All categories

3 years ago

X² + y² = 34x + y =8

Mathematics

1 answer:

Nataly [62]3 years ago

4 0

Step-by-step explanation:

${x}^{2} + {y}^{2} = 34 \\ x + y = 8 \\ x = 8 - y \\ substitute \: x \: in \\ {x}^{2} + {y}^{2} = 34 \\ {(8 - 4)}^{2} + {y}^{2} = 34 \\ {(8 - 4) }^{2} = {y}^{2} - 16y + 64 \\ {y}^{2} - 16y + 64 + {y}^{2} = 34 \\ {2y}^{2} - 16y + 64 = 34 \\ {2y}^{2} - 16y + 64 - 34 = 0 \\ {2y}^{2} - 16y + 30 = 0 \\ 2 {y}^{2} - 6y - 10y + 30 = 0 \\ (2 {y}^{2} - 6y) - (10y + 30) = 0 \\ 2y(y - 3) - 10(y - 3) = 0 \\ (2y - 10)(y - 3) \\ 2y = 10 \\ y = \frac{10}{2} \\ y = 5 \\ or \\ y = 3 \\ if \: y = 5 \: \: x = 3 \\ if \: y = 3 \: \: x = 5$

You might be interested in

Choose an equation for the line that is parallel to the given line and passes through the given point for questions 1 and 2.

lutik1710 [3]

To solve each of the problems, I first looked for the equation that had the same slope value and then plugged in the 'x' coordinate to see if it gave me the correct 'y' coordinate.

1.)
D.) y = 5x + 4
14 = 5(2) + 4
14 = 10 + 4
14 = 13

2.)
C.) y = 1/5x - 19
-16 = 1/5 (15) - 19
-16 = (3) - 19
-16 = -16

3.) First, get 'y' by itself by subtracting 4x from both sides and dividing the whole equation by -12. To find perpendicular lines, the slope must be the opposite reciprocal of the first slope (in this case, flip the fraction to get the three on top). Add the opposite sign to the slope as your final step.

4x - 12y = 2
-4x -4x
-12y = -4x + 2
-12y / -12 = -4x / -12+ 2 / -12
y = 1/3x - 1/6

C.) y = -3x + 29
-1 = -3(10) + 29
-1 = (-30) + 29
- 1 = -1

3 0

4 years ago

NEED ANSWER QUICK WITH STEP BY STEP!!!

Elden [556K]

Y-y(1) = m(x - x(1))

y - 1 = (-3/4)(x + 8)

3 0

3 years ago

How to play tennis is a topic that would be considered<br> too narrow<br> too broad

IRISSAK [1]

Answer:

I would say too broad.

Step-by-step explanation:

8 0

3 years ago

Can someone help with this please

zepelin [54]

Https://www.google.com/?sa=X&ved=0ahUKEwjNk_Sj3v_dAhVfJTQIHW01DQgQOwgC https://www.google.com/?sa=X&ved=0ahUKEwjNk_Sj3v_dAhVfJTQIHW01DQgQOwgC https://www.google.com/?sa=X&ved=0ahUKEwjNk_Sj3v_dAhVfJTQIHW01DQgQOwgC

4 0

3 years ago

3. A rocket is launched from a height of 3 meters with an initial velocity of 15 meters per second.

Vikki [24]

Let the rocket is launched at an angle of \theta with respect to the positive direction of the x-axis with an initial velocity u=15m/s.

Let the initial position of the rocked is at the origin of the cartesian coordinate system where the illustrative path of the rochet has been shown in the figure.

As per assumed sine convention, the physical quantities like displacement, velocity, acceleration, have been taken positively in the positive direction of the x and y-axis.

Let the point P(x,y) be the position of the rocket at any time instant t as shown.

Gravitational force is acting downward, so it will not change the horizontal component of the initial velocity, i.e. $U_x=U cos\theta$ is constant.

So, after time, t, the horizontal component of the position of the rocket is

$x= U \cos\theta t \;\cdots(i)$

The vertical component of the velocity will vary as per the equation of laws of motion,

$s=ut+\frac12at^2\;\cdots(ii)$ , where,s, u and a are the displacement, initial velocity, and acceleration of the object in the same direction.

(a) At instant position P(x,y):

The vertical component of the initial velocity is, $U_y=U sin\theta$ .

Vertical displacement =y

So, s=y

Acceleration due to gravity is $g=9.81 m/s^2$ in the downward direction.

So, $a=-g=-9.81 m/s^2$ (as per sigh convention)

Now, from equation (ii),

$y=U sin\theta t +\frac 12 (-9.81)t^2$

$\Rightarrow y=U \sin\theta \times \frac {x}{U \cos\theta} +\frac 12 (-g)\times \left(\frac {x}{U \cos\theta} \right)^2$

$\Rightarrow y=U^2 \tan\theta-\frac 1 2g U^2 \sec^2 \theta\;\cdots(iii)$

This is the required, quadratic equation, where $U=15 m/s$ and $g=9.81 m/s^2$ .

(b) At the highest point the vertical velocity,v, of the rocket becomes zero.

From the law of motion, $v=u+at$

$\Rightarroe 0=U\sin\theta-gt$

$\Rightarroe t=\frac{U\sin\theta}{g}\cdots(iv)$

The rocket will reach the maximum height at $t= 1.53 \sin\theta s$

So, from equations (ii) and (iv), the maximum height, $y_m$ is

$y_m=U\sin\theta\times \frac{U\sin\theta}{g}-\frac 12 g \left(\frac{U\sin\theta}{g}\right)^2$

$\Rightarrow y_m=23 \sin\theta -11.5 \sin^2\theta$

In the case of vertical launch, $\theta=90^{\circ}$ , and

$\Rightarrow y_m=11.5 m$ and $t=1.53 s$ .

Height from the ground= 11.5+3=14.5 m.

(c) Height of rocket after t=4 s:

$y=15 \sin\theta \times 4- \frac 12 (9.81)\times 4^2$

$\Rightarrow y=15 \sin\theta-78.48$

$\Rightarrow -63.48 m >y> 78.48$

This is the mathematical position of the graph shown which is below ground but there is the ground at $y=-3m$ , so the rocket will be at the ground at t=4 s.