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1 year ago

What are some studying tips because while doing math i feel like dying...

Mathematics

1 answer:

Kazeer [188]1 year ago

3 0

Answer:

– Practicing is key. ...

2 – The formula sheet is your best friend. ...

3 – Understand what your calculator can do. ...

4 – Understanding the derivations of certain formulas. ...

5 – Understand your mistakes and close up any knowledge gaps. ...

6 – Practice on example questions in your textbook.

You might be interested in

Solve for y 31=2y-5

DIA [1.3K]

31=2y-5

Add 5 on both -5+5 which cancels out and 31+5 which is 36 and then divide 36 by 2 which equals to 18.

So y=18 hope this helped (':

8 0

3 years ago

Read 2 more answers

Please help find the missing angle measures

masya89 [10]

Answer:

127

Step-by-step explanation:

(n-2)180

(5-2)180

=540 for total angle

540-90(3)-143

=127

7 0

2 years ago

Read 2 more answers

Find the absolute maximum and absolute minimum values of f on the given interval. f(x) = 16 + 2x − x^2, [0, 5]

RSB [31]

Answer:

<h2>absolute maximum = 16</h2><h2>absolute minimum = 1</h2>

Step-by-step explanation:

To get the absolute maximum and minimum values of the function f(x) = 16 + 2x − x² n the given interval [0,5], we need to get the values of f(x) at the end points. The end points are 0 and 5.

at x = 0;

f(0) = 16 + 2(0) − 0²

f(0) = 16

at the other end point i.e at x = 5;

f(5) = 16 + 2(5) − 5²

f(5) = 16 + 10-25

f(5)= 26-25

f(5) = 1

The absolute minimum value is 1 and occurs at x = 5

The absolute maximum value is 16 and occurs at x = 0

5 0

2 years ago

The point P(1,1/2) lies on the curve y=x/(1+x). (a) If Q is the point (x,x/(1+x)), find the slope of the secant line PQ correct

lukranit [14]

Answer:

See explanation

Step-by-step explanation:

You are given the equation of the curve

$y=\dfrac{x}{1+x}$

Point $P\left(1,\dfrac{1}{2}\right)$ lies on the curve.

Point $Q\left(x,\dfrac{x}{1+x}\right)$ is an arbitrary point on the curve.

The slope of the secant line PQ is

$\dfrac{y_2-y_1}{x_2-x_1}=\dfrac{\frac{x}{1+x}-\frac{1}{2}}{x-1}=\dfrac{\frac{2x-(1+x)}{2(x+1)}}{x-1}=\dfrac{\frac{2x-1-x}{2(x+1)}}{x-1}=\\ \\=\dfrac{\frac{x-1}{2(x+1)}}{x-1}=\dfrac{x-1}{2(x+1)}\cdot \dfrac{1}{x-1}=\dfrac{1}{2(x+1)}\ [\text{When}\ x\neq 1]$