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andrew-mc [135]
2 years ago
9

What is the solution of the system? Use any method. 3x + 2y = 7 y = –3x + 11

Mathematics
1 answer:
Gnom [1K]2 years ago
6 0

Answer:

(5,-4)

Step-by-step explanation:

SOLVE BY SUBSTITUTION:

STEP 1: Replace all occurrences of y with -3x+11 in each equation.

-3x+22=7\\y=-3x+11

STEP 2: Solve for x in the first equation.

x=5\\y=-3x+11

STEP 3: Replace all occurrences of x with 5 in each equation.

y=-4\\x=5

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PLEASE HURRY ITS URGENT!<br>What is distance between A &amp; B?<br>​
Juliette [100K]

Answer:

7.62

Step-by-step explanation:

use the Pythagorean theorem

3^2+7^2=58

the square root of 58 is about 7.62

6 0
2 years ago
(TEN POINTS)
Rainbow [258]
F(-1)= -3 is (-1,-3) and f(2) = 6 is (2,6) where f(x) = y
y=mx + b is the slope-intercept form whereas m equals the slope (rate of change) and b equals the y-intercept (initial amount/what y is when x is 0.)

First, we need to find the slope between the two points (-1,-3) and (2,6). To find the slope we could use one of it's formulas \frac{y^2 - y^1}{x^2 - x^1}.  
1. (-1,-3)
2. (2,6)
\frac{6 - -3}{2 - -1} → \frac{9}{3} → \frac{3}{1}

The slope is 3 (\frac{3}{1}). Thusly, y = 3x + b

To find out the y-intercept, we can reverse the slope. [Note: This \frac{3}{1} is in \frac{rise}{run} where rise is 'y' and run is 'x'. Reversed would be \frac{-3}{-1} ]. Take the second ordered pair and use our reversed slope on it until we get 0 for x. 

(2, 6) ⇒ (2 - 1, 6 -3) ⇒ (1, 3) ⇒ (0,0)

Y-intercept is 0. Therefore, y= 3x + 0 [NOTE: y = f(x), so if you want it in function notation form it's just f(x) = 3x + 0.]
6 0
3 years ago
Find the discriminat of each quadratic equation then state the number and type of solutions
Julli [10]
Hello : 
the discriminat of each quadratic equation : ax²+bx+c=0 ....(a <span>≠ 0) is :
</span><span>Δ = b² -4ac
1 )  </span>Δ > 0  the equation has two reals solutions : x =  (-b±√Δ)/2a
2 ) Δ = 0 : one solution : x = -b/2a
3 ) Δ <span>< 0 : no reals solutions</span>
8 0
3 years ago
Solve y ' ' + 4 y = 0 , y ( 0 ) = 2 , y ' ( 0 ) = 2 The resulting oscillation will have Amplitude: Period: If your solution is A
Vlad [161]

Answer:

y(x)=sin(2x)+2cos(2x)

Step-by-step explanation:

y''+4y=0

This is a homogeneous linear equation. So, assume a solution will be proportional to:

e^{\lambda x} \\\\for\hspace{3}some\hspace{3}constant\hspace{3}\lambda

Now, substitute y(x)=e^{\lambda x} into the differential equation:

\frac{d^2}{dx^2} (e^{\lambda x} ) +4e^{\lambda x} =0

Using the characteristic equation:

\lambda ^2 e^{\lambda x} + 4e^{\lambda x} =0

Factor out e^{\lambda x}

e^{\lambda x}(\lambda ^2 +4) =0

Where:

e^{\lambda x} \neq 0\\\\for\hspace{3}any\hspace{3}\lambda

Therefore the zeros must come from the polynomial:

\lambda^2+4 =0

Solving for \lambda:

\lambda =\pm2i

These roots give the next solutions:

y_1(x)=c_1 e^{2ix} \\\\and\\\\y_2(x)=c_2 e^{-2ix}

Where c_1 and c_2 are arbitrary constants. Now, the general solution is the sum of the previous solutions:

y(x)=c_1 e^{2ix} +c_2 e^{-2ix}

Using Euler's identity:

e^{\alpha +i\beta} =e^{\alpha} cos(\beta)+ie^{\alpha} sin(\beta)

y(x)=c_1 (cos(2x)+isin(2x))+c_2(cos(2x)-isin(2x))\\\\Regroup\\\\y(x)=(c_1+c_2)cos(2x) +i(c_1-c_2)sin(2x)\\

Redefine:

i(c_1-c_2)=c_1\\\\c_1+c_2=c_2

Since these are arbitrary constants

y(x)=c_1sin(2x)+c_2cos(2x)

Now, let's find its derivative in order to find c_1 and c_2

y'(x)=2c_1 cos(2x)-2c_2sin(2x)

Evaluating    y(0)=2 :

y(0)=2=c_1sin(0)+c_2cos(0)\\\\2=c_2

Evaluating     y'(0)=2 :

y'(0)=2=2c_1cos(0)-2c_2sin(0)\\\\2=2c_1\\\\c_1=1

Finally, the solution is given by:

y(x)=sin(2x)+2cos(2x)

5 0
3 years ago
a garden is 18 feet 3 inches long and 10 feet 8 inches wide. the amount of fencing needed to enclose the garden is Need help and
Anna [14]

Answer:

\boxed{57ft 10in}

Step-by-step explanation:

<em>Hey there!</em>

Well if the length is 18 and 3 inches we need to find the sum of both lengths,

18ft 3in + 18ft 3in = 36ft 6in

Width- 10ft 8in + 10ft 8in = 20ft 16 in -> 21ft 4in

l + w = 57ft 10in

<em>Hope this helps :)</em>

5 0
3 years ago
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