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2 years ago

11

Can any of you math wiz help me I am having trouble.

2 answers:

Aleksandr-060686 [28]2 years ago

5 0

Answer:

1 - True

2 - False

3 - False

4 - True

Step-by-step explanation:

N/A

Nesterboy [21]2 years ago

3 0

1) True 2) False 3) False 4)True

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Answer Choice<br> A)72<br> B)64<br> C)50<br> D)42<br> E)13

Alex73 [517]

Answer:

aaaa

Step-by-step explanation:

3x6x4

5 0

3 years ago

Read 2 more answers

A bag contains 6 red marables and 12 blue ones you select one marable at random from the bag what is the probability that you se

-BARSIC- [3]

Answer:

1/3

Step-by-step explanation:

The total number of marbles is

6+12 = 18

P( red) = red/total

= 6/18

=1/3

3 0

3 years ago

Read 2 more answers

Tems11 [23]

QUESTION A

The given multiplication problem is

$\frac{39}{64} \times \frac{8}{13}$

Factor each term to obtain;

$\frac{13\times 3}{8\times8} \times \frac{8}{13}$

Cancel out the common factors to obtain;

$\frac{1\times 3}{8\times1} \times \frac{1}{1}$

Simplify to get;

$\frac{3}{8}$

QUESTION B

The given multiplication problem is

$\frac{2}{3}\times \frac{1}{5}\times \frac{4}{7}$

This the same as

$\frac{2\times 1\times 4}{3\times 5\times 7}$

This simplifies to;

$\frac{8}{105}$

QUESTION C

The given problem is

$\frac{3}{5}\times \frac{10}{12} \times \frac{1}{2}$

This is the same as

$\frac{3}{5}\times \frac{5}{6} \times \frac{1}{2}$

$=\frac{1}{1}\times \frac{1}{2} \times \frac{1}{2}$

This simplifies to

$=\frac{1}{4}$

QUESTION D.

The given expression is

$\frac{4}{9}\times 54$

Factor the 54 to obtain;

$\frac{4}{9}\times 9\times 6$

Cancel the common factors to get;

$\frac{4}{1}\times 1\times 6$

This simplifies to;

$=24$

QUESTION E

The given problem is

$20\times 3\frac{1}{5}$

Convert the mixed numbers to improper fraction to obtain;

$=20\times \frac{16}{5}$

$=4\times5 \times \frac{16}{5}$

Cancel the common factors to get;

$=4\times1 \times \frac{16}{1}$

$=64$

QUESTION F

The multiplication problem is

$11 \times 2 \frac{7}{11}$

Convert the mixed numbers to improper fractions to obtain;

$11 \times \frac{29}{11}$

Cancel out the common factors to get;

$=1 \times \frac{29}{1}$

Simplify;

$=29$

QUESTION G

The given problem is

$5\frac{1}{3}\times 5\frac{1}{8}$

Convert to improper fractions;

$=\frac{16}{3}\times \frac{41}{8}$

Cancel out the common factors to get;

$=\frac{2}{3}\times \frac{41}{1}$

$=\frac{82}{3}$

Convert back to mixed numbers

$=27\frac{1}{3}$

QUESTION H

The given expression is

$10\frac{2}{3} \times 1\frac{3}{8}$

Convert to improper fraction to get;

$\frac{32}{3} \times \frac{11}{8}$

Cancel common factors to get;

$=\frac{4}{3} \times \frac{11}{1}$

Simplify

$=\frac{44}{3}$

Convert back to mixed numbers;

$=14\frac{2}{3}$

7 0

3 years ago

Find the integration of (1-cos2x)/(1+cos2x)

slega [8]

Given:

The expression is:

$\dfrac{1-\cos 2x}{1+\cos 2x}$

To find:

The integration of the given expression.

Solution:

We need to find the integration of $\dfrac{1-\cos 2x}{1+\cos 2x}$ .

Let us consider,

$I=\int \dfrac{1-\cos 2x}{1+\cos 2x}dx$

$I=\int \dfrac{2\sin^2x}{2\cos^2x}dx$ $[\because 1+\cos 2x=2\cos^2x,1-\cos 2x=2\sin^2x]$

$I=\int \dfrac{\sin^2x}{\cos^2x}dx$

$I=\int \tan^2xdx$ $\left[\because \tan \theta =\dfrac{\sin \theta}{\cos \theta}\right]$

It can be written as:

$I=\int (\sec^2x-1)dx$ $[\because 1+\tan^2 \theta =\sec^2 \theta]$

$I=\int \sec^2xdx-\int 1dx$

$I=\tan x-x+C$

Therefore, the integration of $\dfrac{1-\cos 2x}{1+\cos 2x}$ is $I=\tan x-x+C$ .

8 0

3 years ago

If y = 5x – 4, which of the following sets represents possible inputs and outputs of the function, represented as ordered pairs?

lidiya [134]

Answer:

B

Step-by-step explanation:

Substitute the x value into the right side of the function and if the value obtained is equal to the y value of the point then it is a solution.

(0, - 4) → y = 5(0) - 4 = 0 - 4 = -4 ← True

(2, 6) → y = 5(2) - 4 = 10 - 4 = 6 ← True

(4, 20) → 5(4) - 4 = 20 - 4 = 16 ← False

(4, 16) → 5(4) - 4 = 20 - 4 = 16 ← True

(0, - 4), (2, 6), (4, 16) ← possible inputs and outputs

8 0

4 years ago