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3 years ago

Find the student's error in solving the following

Mathematics

2 answers:

Sliva [168]3 years ago

7 0

Answer:

"The student should have switched the direction of the inequality sign to get –5> x for a final answer."

and the second one is:

The student should have added 4 to all parts (left, middle, and right) to get 6 < –3x < 9.

Step-by-step explanation:

i got that on edge

zheka24 [161]3 years ago

5 0

x>7

Step-by-step explanation:

when dividing by a (-)

the inequality sign changes

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Find the area of the composite shape

Alexxandr [17]

Answer:

$13.735$ $ft^{2}$

Step-by-step explanation:

Area of a circle.

$A=\pi r^{2}$

$r=d/2\\r=8/2\\r=4$

$A=\pi \times 4^{2}$

$A=16\pi$

$A \approx 50.2654824574$

Area of a square.

$A=s^{2}$

$A=8^{2}$

$A=64$

Subtract the area of the square with the area of the circle.

$64-50.2654824574$

$= 13.7345175426$

I hope it helps.

Brainliest much appreciated.

8 0

3 years ago

The graph of F ( x ) = 4/x2 - 2x - 3 is shown <br> for which values of X is f ( x ) decreasing

Elza [17]

Hi there!

$\large\boxed{(-\infty, \sqrt[3]{-4}) \text{ and } (0, \infty) }$

We can find the values of x for which f(x) is decreasing by finding the derivative of f(x):

$f(x) = \frac{4}{x^2} - 2x - 3$

Taking the derivative gets:

$f'(x) = -\frac{8}{x^3} - 2$

Find the values for which f'(x) < 0 (less than 0, so f(x) decreasing):

0 = -8/x³ - 2

2 = -8/x³

2x³ = -8

x³ = -4

$x = \sqrt[3]{-4}$

Another critical point is also where the graph has an asymptote (undefined), so at x = 0.

Plug in points into the equation for f'(x) on both sides of each x value to find the intervals for which the graph is less than 0:

f'(1) = -8/1 - 2 = -10 < 0

f'(-1) = -8/(-1) - 2 = 6 > 0

f'(-2) = -8/-8 - 2 = -1 < 0

Thus, the values of x are:

$(-\infty, \sqrt[3]{-4}) \text{ and } (0, \infty)$

6 0

3 years ago

Plss help me solve these 2 !!!!!!!!!!! for a brainlist ;))

lisabon 2012 [21]

Answer:

1. d

2. a

Step-by-step explanation:

Thats how it goes!

Branliest plzzz

3 0

3 years ago

Please help with this worksheet<br><br> real numbers

Rufina [12.5K]

I DO NOT KNOW JUST GIVE ME A SECOND.

6 0

3 years ago

I need help with a problem with solving by square roots in quadratic equation.

Readme [11.4K]

To solve for x, first, we add 3 to the given equation:

$\begin{gathered} 2(x+5)^2-3+3=44+3, \\ 2(x+5)^2=47. \end{gathered}$

Dividing by 2, we get:

$(x+5)^2=\frac{47}{2}.$

Therefore:

$x+5=\pm\sqrt{\frac{47}{2}}.$

Finally, subtracting 5 we get:

$x=\pm\sqrt{\frac{47}{2}}-5.$

Answer:

$x=\operatorname{\pm}\sqrt{\frac{47}{2}}-5.$

4 0

2 years ago