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Galina-37 [17]
2 years ago
9

2(P +1) + 3(P + 2 ) > 2 need help pls

Mathematics
1 answer:
Vadim26 [7]2 years ago
5 0

Answer:

Please check the explanation.

Step-by-step explanation:

Given the expression

2\left(P\:+1\right)\:+\:3\left(P\:+\:2\:\right)\:>\:2

solving the expression

2\left(P\:+1\right)\:+\:3\left(P\:+\:2\:\right)\:>\:2

as

  • 2\left(P+1\right)+3\left(P+2\right)=5P+8

so the expression becomes

5P+8>2

subtracting 8 from both sides

5P+8-8>2-8

5P>-6

Dividing both sides by 5

\frac{5P}{5}>\frac{-6}{5}

P>-\frac{6}{5}

Therefore,

2\left(P+1\right)+3\left(P+2\right)>2\quad :\quad \begin{bmatrix}\mathrm{Solution:}\:&\:P>-\frac{6}{5}\:\\ \:\mathrm{Decimal:}&\:P>-1.2\\ \:\mathrm{Interval\:Notation:}&\:\left(-\frac{6}{5},\:\infty \:\right)\end{bmatrix}

The graph is also attached below.

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What is the domain and range of the relation [(1, -8), (-7,8), (-3, 7). (-3, -5]?​
Nadusha1986 [10]

Step-by-step explanation:

domain = {-7, -3, 1}

range ={-8, -5, 7, 8}

6 0
3 years ago
PLEASE HELP!!! WILL REWARD BRAINLIEST!!!
jonny [76]

Answer:

The approximate monthly rate of growth is 1.24%

Step-by-step explanation:

we have the future value formula

F=P(1+r)^t

we have

r=16\%=16/100=0.16 ----> annual interest rate

so

F=P(1+0.16)^t

F=P(1.16)^t

Find the approximate monthly rate of growth

Remember that

1\ year=12\ months

so

Divide the number of periods t by 12

F=P(1.16)^{\frac{t}{12}}

Applying power rules property

F=P(1.16^\frac{1}{12})^t

F=P(1.0124)^t}

1.0124-1=0.0124

Convert to percentage

0.0124*100=1.24\%

therefore

The approximate monthly rate of growth is 1.24%

6 0
3 years ago
If point A is located at (-7,5) and is rotated around 270 clockwise about the origin, then point A' is located at...
jonny [76]

Answer:

B.(-5,-7)

Step-by-step explanation:

According to the rule of rotation if you were to have a 270* clockwise you would first switch the x and y values. Ex: (-7,5) -> (5,-7). next you would make the original y coordinate a negative. (5,-7) -> (-5,-7).

6 0
3 years ago
Find two unit vectos that are orthogonal to both [0,1,2] and [1,-2,3]
alekssr [168]

Answer:

Let the vectors be

a = [0, 1, 2] and

b = [1, -2, 3]

( 1 ) The cross product of a and b (a x b) is the vector that is perpendicular (orthogonal) to a and b.

Let the cross product be another vector c.

To find the cross product (c) of a and b, we have

\left[\begin{array}{ccc}i&j&k\\0&1&2\\1&-2&3\end{array}\right]

c = i(3 + 4) - j(0 - 2) + k(0 - 1)

c = 7i + 2j - k

c = [7, 2, -1]

( 2 ) Convert the orthogonal vector (c) to a unit vector using the formula:

c / | c |

Where | c | = √ (7)² + (2)² + (-1)²  = 3√6

Therefore, the unit vector is

\frac{[7,2,-1]}{3\sqrt{6} }

or

[ \frac{7}{3\sqrt{6} } , \frac{2}{3\sqrt{6} } , \frac{-1}{3\sqrt{6} } ]

The other unit vector which is also orthogonal to a and b is calculated by multiplying the first unit vector by -1. The result is as follows:

[ \frac{-7}{3\sqrt{6} } , \frac{-2}{3\sqrt{6} } , \frac{1}{3\sqrt{6} } ]

In conclusion, the two unit vectors are;

[ \frac{7}{3\sqrt{6} } , \frac{2}{3\sqrt{6} } , \frac{-1}{3\sqrt{6} } ]

and

[ \frac{-7}{3\sqrt{6} } , \frac{-2}{3\sqrt{6} } , \frac{1}{3\sqrt{6} } ]

<em>Hope this helps!</em>

7 0
3 years ago
Please answer real quick
SCORPION-xisa [38]

Answer:

use symbolab

Step-by-step explanation:

4 0
3 years ago
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