A | math4u.vsb.cz

2010008202

Level:

Identify the solution of the equation.

2^{x} + 2^{x + 2} + 2^{x + 4} = 84

x = 2

x = 1

x = 0

x = 4

2010008006

Level:

Compare the two definite integrals

I_{1} = \int_{0}^{1} (x^{3} - x) d x

and

I_{2} = \int_{1}^{0} (x - x^{3}) d x

I_{1} = I_{2}

I_{1} > I_{2}

I_{1} < I_{2}

These integrals cannot be compared.

2010008005

Level:

Compare the two definite integrals

I_{1} = \int_{1}^{2} (x^{2} - x) d x

and

I_{2} = \int_{2}^{1} (x - x^{2}) d x

I_{1} = I_{2}

I_{1} > I_{2}

I_{1} < I_{2}

These integrals cannot be compared.

2010008004

Level:

Compare the two definite integrals

I_{1} = \int_{0}^{1} (x^{6} \cos^{2} x - 20) d x

and

I_{2} = \int_{0}^{1} (20 - x^{6} \cos^{2} x) d x

I_{1} < I_{2}

I_{1} = I_{2}

I_{1} > I_{2}

These integrals cannot be compared.

2010008003

Level:

Compare the two definite integrals

I_{1} = \int_{0}^{1} (10 - x^{4} \sin^{2} x) d x

and

I_{2} = \int_{0}^{1} (x^{4} \sin^{2} x - 10) d x

I_{1} > I_{2}

I_{1} = I_{2}

I_{1} < I_{2}

These integrals cannot be compared.

2010008002

Level:

Compare the two definite integrals

I_{1} = \int_{0}^{3} \frac{x^{3}}{3^{x}} d x

and

I_{2} = \int_{3}^{0} \frac{x^{3}}{3^{x}} d x

I_{1} > I_{2}

I_{1} = I_{2}

I_{1} < I_{2}

These integrals cannot be compared.

2010008001

Level:

Compare the two definite integrals

I_{1} = \int_{0}^{2} x^{5} \cdot 2^{x} d x

and

I_{2} = \int_{2}^{0} x^{5} \cdot 2^{x} d x

I_{1} > I_{2}

I_{1} = I_{2}

I_{1} < I_{2}

These integrals cannot be compared.

2010007903

Level:

Choose the interval which contains all the solutions of the following quadratic equation.

6 x^{2} + 13 x + 5 = 0

(- 2; - \frac{1}{2}]

[\frac{1}{2}; 2)

(- \frac{3}{2}; \frac{1}{2}]

(- 1; \frac{5}{3}]

2010007802

Level:

Find the domain of the following expression.

\sqrt{(3 x - 2) (4 + 5 x)}

(- \infty; - \frac{4}{5}] \cup [\frac{2}{3}; \infty)

[- \frac{4}{5}; \frac{2}{3}]

(- \infty; - \frac{4}{5}) \cup (\frac{2}{3}; \infty)

(- \frac{4}{5}; \frac{2}{3})

2010007801

Level:

Identify a true statement which concerns to the following equation.

2 \sqrt{x + 5} = x + 2

The solution is in the set

{x \in R : - 1 < x \leq - 5}

The solution is in the set

{x \in R : 5 < x \leq 7}

The solution is in the set

{x \in R : - 4 < x \leq - 1}

The solution is in the set

{x \in R : - 1 < x \leq 2}

« first
‹ previous
…
35
36
37
38
39
40
41
42
43
…
next ›
last »

A

2010008202

2010008006

2010008005

2010008004

2010008003

2010008002

2010008001

2010007903

2010007802

2010007801

Events

Akce

Interests

Zajímavosti

About portal

O portálu