What are the correct answers to this geometry problem?

What is 78 in scientific notation

4vir4ik [10]

Well first you take 78 and turn it to 7.8
then multiply by 10000
therefore 7.8*10000= ?

3 0

3 years ago

Explan how you would graph the line containing a slope<br> of-15 mat goes through the point (1-4).

blsea [12.9K]

Answer:

I wish I knew this maybe use the brainy scanner it could help you or google.

3 0

3 years ago

Select all that applies!

Serhud [2]

Answer:

C.

Step-by-step explanation:

Because there is constant term 4 in the question, but there is no any such constant term in the given options.

7 0

3 years ago

Hey can anyone pls answer these math problems!!

Zepler [3.9K]

Answer:

1={5 and 3} 2={0} 3={0.209 and 4.791} 4={-3 and 4} 5={1 and 9} 6={-4.303 and -0.697}

Step-by-step explanation:

Your probably confused on what a, b, and c means.

The problems are written like this.

$ax^2+bx+c=0$

That formula finds the zeros.

For example 1:

$x^2-8x+15=0\\ax^2-bx-c\\a=1\\b=-8\\c=15$

Then use the formula

$x=\frac{-b+-\sqrt{b^2-4ac} }{2a} \\\\x=\frac{-(-8)+-\sqrt{-8^2-4(1)(15)} }{2} \\x=\frac{8+-\sqrt{64-60} }{2} \\\\x=\frac{8+-2}{2}\\$

If plus then 5, if minus then 3

CHECK:

$x^2-8x+15=0\\5^2-8(5)+15=0\\25-40+15=0\\0=0\\3^2-8(3)+15=0\\9-24+15=0\\0=0$

Hope this helps you understand how to do this by yourself. Now to do the rest ;-;

5 0

2 years ago

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

3 years ago