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4 years ago

For this question What is 4f-9-f+15

Mathematics

1 answer:

Alex_Xolod [135]4 years ago

5 0

Answer:

3f+6

Step-by-step explanation:

To solve this, we combine like terms which are in this case 4f and f. 4f-9-f+15 is (4f-f)+(-9+15), which we simplify to 3f+6.

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Find the 9th term in a geometric sequence whose common ration is 1/3 and first term is 6

sasho [114]

Answer:

Therefore 9th term of the geometric sequence is $\frac{2}{2187}$

Step-by-step explanation:

Given first term(a) = 6

and common ratio = $\frac{1}{3}$

$T_n= ar^{n-1}$

$\therefore T_9=ar^{9-1}$

$\Leftrightarrow T_9=a r^8$

$\Leftrightarrow T_9=6\times ( \frac{1}{3}) ^8$

$\Leftrightarrow T_9=\frac{2}{2187}$

Therefore 9th term of the geometric sequence is $\frac{2}{2187}$

7 0

4 years ago

Is the point (4,-3) a solution for this system?

Oduvanchick [21]

Yes it is.

Step by step explanation:

3 0

3 years ago

In a box of 10 calculators, one is defective. In how many ways can four calculators be selected, if you know one

Juliette [100K]

Given:

Total number of calculators in a box = 10

Defective calculators in the box = 1

To find:

The number of ways in which four calculators be selected and one of the four calculator is defective.

Solution:

We have,

Total calculators = 10

Defective calculators = 1

Then, Non-defective calculator = 10-1 = 9

Out of 4 selected calculators 1 should be defective. So, 3 calculators are selected from 9 non-defective calculators and 1 is selected from the defective calculator.

$\text{Total ways}=^9C_3\times ^1C_1$

$\text{Total ways}=\dfrac{9!}{3!(9-3)!}\times 1$

$\text{Total ways}=\dfrac{9\times 8\times 7\times 6!}{3\times 2\times 1\times 6!}$

$\text{Total ways}=\dfrac{9\times 8\times 7}{6}$

$\text{Total ways}=3\times 4\times 7$

$\text{Total ways}=84$

Therefore, the four calculators can be selected in 84 ways.

7 0

3 years ago

44 divided by 22 divided by 2 divided by 2

Igoryamba

Answer:

Hello, the answer is 0.5

Step-by-step explanation:

44 ÷ 22 = 2

2 ÷ 2 = 1

1 ÷ 2 = 0.5

Hope this helps :)

8 0

3 years ago

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How do you do linear equations

zavuch27 [327]

Step 1: Simplify each side, if needed.

Step 2: Use Add./Sub. Properties to move the variable term to one side and all other terms to the other side.

Step 3: Use Mult./Div. ...

Step 4: Check your answer.
I find this is the quickest and easiest way to approach linear equations.
Example 6: Solve for the variable.

5 0

3 years ago

Read 2 more answers