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Scilla [17]
2 years ago
14

May I please receive help?

Mathematics
2 answers:
Lapatulllka [165]2 years ago
8 0

Are you looking for a acute angle?

ra1l [238]2 years ago
4 0

Answer:

Its right angel

Step-by-step explanation:

Sorry if m wrong

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Work out the area of the trapezium ABDE.<br> B<br> 9 cm<br> 6 cm<br> 8 cm<br> →D<br> E
Evgen [1.6K]

Answer:

Area of the trapezium ABDE = 30 cm²

Step-by-step explanation:

Area of a trapezium = \frac{1}{2}(b_1+b_2)h

Here, b_1 and b_2 are the parallel sides of the trapezium

h = Distance between the parallel sides

From the picture attached,

ΔCAE and ΔCBD are the similar triangles.

So by the property of similarity their sides will be proportional.

\frac{AE}{BD}= \frac{CE}{CD}

\frac{9}{6}=\frac{CE}{8}

CE = \frac{9\times 8}{6}

CE = 12 cm

Therefore, DE = CE - CD

DE = 12 - 8 = 4 cm

Now area of trapezium ABDE = \frac{1}{2}(AE+BD)(DE)

                                                  = \frac{1}{2}(6+9)4

                                                  = 30 cm²

Therefore, area of the trapezium ABDE = 30 cm²

8 0
3 years ago
Cual es el 15% de 36000?
zimovet [89]
15% = 15/100 = 0,15

36000•0,15=5400
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3 years ago
Jessica is planning a trip to South Africa. The following table shows some of her planned expenses for this trip, and
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Step-by-step explanation:

the answer is C on e2020

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Evaluate the line integral c y3 ds,<br> c.x = t3, y = t, 0 ≤ t ≤ 3
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\displaystyle\int_{\mathcal C}y^3\,\mathrm dS=\int_{t=0}^{t=3}t^3\sqrt{1^2+(3t^2)^2}\,\mathrm dt=\int_0^3t^3\sqrt{1+9t^4}\,\mathrm dt

Take u=1+9t^4 so that \mathrm du=36t^3\,\mathrm dt. Then

\displaystyle\int_{\mathcal C}y^3\,\mathrm dS=\frac1{36}\int_{u=1}^{u=730}\sqrt u\,\mathrm du=\frac{730^{3/2}-1}{54}
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3 years ago
Can yall help me with this i forgot how to do this!
ss7ja [257]
Ok so cut the circles into the bottom number! Then which ever is more filled out put a < to it!! Hope this helps tell me if you need more!
7 0
3 years ago
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