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Let f be continuous on [0, a] and differentiable on (0, a), Prove that if f(a)=0 then there is at least one value of x in (0, a)

asambeis [7]

Answer:

See picture attached

Step-by-step explanation:

6 0

3 years ago

(y - 2)2 = y2 – 6y + 4<br> Is this statement true or false?

trasher [3.6K]

<h3>Answer: False</h3>

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Explanation:

I'm assuming you meant to type out

(y-2)^2 = y^2-6y+4

This equation is not true for all real numbers because the left hand side expands out like so

(y-2)^2

(y-2)(y-2)

x(y-2) .... let x = y-2

xy-2x

y(x)-2(x)

y(y-2)-2(y-2) ... replace x with y-2

y^2-2y-2y+4

y^2-4y+4

So if the claim was (y-2)^2 = y^2-4y+4, then the claim would be true. However, the right hand side we're given doesn't match up with y^2-4y+4

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Another approach is to pick some y value such as y = 2 to find that

(y-2)^2 = y^2-6y+4

(2-2)^2 = 2^2 - 6(2) + 4 .... plug in y = 2

0^2 = 2^2 - 6(2) + 4

0 = 4 - 6(2) + 4

0 = 4 - 12 + 4

0 = -4

We get a false statement. This is one counterexample showing the given equation is not true for all values of y.

6 0

2 years ago

Read 2 more answers

Find all solutions of the given system of equations (If the system is infinite many solution, express your answer in terms of x)

lisov135 [29]

Answer:

(a) The system of the equations $\left \{ {2x-3y\:=3} \atop {4x-6y\:=3}} \right.$ has no solution.

(b) The system of the equations $\left \{ {4x-6y\:=10} \atop {16x-24y\:=40}} \right.$ has many solutions $y=\frac{2x}{3}-\frac{5}{3}$

Step-by-step explanation:

(a) To find the solutions of the following system of equations $\left \{ {2x-3y\:=3} \atop {4x-6y\:=3}} \right.$ you must:

Multiply $2x-3y=3$ by 2:

$\begin{bmatrix}4x-6y=6\\ 4x-6y=3\end{bmatrix}$

Subtract the equations

$4x-6y=3\\-\\4x-6y=6\\------\\0=-3$

0 = -3 is false, therefore the system of the equations has no solution.

(b) To find the solutions of the system $\left \{ {4x-6y\:=10} \atop {16x-24y\:=40}} \right.$ you must:

Isolate x for $4x-6y=10$

$x=\frac{5+3y}{2}$

Substitute $x=\frac{5+3y}{2}$ into the second equation

$16\cdot \frac{5+3y}{2}-24y=40\\8\left(3y+5\right)-24y=40\\24y+40-24y=40\\40=40$

The system has many solutions.

Isolate y for $4x-6y=10$

$y=\frac{2x}{3}-\frac{5}{3}$

3 0

3 years ago

A recommendation for wheelchair ramps is that the rise-to-run ratio be no greater than 1:12. This ensures that the ramps are nav

patriot [66]

Answer:

a. 12 feet b. 12 feet 0.5 inches c. 8.33 %

Step-by-step explanation:

a. How far out horizontally on the ground will it protrude from the building?

Since the rise to run ratio is 1:12 and the building is 12 inches off the ground, let x be the horizontal distance the ramp protrudes.

So, by ratios rise/run = 1/12 = 12/x

1/12 = 12/x

x = 12 × 12

x = 144 inches

Since 12 inches = 1 foot, 144 inches = 144 × 1 inch = 144 × 1 foot/12 inches = 12 feet

b. How long should the ramp be?

The length of the ramp, L is gotten from Pythagoras' theorem since the ramp is a right-angled triangle with sides 12 inches and 144 inches respectively.

So, L = √(12² + 144²)

= √[12² + (12² × 12²)]

= 12√(1 + 144)

= 12√145

= 12 × 12.042

= 144.5 inches

Since 12 inches = 1 foot, 144.5 inches = 144 × 1 inch + 0.5 inches = 144 × 1 foot/12 inches + 0.5 inches = 12 feet 0.5 inches

c. What percent grade is the ramp?

The percentage grade of the ramp = rise/run × 100 %

= 12 inches/144 inches × 100 %

= 1/12 × 100 %

= 0.0833 × 100 %

= 8.33 %

6 0

2 years ago

Simplify \root(3)(8x^(6))y^(12)

ExtremeBDS [4]

The simplification form of the provided expression is 2x²y⁴ option first is correct.

<h3>What is an expression?</h3>

It is defined as the combination of constants and variables with mathematical operators.

We have an expression:

$= \rm \sqrt[3]{8x^6y^{12}}$

$\rm =\sqrt[3]{8}\sqrt[3]{x^6}\sqrt[3]{y^{12}}$

$\rm \rm = \rm 2\sqrt[3]{x^6}\sqrt[3]{y^{12}}$

$\rm =2x^2\sqrt[3]{y^{12}}$

$\rm =2x^2y^4$

Thus, the simplification form of the provided expression is 2x²y⁴ option first is correct.

Learn more about the expression here:

brainly.com/question/14083225

#SPJ1

4 0

2 years ago