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2 years ago

Solve the following system. y = (1/2)x 2 + 2x - 1 and 3x - y = 1 The solutions are ( )and ( ) (remember to include the commas)

Mathematics

2 answers:

frozen [14]2 years ago

8 0

$\begin{cases}y=\dfrac{1}{2}x+2x-1\\3x-y=1\end{cases}\\\\\\
\begin{cases}y=\dfrac{5}{2}x-1\3x-y=1\end{cases}\\\\\\
3x-\left(\dfrac{5}{2}x-1\right)=1\\\\3x-\dfrac{5}{2}x+1=1\\\\
3x-\dfrac{5}{2}x=0\\\\\dfrac{x}{2}=0\\\\x=0\\\\3\times0-y=1\\0-y=1\\y=-1\\\\\boxed{(x,y)=(0,-1)}$

The solutions are (0) and (-1)

notsponge [240]2 years ago

5 0

Answer:

The points satisfying the solution is (2,5) and (0,-1).

Step-by-step explanation:

Given : $y=\frac{1}{2}x^2+2x-1$ and $3x-y=1$

To solve : The given system of equations ?

Solution :

Let,

$y=\frac{1}{2}x^2+2x-1$ .....[1]

$3x-y=1$ ......[2]

Now, using substitution method,

Substitute y from [1] into [2]

$3x-(\frac{1}{2}x^2+2x-1)=1$

$3x-\frac{1}{2}x^2-2x+1=1$

$x-\frac{1}{2}x^2=0$

$2-x=0$

$x=2$

Now, substitute the value of x into [2]

$3(2)-y=1$

$6-y=1$

$y=5$

Therefore, One of the solution is (2,5)

Similar way we can substitute [2] into [1] we get another solution (0,-1).

So, for the solutions we can also graph the equations and the intersecting points are the solution of the graph.

Refer the attached figure below.

The points satisfying the solution is (2,5) and (0,-1).

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Which one is right triangle angle a. 45, 55 and 35 b. 52, 48 and 20

Gemiola [76]

<h2>option b. 52, 48 and 20</h2>

Step-by-step explanation:

let's solve:-

To test whether given this triangle is right triangle or not, we can use the Pythagorean Theorem .

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Pythagorean Theorem=

$\bold \green{(leg1)^{2} +(leg2)^{2} =(hypotenuse)^{2} }$

Here:-

$\bold \blue{leg \: 1 = 48} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \bold \blue{ leg2 = 20} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \bold\blue{hypotenuse = 52}$

Putting values:-

${48}^{2} + {20}^{2} = {52}^{2} \: \: \: \: \: \: \\ \\ 2304 + 400 = 2704 \\ \\ 2704 = 2704 \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

4 0

2 years ago

For the following questions find the other half of the fraction.

valentina_108 [34]

Answer:

1) 3/5= 6/10

2) 3/6= 6/12

3) 4/10= 2/5

4) 3/4= 6/8

5) 5/10 = 1/2

6) 4/6= 8/12

7) 5/5= 10/10

8) 1/2= 6/12

4 0

2 years ago

Find the angle between the given vectors. Round your answer, in degrees, to two decimal places. u=⟨2,−6⟩u=⟨2,−6⟩, v=⟨4,−7⟩

NISA [10]

Answer:

$\theta = 108.29$

Step-by-step explanation:

Given

$u =$

$v =$

Required:

Calculate the angle between u and v

The angle $\theta$ is calculated as thus:

$cos\theta = \frac{u.v}{|u|.|v|}$

For a vector

$A =$

$A = a * b$

$cos\theta = \frac{u.v}{|u|.|v|}$ becomes

$cos\theta = \frac{.}{|u|.|v|}$

$cos\theta = \frac{2*6+4*-7}{|u|.|v|}$

$cos\theta = \frac{12-28}{|u|.|v|}$

$cos\theta = \frac{-16}{|u|.|v|}$

For a vector

$A =$

$|A| = \sqrt{a^2 + b^2}$

So;

$|u| = \sqrt{2^2 + 6^2}$

$|u| = \sqrt{4 + 36}$

$|u| = \sqrt{40}$

$|v| = \sqrt{4^2+(-7)^2}$

$|v| = \sqrt{16+49}$

$|v| = \sqrt{65}$

So:

$cos\theta = \frac{-16}{|u|.|v|}$

$cos\theta = \frac{-16}{\sqrt{40}*\sqrt{65}}$

$cos\theta = \frac{-16}{\sqrt{2600}}$

$cos\theta = \frac{-16}{\sqrt{100*26}}$

$cos\theta = \frac{-16}{10\sqrt{26}}$

$cos\theta = \frac{-8}{5\sqrt{26}}$

Take arccos of both sides

$\theta = cos^{-1}(\frac{-8}{5\sqrt{26}})$

$\theta = cos^{-1}(\frac{-8}{5 * 5.0990})$

$\theta = cos^{-1}(\frac{-8}{25.495})$

$\theta = cos^{-1}(-0.31378701706)$

$\theta = 108.288386087$

$\theta = 108.29$ (approximated)

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2 years ago

What is the square root of 9 2/5 cubed?

taurus [48]

Answer:

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Step-by-step explanation:

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Answer ASAP:

Irina-Kira [14]

Answer:

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Step-by-step explanation:

Hope this helps ❤️

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3 years ago