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cricket20 [7]
3 years ago
12

The boundary line on the graph represents the equation

Mathematics
1 answer:
Pachacha [2.7K]3 years ago
8 0

Answer:

x=2(3-y)/5

Step-by-step explanation:

5x=6-2y

x=6-2y/5

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Solve (D*D + 4)y = x + cos²x​
Sedaia [141]

Answer:

(D^2 + 9)y = cos 2x….(1). The corresponding homogeneous equation is (D^2 +9)y= 0,…(2), whose auxiliary equation is m^2 + 9 = 0, which has (+/-)3i as roots. The general solution of (2) is y = A.cos(3x) + B.sin(3x). Now to get a general solution of (1) we have just to add to the above, a particular solution of (1). One such solution is [cos(2x)]/[-2^2 +9] = (1/5).cos 2x. Hence a general solution of the given equation is given by y = A.cos(3x) + B.sin(3x) + (1/5)cos(2x), where A and B are arbitrary constants. The above solution incorporates all the solutions of the given equation.

Step-by-step explanation:

7 0
3 years ago
<img src="https://tex.z-dn.net/?f=f%28x%29%20%3D%20%5Cfrac%7Bx%5E%7B2%7D-x-2%20%7D%7Bx-5%7D" id="TexFormula1" title="f(x) = \fra
Umnica [9.8K]

The given function is

f(x)=\frac{x^2-x-2}{x-5}

i) We factor the numerator to obtain;

f(x)=\frac{(x-2)(x+1)}{x-5}

The vertical asymptote is where the denominator is zero.

V.A: x=5

ii) The roots are the zeros of the numerator;

Roots:

x^2-x-2=0

(x-2)(x+1)=0

x=2,x=-1

iii) At y-intercept x=0.

Put x=0 into the function and solve.

f(0)=\frac{0^2-0-2}{0-5}

f(0)=\frac{-2}{-5}

f(0)=\frac{2}{5}

Y-int: \frac{2}{5}

iv) Horizontal asymptote.

This is an improper rational function.

It has no horizontal asymptote.

v) Holes:

The given rational function has no common factors in both the numerator and the denominator.

Therefore the function has no holes.

6) For oblique Assymptote, we divide using lond division or synthetic division;

     1   -1    -2

5|  <u>      5   20</u>

     1    4    18

The quotient is x+4

The  oblique asymptote is

y=x+4

6 0
3 years ago
What is the value of log2 128?
DedPeter [7]
If you get confused by logarithms, here's an easier way to think about this problem: What power of 2 equals 128? Well, we know 2^3 = 8, then 2^4 must equal 16. Doubling every times, we get 2^5 = 32, 2^6 = 64 and 2^7 = 128. So, the value of log₂128 is 7.
6 0
3 years ago
Read 2 more answers
Please solve this for me using substitution
Vinvika [58]

<h2> <u>Explanation</u><u>:</u></h2>

\bf \underline{★Given-} \\

\textsf{4x + 5y = 8};

\textsf{5x + 4y = 12}

\bf \underline{★To\: find-} \\

\textsf{the value of x and y in equation?}

\bf \underline{★Solution-} \\

\sf \leadsto 4x + 5y = 8 - - - (i)

\sf \leadsto 5x + 4y = 12 - - - (ii)

By first equation,

\sf \leadsto 4x + 5y = 8

\sf \leadsto 4x = 8 - 5y

\sf \leadsto x = \dfrac{8 - 5y}{4}

\textsf{Now, we can find the original value of Y.}\\

\sf \leadsto 5x + 4y = 12

\sf \leadsto 5 \bigg( \dfrac{8 - 5y}{4} \bigg) + 4y = 12

\sf \leadsto \dfrac{40 - 25y}{4} + 4y = 12

\sf \leadsto \dfrac{40 - 25y + 16y}{4} = 12

\sf \leadsto \dfrac{40 - 9y}{4} = 12

\sf \leadsto 40 - 9y = 12(4)

\sf \leadsto 40 - 9y = 48

\sf \leadsto -9y = 48 - 40

\sf \leadsto -9y =  8

\sf \leadsto y = \dfrac{ -8}{9}

\textsf{Now, we can find the original value of X.}\\

\sf \leadsto 4x + 5y = 8

\sf \leadsto  4x + 5 \bigg( \dfrac{ -8}{9} \bigg) = 8

\sf \leadsto 4x - \dfrac{40}{9} = 8

\sf \leadsto \dfrac{36x - 40}{9} = 8

\sf \leadsto 36x - 40 = 8(9)

\sf \leadsto 36x - 40 = 72

\sf \leadsto 36x = 72 + 40

\sf \leadsto 36x = 112

\sf \leadsto x = \dfrac{112}{36}

\underline{\textsf{Answer-}}\\

Therefore, the values of x and y are -8/9 and 112/36 respectively.

6 0
2 years ago
How many solutions does the following equation have?
Masteriza [31]

Answer:

One solution

Step-by-step explanation:

Hi,

3(x + 5) = -4x + 8

3x + 15 = -4x + 8

7x + 15 = 8

7x = -7

x = -1

One solution :)

8 0
3 years ago
Read 2 more answers
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