Someone please help:)))

Does anyone know how to factor y^2+49?

Delvig [45]

Answer:

(y + 7i)(y - 7i)

Step-by-step explanation:

It cannot be factored using real numbers, but consider

7i × - 7i

= -49i² and i² = - 1

= 49

The factoring as a difference of squares to obtain

y² + 49 = (y + 7i)(y - 7i)

5 0

2 years ago

Read 2 more answers

5/8(x-1/3)=5/12 can someone help me figure out what x is?

eimsori [14]

Solve for x:(5 (x - 1/3))/(8) = 5/12
Put each term in x - 1/3 over the common denominator 3: x - 1/3 = (3 x)/3 - 1/3:(5 (3 x)/3 - 1/3)/(8) = 5/12
(3 x)/3 - 1/3 = (3 x - 1)/3:(5 (3 x - 1)/3)/(8) = 5/12
3×8 = 24:(5 (3 x - 1))/24 = 5/12
Multiply both sides of (5 (3 x - 1))/24 = 5/12 by 24/5:(24×5 (3 x - 1))/(5×24) = (24×5)/(5×12)
(24×5)/(5×12) = (24×5)/(5×12):(24×5 (3 x - 1))/(5×24) = (24×5)/(5×12)
(24×5 (3 x - 1))/(5×24) = (5×24)/(5×24)×(3 x - 1) = 3 x - 1:3 x - 1 = (24×5)/(5×12)
(24×5)/(5×12) = 5/5×24/12 = 24/12:3 x - 1 = 24/12
The gcd of 24 and 12 is 12, so 24/12 = (12×2)/(12×1) = 12/12×2 = 2:3 x - 1 = 2
Add 1 to both sides:3 x + (1 - 1) = 1 + 2
1 - 1 = 0:3 x = 2 + 1
2 + 1 = 3:3 x = 3
Divide both sides of 3 x = 3 by 3:(3 x)/3 = 3/3
3/3 = 1:x = 3/3
3/3 = 1:Answer: x = 1

3 0

2 years ago

6.8 dollars. How do you write the decimal 6.8 when it refers ti money

Rainbow [258]

$ 6.80.

hope that helped

7 0

2 years ago

Read 2 more answers

Find the general solution to each of the following ODEs. Then, decide whether or not the set of solutions form a vector space. E

Ipatiy [6.2K]

Answer:

(A) $y=ke^{2t}$ with $k\in\mathbb{R}$ .

(B) $y=ke^{2t}/2-1/2$ with $k\in\mathbb{R}$

(C) $y=k_1e^{2t}+k_2e^{-2t}$ with $k_1,k_2\in\mathbb{R}$

(D) $y=k_1e^{2t}+k_2e^{-2t}+e^{3t}/5$ with $k_1,k_2\in\mathbb{R}$ ,

Step-by-step explanation

(A) We can see this as separation of variables or just a linear ODE of first grade, then $0=y'-2y=\frac{dy}{dt}-2y\Rightarrow \frac{dy}{dt}=2y \Rightarrow \frac{1}{2y}dy=dt \ \Rightarrow \int \frac{1}{2y}dy=\int dt \Rightarrow \ln |y|^{1/2}=t+C \Rightarrow |y|^{1/2}=e^{\ln |y|^{1/2}}=e^{t+C}=e^{C}e^t} \Rightarrow y=ke^{2t}$ . With this answer we see that the set of solutions of the ODE form a vector space over, where vectors are of the form $e^{2t}$ with $t$ real.

(B) Proceeding and the previous item, we obtain $1=y'-2y=\frac{dy}{dt}-2y\Rightarrow \frac{dy}{dt}=2y+1 \Rightarrow \frac{1}{2y+1}dy=dt \ \Rightarrow \int \frac{1}{2y+1}dy=\int dt \Rightarrow 1/2\ln |2y+1|=t+C \Rightarrow |2y+1|^{1/2}=e^{\ln |2y+1|^{1/2}}=e^{t+C}=e^{C}e^t \Rightarrow y=ke^{2t}/2-1/2$ . Which is not a vector space with the usual operations (this is because $-1/2$ ), in other words, if you sum two solutions you don't obtain a solution.

(C) This is a linear ODE of second grade, then if we set $y=e^{mt} \Rightarrow y''=m^2e^{mt}$ and we obtain the characteristic equation $0=y''-4y=m^2e^{mt}-4e^{mt}=(m^2-4)e^{mt}\Rightarrow m^{2}-4=0\Rightarrow m=\pm 2$ and then the general solution is $y=k_1e^{2t}+k_2e^{-2t}$ with $k_1,k_2\in\mathbb{R}$ , and as in the first items the set of solutions form a vector space.

(D) Using C, let be $y=me^{3t}$ we obtain that it must satisfies $3^2m-4m=1\Rightarrow m=1/5$ and then the general solution is $y=k_1e^{2t}+k_2e^{-2t}+e^{3t}/5$ with $k_1,k_2\in\mathbb{R}$ , and as in (B) the set of solutions does not form a vector space (same reason! as in (B)).

4 0

2 years ago

Quick help, please.

sweet [91]

Answer:

Step-by-step explanation:

6 0

2 years ago