Since AB=AD, the triangle on the left is isosceles and has two 35 degree angles. Since the sum of all the interior angles is 180 deg,

x = 180 deg - 2(35 deg) = 110 deg (answer)

3 0

3 years ago

Find the area under the curve y = 27/x3 from x = 1 to x = t. Evaluate the area under this curve for t = 10, t = 100, and t = 100

Darina [25.2K]

Answer:

Step-by-step explanation:

given is a function as

$y=\frac{27}{x^3}$

We are to find the area form x=1 to x=t

The curve from x=1 lies in the I quadratnt.

So area above x axis is to be calculated

Area = $\int\limits^t_1 {\frac{27}{x^3} } \, dx \\=\frac{-27}{2x^2} \\= \frac{-27}{2t^2}-\frac{-27}{2}\\=\frac{27}{2}(1-\frac{1}{t^2} )$

a) When t =10,

area = $\frac{27}{2} (1-\frac{1}{10^2} )\\= 13.365$

b) t=100

area = $\frac{27}{2} (1-\frac{1}{100^2} )\\= 13.49865$

c) t=10000

area = $\frac{27}{2} (1-\frac{1}{10000^2} )\\= 13.5$

6 0

3 years ago

What is the exact distance from (−1, 4) to (6, −2)?

Studentka2010 [4]

Same as last time:

d^2=(x2-x1)^2+(y2-y1)^2

d^2=(6--1)^2+(-2-4)^2

d^2=7^2+(-6)^2

d^2=49+36

d^2=85

d=√85 (85=5*7 and both 5 and 7 are prime so √85 cannot be simplified any further)

The exact distance is √85 units.

8 0

3 years ago

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