:

(16, 6 )

Step-by-step explanation:

Basically you are expressing the points (dots ) in coordinate form

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Then go vertically up to the point at (16, 6 )

5 0
2 years ago
Read 2 more answers
Consider the region, R, bounded above by f(x)=x2−6x+9 and g(x)=−3x+27 and bounded below by the x-axis over the interval [3,9]. F
Salsk061 [2.6K]

Answer:

22.5

Step-by-step explanation:

The region R contains every point of the plane with coordinate x between 3 and 9, and with coordinate y positive such that y < f(x) and y < g(x).

We can note that both f and g are positive on [3,9] because g is a decreasing linear function and g(9) = 0, thus g is positive in every other point of the interval, and f(x) = (x-3)^2 is always positive excpept when x = 3, where it reaches the value 0.

The interception of the graphs takes place for a value x such that f(x) = g(x).

We compute x^2-6x+9 = -3x + 27, thus x^2-6x+9-(-3x + 27) = x^2-3x -18 = 0.

The roots of that quadratic function are

r_1, r_2 = \frac{3 ^+_- \sqrt { 9 +72}}{2} = \frac{3^+_-9}{2} , thus r1 = 6, r2 = -3. We dont care about -3 because it is outside the interval, but we know that f and g graphs intersects on x = 6. Thus, we obtain, due to Bolzano Theorem:

  • On the interval [3,6), the function f in smaller because it takes the value 0 on x=3, while g is always positive.
  • On the interval (6,9]. the function g is smaller because it takes the value 0 on x=9, while f is always positive

Hence, the upper bound is f on the interval [3,6) and g on the interval (6,9]. While the lower bound is the 0 function.

We need to calculate the following integral, using Barrow's rule

\int\limits^6_3 {x^2-6x+9} \, dx + \int\limits^9_6 {-3x+27} \, dx = (\frac{x^3}{3} - 3x^2 + 9x) |^6_3 + (\frac{-3x^2}{2} + 27x)|^9_6 = \\  (18 - 9) + (121.5-108) = 22.5

As a result, the area of the region R is 22.5

6 0
3 years ago
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