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astra-53 [7]
3 years ago
7

What is the length of Line segment G D? GD =___

Mathematics
2 answers:
olya-2409 [2.1K]3 years ago
4 0
8/6 = 14/x
8x = 84
x = 10.5
The length of GD = 10.5
** I think that is right not positive
andrey2020 [161]3 years ago
3 0

Answer: 10.5 units

Step-by-step explanation:

Basic proportionality theorem says that if a line is drawn parallel to one side of a triangle intersecting the other two sides then it divides the two sides in the same ratio.

In the given triangle DEF , Line segment GH is drawn parallel to segment EF also GH is intersecting the other two sides ED and FH at G and H respectively.

GE= 6 units  ,  DH = 14 units ,   HF = 8 units.

Then by Basic proportionality theorem , we have

\dfrac{DG}{GE}=\dfrac{DH}{HF}\\\\\Rightarrow\dfrac{DG}{6}=\dfrac{14}{8}\\\\\Rightarrow\ DG=\dfrac{14}{8}\times6=10.5

Hence, the length of Line segment GD  = 10.5 units.

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A store sells small notebooks for ​$6 and large notebooks for ​$10. If a student buys 6 notebooks and spends ​$52​, how many of
NISA [10]

Answer:

you can say he bought 2 small notebooks and 4 big notebooks

Step-by-step explanation:

we have 52 dollars here

so if we have 4 large notebooks that will cost 40 dollars and if we have 2 small notebooks on addiction to that, that will cost 12 more dollars

now adding up 40 and 12 we get $52 !

3 0
2 years ago
Suppose m<3 = 105. find m<6
tatyana61 [14]
If you mean the m^3=105, then m^6 will be ( \sqrt[3]{105}) ^6 which is 105^2 which is 11025
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3 years ago
Can someone tell me how to work it out or the answer please
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Answer:

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8 0
2 years ago
Consider the differential equation:
Wewaii [24]

(a) Take the Laplace transform of both sides:

2y''(t)+ty'(t)-2y(t)=14

\implies 2(s^2Y(s)-sy(0)-y'(0))-(Y(s)+sY'(s))-2Y(s)=\dfrac{14}s

where the transform of ty'(t) comes from

L[ty'(t)]=-(L[y'(t)])'=-(sY(s)-y(0))'=-Y(s)-sY'(s)

This yields the linear ODE,

-sY'(s)+(2s^2-3)Y(s)=\dfrac{14}s

Divides both sides by -s:

Y'(s)+\dfrac{3-2s^2}sY(s)=-\dfrac{14}{s^2}

Find the integrating factor:

\displaystyle\int\frac{3-2s^2}s\,\mathrm ds=3\ln|s|-s^2+C

Multiply both sides of the ODE by e^{3\ln|s|-s^2}=s^3e^{-s^2}:

s^3e^{-s^2}Y'(s)+(3s^2-2s^4)e^{-s^2}Y(s)=-14se^{-s^2}

The left side condenses into the derivative of a product:

\left(s^3e^{-s^2}Y(s)\right)'=-14se^{-s^2}

Integrate both sides and solve for Y(s):

s^3e^{-s^2}Y(s)=7e^{-s^2}+C

Y(s)=\dfrac{7+Ce^{s^2}}{s^3}

(b) Taking the inverse transform of both sides gives

y(t)=\dfrac{7t^2}2+C\,L^{-1}\left[\dfrac{e^{s^2}}{s^3}\right]

I don't know whether the remaining inverse transform can be resolved, but using the principle of superposition, we know that \frac{7t^2}2 is one solution to the original ODE.

y(t)=\dfrac{7t^2}2\implies y'(t)=7t\implies y''(t)=7

Substitute these into the ODE to see everything checks out:

2\cdot7+t\cdot7t-2\cdot\dfrac{7t^2}2=14

5 0
3 years ago
In the diagram shown, chords AB and CD intersect at E.
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Check the picture below.

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