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3 years ago

Solve the system of equations. Write you answer as an ordered pair.

Mathematics

1 answer:

lawyer [7]3 years ago

6 0

Answer:

x = -4 y = -5

Step-by-step explanation:

2x + 3 = -x - 9

3x + 3 = -9

3x = -12

x = -4

y = - (-4) - 9

y = 4 - 9

y = -5

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Ring the number that is 10 more than 36

erik [133]

36+10=46
=46
(20 characterssss)

3 0

3 years ago

Can someone help me :(

k0ka [10]

Answer:

x = 20

Step-by-step explanation:

19 < x < 25

x had to be higher than 19 but lower than 25

18 cant be the answer because it is less than 19

27 cant be the answer because it is higher than 25

that leaves 20, which is the right answer

19 < 20 < 25

4 0

3 years ago

Given the position of the particle, what the position(s) of the particle when it’s at rest

choli [55]

The position function of a particle is given by:

$X\mleft(t\mright)=\frac{2}{3}t^3-\frac{9}{2}t^2-18t$

The velocity function is the derivative of the position:

$\begin{gathered} V(t)=\frac{2}{3}(3t^2)-\frac{9}{2}(2t)-18 \\ \text{Simplifying:} \\ V(t)=2t^2-9t-18 \end{gathered}$

The particle will be at rest when the velocity is 0, thus we solve the equation:

$2t^2-9t-18=0$

The coefficients of this equation are: a = 2, b = -9, c = -18

Solve by using the formula:

$t=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}$

Substituting:

$\begin{gathered} t=\frac{9\pm\sqrt[]{81-4(2)(-18)}}{2(2)} \\ t=\frac{9\pm\sqrt[]{81+144}}{4} \\ t=\frac{9\pm\sqrt[]{225}}{4} \\ t=\frac{9\pm15}{4} \end{gathered}$

We have two possible answers:

$\begin{gathered} t=\frac{9+15}{4}=6 \\ t=\frac{9-15}{4}=-\frac{3}{2} \end{gathered}$

We only accept the positive answer because the time cannot be negative.

Now calculate the position for t = 6:

$undefined$

6 0

1 year ago

Science pls help this is due today

nadezda [96]

Answer:

last one

Step-by-step explanation:

Brainliest please

4 0

3 years ago

Read 2 more answers

PRECALC QUESTION! Convert the polar representation of this complex number into its standard form: z = 2(cos 11pi/6+ i sin 11pi/6

Tanzania [10]

$\bf z=2\left[ cos\left( \frac{11\pi }{6} \right) + i\ sin\left( \frac{11\pi }{6} \right) \right]\qquad 
\begin{cases}
r=2\\
\theta=\frac{11\pi }{6}\\
-----\\
a=x=rcos(\theta)\\
b=y=rsin(\theta)
\end{cases}\implies a+bi
\\\\\\
2cos\left( \frac{11\pi }{6} \right)+2sin\left( \frac{11\pi }{6} \right)\ i\implies 2\cdot \cfrac{\sqrt{3}}{2}+2\cdot \cfrac{-1}{2}\ i\implies \sqrt{3}-1i
\\\\\\
\boxed{\sqrt{3}-i}$

7 0

3 years ago

Read 2 more answers