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3 years ago

Solve the following system using any method you choose:

Mathematics

2 answers:

ahrayia [7]3 years ago

5 0

Answer:

(4,2)

Step-by-step explanation:

(4)-4(2)=-4

5(4)-4(2)=12

krok68 [10]3 years ago

4 0

ANSWER: (4,2) it satisfies both equations

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Determine the solution to f(x) = g(x) using the following system of equations: (5 points) f(x) = 3x − 23 g(x) = −4.5x + 7

jeka94

Answer:

(The solution is (4, -11).

Step-by-stp explanation:

Let f(x) = g(x) = y:

y = 3x − 23

y = -4.5x + 7 Subtract the second equation from the first to eliminate y:

0 = 7.5x - 30

7.5x = 30

x = 4

Plug this into the first equation:

y = 3(4) - 23

y = -11.

5 0

3 years ago

What are two equivalent forms for the polynomial 3x - 2x -2x? How do you get them?

Lynna [10]

Simplify 3x -2x -2x = 3x + -2x + -2x
Combine like terms: 3x + -2x + -2x = (3x + -2x +-2x) = -x
answer is: -x

5 0

3 years ago

A circle has the order pairs (-1, 2) (0, 1) (-2, -1) what is the equation . Show your work.

olga55 [171]

We know that:

$(x-a)^2+(y-b)^2=r^2$

is an equation of a circle.

When we substitute x and y (from the pairs we have), we'll get a system of equations:

$\begin{cases}(-1-a)^2+(2-b)^2=r^2\\(0-a)^2+(1-b)^2=r^2\\(-2-a)^2+(-1-b)^2=r^2\end{cases}$

and all we have to do is solve it for a, b and r.

There will be:

$\begin{cases}(-1-a)^2+(2-b)^2=r^2\\(0-a)^2+(1-b)^2=r^2\\(-2-a)^2+(-1-b)^2=r^2\end{cases}\\\\\\
\begin{cases}1+2a+a^2+4-4b+b^2=r^2\\a^2+1-2b+b^2=r^2\\4+4a+a^2+1+2b+b^2=r^2\end{cases}\\\\\\
\begin{cases}a^2+b^2+2a-4b+5=r^2\\a^2+b^2-2b+1=r^2\\a^2+b^2+4a+2b+5=r^2\end{cases}\\\\\\
$

From equations (II) and (III) we have:

$\begin{cases}a^2+b^2-2b+1=r^2\\a^2+b^2+4a+2b+5=r^2\end{cases}\\--------------(-)\\\\a^2+b^2-2b+1-a^2-b^2-4a-2b-5=r^2-r^2\\\\-4a-4b-4=0\qquad|:(-4)\\\\\boxed{-a-b-1=0}$

and from (I) and (II):

$\begin{cases}a^2+b^2+2a-4b+5=r^2\\a^2+b^2-2b+1=r^2\end{cases}\\--------------(-)\\\\a^2+b^2+2a-4b+5-a^2-b^2+2b-1=r^2-r^2\\\\2a-2b+4=0\qquad|:2\\\\\boxed{a-b+2=0}$

Now we can easly calculate a and b:

$\begin{cases}-a-b-1=0\\a-b+2=0\end{cases}\\--------(+)\\\\-a-b-1+a-b+2=0+0\\\\-2b+1=0\\\\-2b=-1\qquad|:(-2)\\\\\boxed{b=\frac{1}{2}}\\\\\\\\a-b+2=0\\\\\\a-\dfrac{1}{2}+2=0\\\\\\a+\dfrac{3}{2}=0\\\\\\\boxed{a=-\frac{3}{2}}$

Finally we calculate $r^2$ :

$a^2+b^2-2b+1=r^2\\\\\\\left(-\dfrac{3}{2}\right)^2+\left(\dfrac{1}{2}\right)^2-2\cdot\dfrac{1}{2}+1=r^2\\\\\\\dfrac{9}{4}+\dfrac{1}{4}-1+1=r^2\\\\\\\dfrac{10}{4}=r^2\\\\\\\boxed{r^2=\frac{5}{2}}$

And the equation of the circle is:

$(x-a)^2+(y-b)^2=r^2\\\\\\\left(x-\left(-\dfrac{3}{2}\right)\right)^2+\left(y-\dfrac{1}{2}\right)^2=\dfrac{5}{2}\\\\\\\boxed{\left(x+\dfrac{3}{2}\right)^2+\left(y-\dfrac{1}{2}\right)^2=\dfrac{5}{2}}$

7 0

3 years ago

What is 593,204 rounded to the nearest hundred thousand

Alina [70]

Answer:

the answer is 600,000

Step-by-step explanation:

the 9 in the ten thousandth place gets rounded hire to 600,000

5 0

2 years ago

Help me please, I need help I don’t understand this problem

jarptica [38.1K]

The maximum speed of a boat at 30 feet length of water is 0.093 nautical miles/hour or knots.

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

The equation for the maximum speed, s is given by s²= (16/9)x
where, x is the length of the water line in feet.

It is given that, the modeled equation s²= (16/9)x is used to find the maximum speed in knots or nautical miles per hour.

The question is asked to find the maximum speed when the length of the water is 30 feet.

Therefore, to find the maximum speed in 30 feet water, the given modeled equation is used. So, substitute the 30 feet in place of x.

<u>Now, calculating the maximum speed :</u>

s² = (16/9)(30)

s² = 480 / 9

s² = 53.3

Taking square root on both sides,

s = √53.3

s = 7.3

The maximum speed of a boat at 30 feet length of water is 7.3 nautical miles/hour or knots.

6 0

3 years ago