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3 years ago

A football team lost 5 yards on one play and gained 12 yards on the next play. Find the overall change in field position.

Mathematics

1 answer:

MrRissso [65]3 years ago

6 0

Answer:

Equation: -5 + 12

Overall change: 7 yards

Step-by-step explanation:

In the problem, it says "lost 5 yards". "Lost" is another way of saying subtraction or negative. Therefore, our starting number is -5.

The problem also says "gained 12 yards". "Gained" is another was of saying addition or positive. Therefore, our second number is positive 12.

Together, these two numbers look like this: -5 + 12. To add them, go 12 to the right of -5 if on a number line. That would be like this: -5, -4, -3, -2, -1, 0, etc until you get to 7 which should be 12 more than -5.

Therefore, the answer is 7 yards.

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Megan went to the beach 34 times last summer. She went 6 more times in July than she did in June. She went twice as many times i

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<span>Megan went to beach 34 times last summer.
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3 years ago

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3 years ago

Two semicircles are drawn inside of a circle with radius 8 inches as shown below. Find the area of the shaded region. Round your

Kazeer [188]

Answer:

Step-by-step explanation:

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3 years ago

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