If y varies inversely as x, when y=20 and x=2 find the equation for y. then, find y when x=4

Question 2 Not yet answered Points out of 2

nata0808 [166]

Answer:

Trapezoid

Step-by-step explanation:

perpendicular: They are perpendicular due to the fact they cross at right angles to each other. Another true instance of perpendicular lines we will see in nature is a football field. All four corners of a football field are perpendicular to every other.

In the trapezoid attached there aren't any perpendicular sides. A and B are both obtuse angles in this diagram, and C and D are both acute angles. However, now not all trapezoids appear like this. ... Definition B: A trapezoid is any quadrilateral with at the least one pair of parallel sides.

please mark me brainliest!

3 0

2 years ago

Which of the following is most likely the next step in the series?

zysi [14]

Answer:

it's D

Step-by-step explanation:

3 0

3 years ago

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Solve for x.<br><br><br><br> 5+x+(−2)=−8

levacccp [35]

5+х-2+8=0
х+5-2+8=0
х+11=0
х=0-11
х=-11

7 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

A polygon with four sides

Misha Larkins [42]

A polygon with four sides is known as a quadrilateral.
Quadrilaterals have four sides and four angles as well.
There's many different types of quadrilaterals:
square
rectangle
rhombus
trapezoid
The trapezoid has to meet qualifications to be considered a quadrilateral, such as a pair of parellel lines.

7 0

3 years ago

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