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3 years ago

Find the minimum or maximum of the function y=−x^2+4x−5.

Mathematics

1 answer:

strojnjashka [21]3 years ago

5 0

Answer:

maximum value : y = - 1 at (2, -1)

No local minimum

Step-by-step explanation:

The function y=−x^2+4x−5. is a downward opening parabola.

The local maximum is at the vertex

y = - (xx - 4x ) - 5

y = - (xx - 4x +4 - 4) - 5

y = - (xx - 4x + 4) + 4 - 5

y = -(x - 2)^2 - 1

vertex at (2, -1)

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Over the summer, Gabriela read 10 times as

sukhopar [10]

Answer:

i’m pretty sure it’s 6,039

Step-by-step explanation:

12,078 divided by 2

4 0

1 year ago

How much work is done when you lift an object that weighs 150 N to a height of 5 meters

lara31 [8.8K]

Answer:

Work done = 150 x 5 = 750 J

Step-by-step explanation:

Work done is the measure of energy transferred to an object when a force is added causing the object to move a distance in the direction of the force.

The equation for work done is force time the distance and is measured in Joules. Therefore the formula is W = F x D

work done (W) is measured in joules (J)

force (F) is measured in newtons (N)

distance (d) is in the same direction as the force and is measured in

metres (m)

W = F x D

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7 0

3 years ago

Be sure to answer all parts. List the evaluation points corresponding to the midpoint of each subinterval to three decimal place

gayaneshka [121]

Answer:

The Riemann Sum for $\int\limits^5_4 {x^2+4} \, dx$ with n = 4 using midpoints is about 24.328125.

Step-by-step explanation:

We want to find the Riemann Sum for $\int\limits^5_4 {x^2+4} \, dx$ with n = 4 using midpoints.

The Midpoint Sum uses the midpoints of a sub-interval:

$\int_{a}^{b}f(x)dx\approx\Delta{x}\left(f\left(\frac{x_0+x_1}{2}\right)+f\left(\frac{x_1+x_2}{2}\right)+f\left(\frac{x_2+x_3}{2}\right)+...+f\left(\frac{x_{n-2}+x_{n-1}}{2}\right)+f\left(\frac{x_{n-1}+x_{n}}{2}\right)\right)$

where $\Delta{x}=\frac{b-a}{n}$

We know that a = 4, b = 5, n = 4.

Therefore, $\Delta{x}=\frac{5-4}{4}=\frac{1}{4}$

Divide the interval [4, 5] into n = 4 sub-intervals of length $\Delta{x}=\frac{1}{4}$

$\left[4, \frac{17}{4}\right], \left[\frac{17}{4}, \frac{9}{2}\right], \left[\frac{9}{2}, \frac{19}{4}\right], \left[\frac{19}{4}, 5\right]$

Now, we just evaluate the function at the midpoints:

$f\left(\frac{x_{0}+x_{1}}{2}\right)=f\left(\frac{\left(4\right)+\left(\frac{17}{4}\right)}{2}\right)=f\left(\frac{33}{8}\right)=\frac{1345}{64}=21.015625$

$f\left(\frac{x_{1}+x_{2}}{2}\right)=f\left(\frac{\left(\frac{17}{4}\right)+\left(\frac{9}{2}\right)}{2}\right)=f\left(\frac{35}{8}\right)=\frac{1481}{64}=23.140625$

$f\left(\frac{x_{2}+x_{3}}{2}\right)=f\left(\frac{\left(\frac{9}{2}\right)+\left(\frac{19}{4}\right)}{2}\right)=f\left(\frac{37}{8}\right)=\frac{1625}{64}=25.390625$

$f\left(\frac{x_{3}+x_{4}}{2}\right)=f\left(\frac{\left(\frac{19}{4}\right)+\left(5\right)}{2}\right)=f\left(\frac{39}{8}\right)=\frac{1777}{64}=27.765625$