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alexandr1967 [171]
2 years ago
9

Solve x + 2y = 3 x - y = 6 x = ? y = ?

Mathematics
1 answer:
Airida [17]2 years ago
6 0
Y=1

x=7

that's the answer !!
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There is large clock on the wall so student can check the time.Gym class starts at 1:15pm and end at 2:00 pm
lys-0071 [83]
It's 35 minutes passed
7 0
2 years ago
Read 2 more answers
Please help!!!!! <br> Which defines the piecewise function shown?
denis-greek [22]
<h2>Hello!</h2>

The answer is:

The second piecewise function,

f(x)=\left \{ {{-x-2,x

<h2>Why?</h2>

To answer the question, we need to find which piecewise function contains the graphs shown in the picture, we have that first line shown, is decreasing and exists from the negative numbers to 0,  and the line cuts the x-axis and the y-axis at "-2", for the second line shown,  we have that is decreasing but exists from 0 (including it) to the the positive numbers.

So, finding which piecewise satisfies the function shown, we have:

- First piecewise function:

f(x)=\left \{ {{-x-2,x

First line,

We have the first line,

-x-2

The variable has a negative coefficient (-1), meaning that the function (line) is decreasing, also we can see that the function does exists from all the numbers less than 0, to 0, and it cuts the x-axis at -2 and the y-axis at -2.

Second line,

\frac{x}{2}\geq 0

The variable has positive coefficient, meaning that the function (line) is increasing.

Hence, since the second line is not decreasing, the piecewise function is not the piecewise function shown in the picture.

- Second piecewise function,

f(x)=\left \{ {{-x-2,x

First line,

We have the first line,

-x-2

The variable has a negative coefficient (-1), meaning that the function (line) is decreasing, also we can see that the function does exist from all the numbers less than 0, to 0.

Second line,

-\frac{x}{2}\geq 0

The variable has a negative coefficient ((-\frac{1}{2}), meaning the the unction (line) is decreasing, also we can see that the function does exists from 0 (including it) to the positive numbers.

Hence, we have that the correct option is the second piecewise function,

f(x)=\left \{ {{-x-2,x

Note: I have attached a picture for better understanding.

Have a nice day!

3 0
3 years ago
Read 2 more answers
Find the value of x in (4x -5) = 7(x -5)​
Vlad1618 [11]

Answer: x = 10

Step-by-step explanation:

( 4 x − 5 ) = 7 ( x − 5 )

Simplify  7 ( x − 5 ) .

4 x − 5 = 7 x − 35

Move all terms containing  x  to the left side of the equation.

− 3 x − 5 = − 35

Move all terms not containing  x  to the right side of the equation.  

− 3 x = − 30

Divide each term by   − 3  and simplify.

x = 10

6 0
3 years ago
Which statement describes the inverse of m(x) = x2 – 17x?
stealth61 [152]

Answer:

The correct option is;

The \ domain \ restriction \ x \geq \dfrac{17}{2} \ results \ in \ m^{-1}(x) = \dfrac{17}{2} \pm \sqrt{x + \dfrac{289}{4} }}

Step-by-step explanation:

The given information is that m(x) = x² - 17·x

The above equation can be written in the form;

y = x² - 17·x

Therefore;

0 = x² - 17·x - y

From the general solution of a quadratic equation, 0 = a·x² + b·x + c we have;

x = \dfrac{-b\pm \sqrt{b^{2}-4\cdot a\cdot c}}{2\cdot a}

By comparison to the equation,0 = x² - 17·x - y, we have;

a = 1, b = -17, and c = -y

Substituting the values of a, b and c into the formula for the general solution of a quadratic equation, we have;

x = \dfrac{-(-17)\pm \sqrt{(-17)^{2}-4\times (1) \times (-y)}}{2\times (1)} = \dfrac{17\pm \sqrt{289+4\cdot y}}{2}

Which can be simplified as follows;

x =  \dfrac{17\pm \sqrt{289+4\cdot y}}{2}= \dfrac{17}{2} \pm \dfrac{1}{2}  \times \sqrt{289+4\cdot y}} = \dfrac{17}{2} \pm \sqrt{\dfrac{289}{4} +\dfrac{4\cdot y}{4} }}

And further simplified as follows;

x = \dfrac{17}{2} \pm \sqrt{\dfrac{289}{4} +y }} = \dfrac{17}{2} \pm \sqrt{y + \dfrac{289}{4} }}

Interchanging x and y in the function of the inverse, m⁻¹(x), we have;

m^{-1}(x) = \dfrac{17}{2} \pm \sqrt{x + \dfrac{289}{4} }}

We note that the maximum or minimum point of the function, m(x) = x² - 17·x found by differentiating the function and equating the result to zero, gives;

m'(x) = 2·x - 17 = 0

x = 17/2

Similarly, the second derivative is taken to determine if the given point is a maximum or minimum point as follows;

m''(x) = 2 > 0, therefore, the point is a minimum point on the graph

Therefore, as x increases past the minimum point of 17/2, m⁻¹(x) increases to give;

The \ domain \ restriction \ x \geq \dfrac{17}{2} \ results \ in \ m^{-1}(x) = \dfrac{17}{2} \pm \sqrt{x + \dfrac{289}{4} }} to increase m⁻¹(x) above the minimum.

8 0
2 years ago
A student reading his physics book on a lake dock notices that the distance between two incoming wave crests is about 2.4 m and
mojhsa [17]

Answer:

v=2.4/1.6=1.5 m/s

Hope it helps

Have a good day

4 0
2 years ago
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