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ella [17]
3 years ago
8

Please help this is for my geometry final i will appreciate it so much :)

Mathematics
2 answers:
ludmilkaskok [199]3 years ago
8 0
True
Hope this helps:)
Sliva [168]3 years ago
7 0

Answer and Step-by-step explanation:

<u>The answer is True.</u>

This is because rain is falling water (precipitating water), and water makes things wet, so because it is raining, the outside is wet from the falling water.

<u><em>#teamtrees #PAW (Plant And Water)</em></u>

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Find the value of x<br> (x +40)<br> x
gayaneshka [121]

Answer: x^2 + 40x

Nothing can further be done with the topic.....

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3 years ago
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brilliants [131]

Answer:

<em>(a) x=2, y=-1</em>

<em>(b)  x=2, y=2</em>

<em>(c)</em> \displaystyle x=\frac{5}{2}, y=\frac{5}{4}

<em>(d) x=-2, y=-7</em>

Step-by-step explanation:

<u>Cramer's Rule</u>

It's a predetermined sequence of steps to solve a system of equations. It's a preferred technique to be implemented in automatic digital solutions because it's easy to structure and generalize.

It uses the concept of determinants, as explained below. Suppose we have a 2x2 system of equations like:

\displaystyle \left \{ {{ax+by=p} \atop {cx+dy=q}} \right.

We call the determinant of the system

\Delta=\begin{vmatrix}a &b \\c  &d \end{vmatrix}

We also define:

\Delta_x=\begin{vmatrix}p &b \\q  &d \end{vmatrix}

And

\Delta_y=\begin{vmatrix}a &p \\c  &q \end{vmatrix}

The solution for x and y is

\displaystyle x=\frac{\Delta_x}{\Delta}

\displaystyle y=\frac{\Delta_y}{\Delta}

(a) The system to solve is

\displaystyle \left \{ {{x+y=1} \atop {x-2y=4}} \right.

Calculating:

\Delta=\begin{vmatrix}1 &1 \\1  &-2 \end{vmatrix}=-2-1=-3

\Delta_x=\begin{vmatrix}1 &1 \\4  &-2 \end{vmatrix}=-2-4=-6

\Delta_y=\begin{vmatrix}1 &1 \\1  &4 \end{vmatrix}=4-3=3

\displaystyle x=\frac{\Delta_x}{\Delta}=\frac{-6}{-3}=2

\displaystyle y=\frac{\Delta_y}{\Delta}=\frac{3}{-3}=-1

The solution is x=2, y=-1

(b) The system to solve is

\displaystyle \left \{ {{4x-y=6} \atop {x-y=0}} \right.

Calculating:

\Delta=\begin{vmatrix}4 &-1 \\1  &-1 \end{vmatrix}=-4+1=-3

\Delta_x=\begin{vmatrix}6 &-1 \\0  &-1 \end{vmatrix}=-6-0=-6

\Delta_y=\begin{vmatrix}4 &6 \\1  &0 \end{vmatrix}=0-6=-6

\displaystyle x=\frac{\Delta_x}{\Delta}=\frac{-6}{-3}=2

\displaystyle y=\frac{\Delta_y}{\Delta}=\frac{-6}{-3}=2

The solution is x=2, y=2

(c) The system to solve is

\displaystyle \left \{ {{-x+2y=0} \atop {x+2y=5}} \right.

Calculating:

\Delta=\begin{vmatrix}-1 &2 \\1  &2 \end{vmatrix}=-2-2=-4

\Delta_x=\begin{vmatrix}0 &2 \\5  &2 \end{vmatrix}=0-10=-10

\Delta_y=\begin{vmatrix}-1 &0 \\1  &5 \end{vmatrix}=-5-0=-5

\displaystyle x=\frac{\Delta_x}{\Delta}=\frac{-10}{-4}=\frac{5}{2}

\displaystyle y=\frac{\Delta_y}{\Delta}=\frac{-5}{-4}=\frac{5}{4}

The solution is

\displaystyle x=\frac{5}{2}, y=\frac{5}{4}

(d) The system to solve is

\displaystyle \left \{ {{6x-y=-5} \atop {4x-2y=6}} \right.

Calculating:

\Delta=\begin{vmatrix}6 &-1 \\4  &-2 \end{vmatrix}=-12+4=-8

\Delta_x=\begin{vmatrix}-5 &-1 \\6  &-2 \end{vmatrix}=10+6=16

\Delta_y=\begin{vmatrix}6 &-5 \\4  &6 \end{vmatrix}=36+20=56

\displaystyle x=\frac{\Delta_x}{\Delta}=\frac{16}{-8}=-2

\displaystyle y=\frac{\Delta_y}{\Delta}=\frac{56}{-8}=-7

The solution is x=-2, y=-7

4 0
3 years ago
Compare the functions. f(x)=10x g(x)=5^x h(x)=4x^2+5x select the true statement
antiseptic1488 [7]

Answer:

equals 56

Step-by-step explanation:

7 0
3 years ago
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The perimeters of square region S and rectangular region R are equal. If the sides of R are in the ratio 2 : 3, what is the rati
Ksivusya [100]
<h2>Answer:</h2>

The ratio of the area of region R to the area of region S is:

                    \dfrac{24}{25}

<h2>Step-by-step explanation:</h2>

The sides of R are in the ratio : 2:3

Let the length of R be: 2x

and the width of R be: 3x

i.e. The perimeter of R is given by:

Perimeter\ of\ R=2(2x+3x)

( Since, the perimeter of a rectangle with length L and breadth or width B is given by:

Perimeter=2(L+B) )

Hence, we get:

Perimeter\ of\ R=2(5x)

i.e.

Perimeter\ of\ R=10x

Also, let " s " denote the side of the square region.

We know that the perimeter of a square with side " s " is given by:

\text{Perimeter\ of\ square}=4s

Now, it is given that:

The perimeters of square region S and rectangular region R are equal.

i.e.

4s=10x\\\\i.e.\\\\s=\dfrac{10x}{4}\\\\s=\dfrac{5x}{2}

Now, we know that the area of a square is given by:

\text{Area\ of\ square}=s^2

and

\text{Area\ of\ Rectangle}=L\times B

Hence, we get:

\text{Area\ of\ square}=(\dfrac{5x}{2})^2=\dfrac{25x^2}{4}

and

\text{Area\ of\ Rectangle}=2x\times 3x

i.e.

\text{Area\ of\ Rectangle}=6x^2

Hence,

Ratio of the area of region R to the area of region S is:

=\dfrac{6x^2}{\dfrac{25x^2}{4}}\\\\=\dfrac{6x^2\times 4}{25x^2}\\\\=\dfrac{24}{25}

6 0
3 years ago
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Answer:

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Step-by-step explanation:

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