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3 years ago

ANSWER PART B

Mathematics

2 answers:

Maurinko [17]3 years ago

5 0

Answer:

Answer:Back 2

Step-by-step explanation:Blue + Red=-2

Step-by-step explanation:

Talja [164]3 years ago

4 0

Answer:

naannansnccjdksjsjdjdhdhd

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Ray MN and ray ML are two sides of an angle. What is a name of this angle?

Setler [38]

It is (c) NML
Because MN and ML is the vertex and the point meeting is M which is the angle.
The M is the angle because the common letter of both vertex is the letter M

7 0

3 years ago

The linear equation of the initial (red ) incline would be y = __x +__

RideAnS [48]

Answer:

$y=\underline {1.5}x+\underline 0$

Step-by-step explanation:

From the graph,

The initial red incline is a straight line passing through the origin.

So, a straight line passing through the origin is of the form:

$y=mx+0$

Where, 'm' is the slope.

Now, the slope of a given as the change in y value to change in x value.

Consider the two end points of the red line (0, 0) and (4, 6).

The slope with two points $(x_1,y_1)\ and\ (x_2,y_2)$ on the line is given as:

$m=\frac{(y_2-y_1)}{(x_2-x_1)}$

Plug in $x_1=0,y_1=0,x_2=4,y_2=6$ . Therefore, slope is:

$m=\frac{6-0}{4-0}\\m=\frac{6}{4}=1.5$

Hence, the equation of the initial red incline is:

$y=\underline {1.5}x+\underline 0$

3 0

3 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 $k\in\mathbb{R}$

(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

determine if it can be proven that the triangles are congruent if they are the choose a theorem that proves its congruent

Lostsunrise [7]

Answer:

These triangles are congruent by SAS

Step-by-step explanation:

They share the same angle which is in the same angle which is the vertical angle with congruent sides on either side of the angle, therefore this is congruent by SAS.

7 0

3 years ago

46.8 divided by 0.002

zmey [24]

46.8/0.002 = 23,400
Hope this helps!

4 0

3 years ago

Read 2 more answers