What is the area of this trapezoid?

Mathematics

1 answer:

son4ous [18]3 years ago

5 0

Its C)126in I think because top base is 16, bottom base is 12 and the height is 9.

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What is the image of (-4,-5) after a dilation by a scale factor of 3 centered at the origin?

prohojiy [21]

Answer:

I HOPE THIS HELPS I KINDA SUCK AT MATH

Step-by-step explanation:

B': (-12,15)

Step-by-step explanation:

B': (3×-4, 3×5)

B': (-12,15)

7 0

3 years ago

Help I am very bad at math I need answers by tomorrow if you can help please do

d1i1m1o1n [39]

Answer:

Step-by-step explanation:

(a - b)(a +b) = a² - b²

1 - Sin² A = Cos² A

$LHS = \frac{1}{1- Sin A} + \frac{1}{1 + Sin A}\\\\= \frac{1*(1 + Sin A)}{(1- Sin A)(1 + Sin A)} + \frac{1*(1- Sin A)}{(1 + Sin A)(1- Sin A)}\\\\= \frac{1 + Sin A+ 1 - Sin A}{1^{2}- Sin^{2} A}\\\\= \frac{2}{1 - Sin^{2} A}\\\\= \frac{2}{Cos^{2} A}\\\\= 2 Sec^{2} A$

2) Sec² A - Tan² A = 1

$LHS = \frac{1}{Sec A - Tan A}\\\\=\frac{1*(Sec A + Tan A)}{(Sec A - Tan A)(Sec A + Tan A)}\\\\=\frac{Sec A + Tan A}{Sec^{2} A - Tan^{2} A}\\\\=\frac{Sec A + Tan A }{1}\\\\= Sec A + Tan A = RHS\\\\\\$

3) LHS = Cosec² A + Cot² A

= Cosec² A + Cosec² A - 1

= 2Cosec² A - 1 = RHS

$4) LHS = \frac{Sec A}{Cos A}- \frac{Tan A}{Cot A}\\\\ = Sec A*\frac{1}{Cos A}-Tan A*\frac{1}{Cot A}\\\\ = Sec A * Sec A - Tan A * Tan A\\\\= Sec^{2} A - Tan^{2} A \\\\= 1$