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Studentka2010 [4]

3 years ago

8

7.

1 answer:

ra1l [238]3 years ago

5 0

Step-by-step explanation:

if n=5

5-10+9(5)-3

-5+45-3

45-8=37

if n=-9

-9-10+9( -9)-3

-19-81-3=-103

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Show why it's not and find the 10 combinations that will be in both groups?

hammer [34]

Answer:

Step-by-step explanation:

Given that there are 3 events as

rolling 3 dice, are the events A: sum divisible by 3 and B: sum divisible 5

The sample space will have

(1,1,1)...(6,6,6)

Sum will start with 3 and end with 18

Sum divisible by 3 are 3,6,9..18

Sum divisible by 5 are 5,10,....15

The common element is 15

Hence these two events are not mutually exclusive.

3 0

3 years ago

Vasko was renting a high performance bicycle for his Olympic training. Bicycle A is $25 per month and requires a down payment of

Agata [3.3K]

You need an equation system.

First, translate the text into equations :

y = total price for the bike
x = number of months

first bike :
y = 25x + 500

second bike :
y = 50x + 200

Then you have to solve the system by placing the two equations side by side and comparing them.

25x + 500 = 50x + 200

Solve to find the value of x

7 0

2 years ago

Please answer, thank you

cupoosta [38]

Answer:

Step-by-step explanation:

Since y = -3 we can plug straight into the other equation

--> -x + 2y = -6

-x + 2(-3) = -6

-x - 6 = -6

-x = 0

x = 0

To graph this draw a horizontal like across the page at y = -3 and a vertical line down the page to represent x = 0

The point where these lines intersect is (0,-3)

So that would be the ordered pair solution.

4 0

3 years ago

Sort the ratios listed at the right into bins so that equivalent ratios are grouped together:

cestrela7 [59]

Step $1$

<u>Find the irreducible fraction in each ratio</u>

<u>case 1)</u> $\frac{40}{64}$

Divide by $8$ boths numerator and denominator

$\frac{40}{64}=\frac{5}{8}$

<u>case 2)</u> $\frac{5}{75}$

Divide by $5$ boths numerator and denominator

$\frac{5}{75}=\frac{1}{15}$

<u>case 3)</u> $\frac{14}{35}$

Divide by $7$ boths numerator and denominator

$\frac{14}{35}=\frac{2}{5}$

<u>case 4)</u> $\frac{24}{64}$

Divide by $8$ boths numerator and denominator

$\frac{24}{64}=\frac{3}{8}$

<u>case 5)</u> $\frac{6}{15}$

Divide by $3$ boths numerator and denominator

$\frac{6}{15}=\frac{2}{5}$

<u>case 6)</u> $\frac{65}{104}$

Divide by $13$ boths numerator and denominator

$\frac{65}{104}=\frac{5}{8}$

<u>case 7)</u> $\frac{66}{176}$

Divide by $22$ boths numerator and denominator

$\frac{66}{176}=\frac{3}{8}$

<u>case 8)</u> $\frac{12}{30}$

Divide by $6$ boths numerator and denominator

$\frac{12}{30}=\frac{2}{5}$

<u>case 9)</u> $\frac{15}{225}$

Divide by $15$ boths numerator and denominator

$\frac{15}{225}=\frac{1}{15}$

<u>case 10)</u> $\frac{6}{16}$

Divide by $2$ boths numerator and denominator

$\frac{6}{16}=\frac{3}{8}$

<u>case 11)</u> $\frac{15}{24}$

Divide by $3$ boths numerator and denominator

$\frac{15}{24}=\frac{5}{8}$

<u>case 12)</u> $\frac{48}{128}$

Divide by $16$ boths numerator and denominator

$\frac{48}{128}=\frac{3}{8}$

Step $2$

<u>Sort the ratios into bins</u>

1<u>) First Bin</u>

<u> $Ratio=\frac{5}{8}$ </u>

$\frac{40}{64}$

$\frac{65}{104}$

$\frac{15}{24}$

<u>2) Second Bin </u>

<u> $Ratio=\frac{1}{15}$ </u>

$\frac{5}{75}$

$\frac{15}{225}$

<u>3) Third Bin</u>

$Ratio=\frac{2}{5}$

$\frac{14}{35}$

$\frac{6}{15}$

$\frac{12}{30}$

4<u>) Fourth Bin</u>

<u> $Ratio=\frac{3}{8}$ </u>

$\frac{24}{64}$

$\frac{66}{176}$

$\frac{6}{16}$

$\frac{48}{128}$

3 0

2 years ago

Read 2 more answers

Prove that . 1+cot2 theta = cosec2theta

igor_vitrenko [27]

Answer:

see explanation

Step-by-step explanation:

Using the Pythagorean identity

sin²θ + cos²θ = 1 ( divide terms by sin²θ )

$\frac{sin^20}{sin^20}$ + $\frac{cos^20}{sin^20}$ = $\frac{1}{sin^20}$ , that is

1 + cot²θ = cosec²θ ← as required

6 0

2 years ago