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PtichkaEL [24]
3 years ago
8

What are the measures of the two angles in the figure below?

Mathematics
1 answer:
ss7ja [257]3 years ago
4 0
Where is the figure???
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You start at (7, 2). You move up 5 units. Where do you end?
tankabanditka [31]

Answer:

(7,7)

Step-by-step explanation:

Because going up would be moving along the y-axis, then you would add 5 to the Y value which was 2 so therefore, if you move up 5 you'll be at (7,7).:)

5 0
3 years ago
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What is the axis of symmetry equation from f(x)=-x^(2)-6x
Sergeeva-Olga [200]

Answer:

  • \lim_{n \to \infty} a_n13

Step-by-step explanation:

6 0
3 years ago
A line segment of length k is divided into 3 equal parts what is the distance between midpoints of the first and third segments
anastassius [24]

Line segment of length k is divided into 3 equal parts.

so first segment is 0-k/3 and third segment is 2/3k-k

so mid-pt of 1st = k/6 and 3rd = 5/6k

so the distance in between = 5/6k-k/6 = 4/6k = 2/3k

4 0
3 years ago
HCF of 405 and 1605 answer it fast
Rashid [163]

Answer:

hcf=(1,605; 600) = 3 × 5

Step-by-step explanation:

Prime Factorization of a number: finding the prime numbers that multiply together to make that number.

1,605 = 3 × 5 × 107;

1,605 is not a prime, is a composite number;

600 = 23 × 3 × 52;

600 is not a prime, is a composite number;

* Positive integers that are only dividing by themselves and 1 are called prime numbers. A prime number has only two factors: 1 and itself.

* A composite number is a positive integer that has at least one factor (divisor) other than 1 and itself.

Multiply all the common prime factors, by the lowest exponents (if any).

gcf, hcf, gcd (1,605; 600) = 3 × 5

gcf, hcf, gcd (1,605; 600) = 3 × 5 = 15;

The numbers have common prime factors.

Must mark brainliest for more answers.

And you should friend me.

7 0
3 years ago
Apply the method of undetermined coefficients to find a particular solution to the following system.wing system.
jarptica [38.1K]
  • y''-y'+y=\sin x

The corresponding homogeneous ODE has characteristic equation r^2-r+1=0 with roots at r=\dfrac{1\pm\sqrt3}2, thus admitting the characteristic solution

y_c=C_1e^x\cos\dfrac{\sqrt3}2x+C_2e^x\sin\dfrac{\sqrt3}2x

For the particular solution, assume one of the form

y_p=a\sin x+b\cos x

{y_p}'=a\cos x-b\sin x

{y_p}''=-a\sin x-b\cos x

Substituting into the ODE gives

(-a\sin x-b\cos x)-(a\cos x-b\sin x)+(a\sin x+b\cos x)=\sin x

-b\cos x+a\sin x=\sin x

\implies a=1,b=0

Then the general solution to this ODE is

\boxed{y(x)=C_1e^x\cos\dfrac{\sqrt3}2x+C_2e^x\sin\dfrac{\sqrt3}2x+\sin x}

  • y''-3y'+2y=e^x\sin x

\implies r^2-3r+2=(r-1)(r-2)=0\implies r=1,r=2

\implies y_c=C_1e^x+C_2e^{2x}

Assume a solution of the form

y_p=e^x(a\sin x+b\cos x)

{y_p}'=e^x((a+b)\cos x+(a-b)\sin x)

{y_p}''=2e^x(a\cos x-b\sin x)

Substituting into the ODE gives

2e^x(a\cos x-b\sin x)-3e^x((a+b)\cos x+(a-b)\sin x)+2e^x(a\sin x+b\cos x)=e^x\sin x

-e^x((a+b)\cos x+(a-b)\sin x)=e^x\sin x

\implies\begin{cases}-a-b=0\\-a+b=1\end{cases}\implies a=-\dfrac12,b=\dfrac12

so the solution is

\boxed{y(x)=C_1e^x+C_2e^{2x}-\dfrac{e^x}2(\sin x-\cos x)}

  • y''+y=x\cos(2x)

r^2+1=0\implies r=\pm i

\implies y_c=C_1\cos x+C_2\sin x

Assume a solution of the form

y_p=(ax+b)\cos(2x)+(cx+d)\sin(2x)

{y_p}''=-4(ax+b-c)\cos(2x)-4(cx+a+d)\sin(2x)

Substituting into the ODE gives

(-4(ax+b-c)\cos(2x)-4(cx+a+d)\sin(2x))+((ax+b)\cos(2x)+(cx+d)\sin(2x))=x\cos(2x)

-(3ax+3b-4c)\cos(2x)-(3cx+3d+4a)\sin(2x)=x\cos(2x)

\implies\begin{cases}-3a=1\\-3b+4c=0\\-3c=0\\-4a-3d=0\end{cases}\implies a=-\dfrac13,b=c=0,d=\dfrac49

so the solution is

\boxed{y(x)=C_1\cos x+C_2\sin x-\dfrac13x\cos(2x)+\dfrac49\sin(2x)}

7 0
3 years ago
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