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maw [93]
3 years ago
15

Derivative of tanx-cscx/secx-cotx

Mathematics
1 answer:
Ksenya-84 [330]3 years ago
6 0
I hope this is right! :)

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1st to answer will get Brainliest 
storchak [24]
Find current interest:
=$2543.32 * 17.4%
=2543.32 * 0.174
=442.537

Add the interest to the current balance:
=2543.32 + 442.537
=2985.857

Add new transaction:
=2985.857 + 320
=3305.857 rounded to 3305.86

The answer is C) $3305.86

Hope this helps! :)
4 0
3 years ago
Read 2 more answers
After page 182 in a book came page 187. How many pages were missing ?
WITCHER [35]
5 pages because 187-182=5
6 0
3 years ago
Read 2 more answers
Write the equation of the circle graphed below
Papessa [141]

Answer:

(x +2)^2 + (y +2)^2 = .5625

Step-by-step explanation:

(x - h)^2 + (y - k)^2 = r^2

(x + 2)^2 + (y + 2)^2 = .75^2

5 0
2 years ago
Can some explain how to do this?
zepelin [54]

Answer:

  • Exact Area = 210.25pi - 210
  • Approximate Area = 450.185

The units for the area are in square inches or in^2. The approximate value shown above is when using pi = 3.14

=============================================================

Explanation:

Use the pythagorean theorem to find the length of the hypotenuse

  • a = 20
  • b = 21
  • c = unknown

a^2 + b^2 = c^2

20^2 + 21^2 = c^2

400 + 441 = c^2

c^2 = 841

c = sqrt(841)

c = 29

The hypotenuse is 29 inches long. This is the diameter of the circle. Half of that is the radius at r = d/2 = 29/2 = 14.5 inches.

The area of the circle is...

A = pi*r^2

A = pi*(14.5)^2

A = pi*210.25

A = 210.25pi

Which is exact in terms of pi

We'll subtract off the triangular region as this isn't shaded in. The area of the triangle is base*height/2 = 20*21/2 = 420/2 = 210 square inches.

So the shaded region is therefore 210.25pi - 210 square inches

This approximates to 210.25*3.14 - 210 = 450.185 when using the approximation pi = 3.14; use more decimal digits of pi to get a more accurate value.

4 0
3 years ago
If fg + gh = fh,which of the following statements must be true?
Oksanka [162]
The only safe conclusion is that point G lies on line FH or that point G lies somewhere between line FH. We cannot conclude that point G is the midpoint of line FH eventhough by virtue of definition of midpoint, the given equation is a proof equation. If G were to be midpoint, segment FG must be equal to segment GH in line FH. 
5 0
3 years ago
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